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r""" 

This module contains :py:meth:`~sympy.solvers.ode.dsolve` and different helper 

functions that it uses. 

 

:py:meth:`~sympy.solvers.ode.dsolve` solves ordinary differential equations. 

See the docstring on the various functions for their uses.  Note that partial 

differential equations support is in ``pde.py``.  Note that hint functions 

have docstrings describing their various methods, but they are intended for 

internal use.  Use ``dsolve(ode, func, hint=hint)`` to solve an ODE using a 

specific hint.  See also the docstring on 

:py:meth:`~sympy.solvers.ode.dsolve`. 

 

**Functions in this module** 

 

    These are the user functions in this module: 

 

    - :py:meth:`~sympy.solvers.ode.dsolve` - Solves ODEs. 

    - :py:meth:`~sympy.solvers.ode.classify_ode` - Classifies ODEs into 

      possible hints for :py:meth:`~sympy.solvers.ode.dsolve`. 

    - :py:meth:`~sympy.solvers.ode.checkodesol` - Checks if an equation is the 

      solution to an ODE. 

    - :py:meth:`~sympy.solvers.ode.homogeneous_order` - Returns the 

      homogeneous order of an expression. 

    - :py:meth:`~sympy.solvers.ode.infinitesimals` - Returns the infinitesimals 

      of the Lie group of point transformations of an ODE, such that it is 

      invariant. 

    - :py:meth:`~sympy.solvers.ode_checkinfsol` - Checks if the given infinitesimals 

      are the actual infinitesimals of a first order ODE. 

 

    These are the non-solver helper functions that are for internal use.  The 

    user should use the various options to 

    :py:meth:`~sympy.solvers.ode.dsolve` to obtain the functionality provided 

    by these functions: 

 

    - :py:meth:`~sympy.solvers.ode.odesimp` - Does all forms of ODE 

      simplification. 

    - :py:meth:`~sympy.solvers.ode.ode_sol_simplicity` - A key function for 

      comparing solutions by simplicity. 

    - :py:meth:`~sympy.solvers.ode.constantsimp` - Simplifies arbitrary 

      constants. 

    - :py:meth:`~sympy.solvers.ode.constant_renumber` - Renumber arbitrary 

      constants. 

    - :py:meth:`~sympy.solvers.ode._handle_Integral` - Evaluate unevaluated 

      Integrals. 

 

    See also the docstrings of these functions. 

 

**Currently implemented solver methods** 

 

The following methods are implemented for solving ordinary differential 

equations.  See the docstrings of the various hint functions for more 

information on each (run ``help(ode)``): 

 

  - 1st order separable differential equations. 

  - 1st order differential equations whose coefficients or `dx` and `dy` are 

    functions homogeneous of the same order. 

  - 1st order exact differential equations. 

  - 1st order linear differential equations. 

  - 1st order Bernoulli differential equations. 

  - Power series solutions for first order differential equations. 

  - Lie Group method of solving first order differential equations. 

  - 2nd order Liouville differential equations. 

  - Power series solutions for second order differential equations 

    at ordinary and regular singular points. 

  - `n`\th order linear homogeneous differential equation with constant 

    coefficients. 

  - `n`\th order linear inhomogeneous differential equation with constant 

    coefficients using the method of undetermined coefficients. 

  - `n`\th order linear inhomogeneous differential equation with constant 

    coefficients using the method of variation of parameters. 

 

**Philosophy behind this module** 

 

This module is designed to make it easy to add new ODE solving methods without 

having to mess with the solving code for other methods.  The idea is that 

there is a :py:meth:`~sympy.solvers.ode.classify_ode` function, which takes in 

an ODE and tells you what hints, if any, will solve the ODE.  It does this 

without attempting to solve the ODE, so it is fast.  Each solving method is a 

hint, and it has its own function, named ``ode_<hint>``.  That function takes 

in the ODE and any match expression gathered by 

:py:meth:`~sympy.solvers.ode.classify_ode` and returns a solved result.  If 

this result has any integrals in it, the hint function will return an 

unevaluated :py:class:`~sympy.integrals.Integral` class. 

:py:meth:`~sympy.solvers.ode.dsolve`, which is the user wrapper function 

around all of this, will then call :py:meth:`~sympy.solvers.ode.odesimp` on 

the result, which, among other things, will attempt to solve the equation for 

the dependent variable (the function we are solving for), simplify the 

arbitrary constants in the expression, and evaluate any integrals, if the hint 

allows it. 

 

**How to add new solution methods** 

 

If you have an ODE that you want :py:meth:`~sympy.solvers.ode.dsolve` to be 

able to solve, try to avoid adding special case code here.  Instead, try 

finding a general method that will solve your ODE, as well as others.  This 

way, the :py:mod:`~sympy.solvers.ode` module will become more robust, and 

unhindered by special case hacks.  WolphramAlpha and Maple's 

DETools[odeadvisor] function are two resources you can use to classify a 

specific ODE.  It is also better for a method to work with an `n`\th order ODE 

instead of only with specific orders, if possible. 

 

To add a new method, there are a few things that you need to do.  First, you 

need a hint name for your method.  Try to name your hint so that it is 

unambiguous with all other methods, including ones that may not be implemented 

yet.  If your method uses integrals, also include a ``hint_Integral`` hint. 

If there is more than one way to solve ODEs with your method, include a hint 

for each one, as well as a ``<hint>_best`` hint.  Your ``ode_<hint>_best()`` 

function should choose the best using min with ``ode_sol_simplicity`` as the 

key argument.  See 

:py:meth:`~sympy.solvers.ode.ode_1st_homogeneous_coeff_best`, for example. 

The function that uses your method will be called ``ode_<hint>()``, so the 

hint must only use characters that are allowed in a Python function name 

(alphanumeric characters and the underscore '``_``' character).  Include a 

function for every hint, except for ``_Integral`` hints 

(:py:meth:`~sympy.solvers.ode.dsolve` takes care of those automatically). 

Hint names should be all lowercase, unless a word is commonly capitalized 

(such as Integral or Bernoulli).  If you have a hint that you do not want to 

run with ``all_Integral`` that doesn't have an ``_Integral`` counterpart (such 

as a best hint that would defeat the purpose of ``all_Integral``), you will 

need to remove it manually in the :py:meth:`~sympy.solvers.ode.dsolve` code. 

See also the :py:meth:`~sympy.solvers.ode.classify_ode` docstring for 

guidelines on writing a hint name. 

 

Determine *in general* how the solutions returned by your method compare with 

other methods that can potentially solve the same ODEs.  Then, put your hints 

in the :py:data:`~sympy.solvers.ode.allhints` tuple in the order that they 

should be called.  The ordering of this tuple determines which hints are 

default.  Note that exceptions are ok, because it is easy for the user to 

choose individual hints with :py:meth:`~sympy.solvers.ode.dsolve`.  In 

general, ``_Integral`` variants should go at the end of the list, and 

``_best`` variants should go before the various hints they apply to.  For 

example, the ``undetermined_coefficients`` hint comes before the 

``variation_of_parameters`` hint because, even though variation of parameters 

is more general than undetermined coefficients, undetermined coefficients 

generally returns cleaner results for the ODEs that it can solve than 

variation of parameters does, and it does not require integration, so it is 

much faster. 

 

Next, you need to have a match expression or a function that matches the type 

of the ODE, which you should put in :py:meth:`~sympy.solvers.ode.classify_ode` 

(if the match function is more than just a few lines, like 

:py:meth:`~sympy.solvers.ode._undetermined_coefficients_match`, it should go 

outside of :py:meth:`~sympy.solvers.ode.classify_ode`).  It should match the 

ODE without solving for it as much as possible, so that 

:py:meth:`~sympy.solvers.ode.classify_ode` remains fast and is not hindered by 

bugs in solving code.  Be sure to consider corner cases.  For example, if your 

solution method involves dividing by something, make sure you exclude the case 

where that division will be 0. 

 

In most cases, the matching of the ODE will also give you the various parts 

that you need to solve it.  You should put that in a dictionary (``.match()`` 

will do this for you), and add that as ``matching_hints['hint'] = matchdict`` 

in the relevant part of :py:meth:`~sympy.solvers.ode.classify_ode`. 

:py:meth:`~sympy.solvers.ode.classify_ode` will then send this to 

:py:meth:`~sympy.solvers.ode.dsolve`, which will send it to your function as 

the ``match`` argument.  Your function should be named ``ode_<hint>(eq, func, 

order, match)`.  If you need to send more information, put it in the ``match`` 

dictionary.  For example, if you had to substitute in a dummy variable in 

:py:meth:`~sympy.solvers.ode.classify_ode` to match the ODE, you will need to 

pass it to your function using the `match` dict to access it.  You can access 

the independent variable using ``func.args[0]``, and the dependent variable 

(the function you are trying to solve for) as ``func.func``.  If, while trying 

to solve the ODE, you find that you cannot, raise ``NotImplementedError``. 

:py:meth:`~sympy.solvers.ode.dsolve` will catch this error with the ``all`` 

meta-hint, rather than causing the whole routine to fail. 

 

Add a docstring to your function that describes the method employed.  Like 

with anything else in SymPy, you will need to add a doctest to the docstring, 

in addition to real tests in ``test_ode.py``.  Try to maintain consistency 

with the other hint functions' docstrings.  Add your method to the list at the 

top of this docstring.  Also, add your method to ``ode.rst`` in the 

``docs/src`` directory, so that the Sphinx docs will pull its docstring into 

the main SymPy documentation.  Be sure to make the Sphinx documentation by 

running ``make html`` from within the doc directory to verify that the 

docstring formats correctly. 

 

If your solution method involves integrating, use :py:meth:`Integral() 

<sympy.integrals.integrals.Integral>` instead of 

:py:meth:`~sympy.core.expr.Expr.integrate`.  This allows the user to bypass 

hard/slow integration by using the ``_Integral`` variant of your hint.  In 

most cases, calling :py:meth:`sympy.core.basic.Basic.doit` will integrate your 

solution.  If this is not the case, you will need to write special code in 

:py:meth:`~sympy.solvers.ode._handle_Integral`.  Arbitrary constants should be 

symbols named ``C1``, ``C2``, and so on.  All solution methods should return 

an equality instance.  If you need an arbitrary number of arbitrary constants, 

you can use ``constants = numbered_symbols(prefix='C', cls=Symbol, start=1)``. 

If it is possible to solve for the dependent function in a general way, do so. 

Otherwise, do as best as you can, but do not call solve in your 

``ode_<hint>()`` function.  :py:meth:`~sympy.solvers.ode.odesimp` will attempt 

to solve the solution for you, so you do not need to do that.  Lastly, if your 

ODE has a common simplification that can be applied to your solutions, you can 

add a special case in :py:meth:`~sympy.solvers.ode.odesimp` for it.  For 

example, solutions returned from the ``1st_homogeneous_coeff`` hints often 

have many :py:meth:`~sympy.functions.log` terms, so 

:py:meth:`~sympy.solvers.ode.odesimp` calls 

:py:meth:`~sympy.simplify.simplify.logcombine` on them (it also helps to write 

the arbitrary constant as ``log(C1)`` instead of ``C1`` in this case).  Also 

consider common ways that you can rearrange your solution to have 

:py:meth:`~sympy.solvers.ode.constantsimp` take better advantage of it.  It is 

better to put simplification in :py:meth:`~sympy.solvers.ode.odesimp` than in 

your method, because it can then be turned off with the simplify flag in 

:py:meth:`~sympy.solvers.ode.dsolve`.  If you have any extraneous 

simplification in your function, be sure to only run it using ``if 

match.get('simplify', True):``, especially if it can be slow or if it can 

reduce the domain of the solution. 

 

Finally, as with every contribution to SymPy, your method will need to be 

tested.  Add a test for each method in ``test_ode.py``.  Follow the 

conventions there, i.e., test the solver using ``dsolve(eq, f(x), 

hint=your_hint)``, and also test the solution using 

:py:meth:`~sympy.solvers.ode.checkodesol` (you can put these in a separate 

tests and skip/XFAIL if it runs too slow/doesn't work).  Be sure to call your 

hint specifically in :py:meth:`~sympy.solvers.ode.dsolve`, that way the test 

won't be broken simply by the introduction of another matching hint.  If your 

method works for higher order (>1) ODEs, you will need to run ``sol = 

constant_renumber(sol, 'C', 1, order)`` for each solution, where ``order`` is 

the order of the ODE.  This is because ``constant_renumber`` renumbers the 

arbitrary constants by printing order, which is platform dependent.  Try to 

test every corner case of your solver, including a range of orders if it is a 

`n`\th order solver, but if your solver is slow, such as if it involves hard 

integration, try to keep the test run time down. 

 

Feel free to refactor existing hints to avoid duplicating code or creating 

inconsistencies.  If you can show that your method exactly duplicates an 

existing method, including in the simplicity and speed of obtaining the 

solutions, then you can remove the old, less general method.  The existing 

code is tested extensively in ``test_ode.py``, so if anything is broken, one 

of those tests will surely fail. 

 

""" 

from __future__ import print_function, division 

 

from collections import defaultdict 

from itertools import islice 

 

from sympy.core import Add, S, Mul, Pow, oo 

from sympy.core.compatibility import ordered, iterable, is_sequence, range 

from sympy.core.containers import Tuple 

from sympy.core.exprtools import factor_terms 

from sympy.core.expr import AtomicExpr, Expr 

from sympy.core.function import (Function, Derivative, AppliedUndef, diff, 

    expand, expand_mul, Subs, _mexpand) 

from sympy.core.multidimensional import vectorize 

from sympy.core.numbers import NaN, zoo, I, Number 

from sympy.core.relational import Equality, Eq 

from sympy.core.symbol import Symbol, Wild, Dummy, symbols 

from sympy.core.sympify import sympify 

 

from sympy.logic.boolalg import BooleanAtom 

from sympy.functions import cos, exp, im, log, re, sin, tan, sqrt, \ 

    atan2, conjugate 

from sympy.functions.combinatorial.factorials import factorial 

from sympy.integrals.integrals import Integral, integrate 

from sympy.matrices import wronskian, Matrix, eye, zeros 

from sympy.polys import Poly, RootOf, terms_gcd, PolynomialError, lcm 

from sympy.polys.polyroots import roots_quartic 

from sympy.polys.polytools import cancel, degree, div 

from sympy.series import Order 

from sympy.series.series import series 

from sympy.simplify import collect, logcombine, powsimp, separatevars, \ 

    simplify, trigsimp, denom, posify, cse 

from sympy.simplify.simplify import collect_const, powdenest 

from sympy.solvers import solve 

from sympy.solvers.pde import pdsolve 

 

from sympy.utilities import numbered_symbols, default_sort_key, sift 

from sympy.solvers.deutils import _preprocess, ode_order, _desolve 

 

#: This is a list of hints in the order that they should be preferred by 

#: :py:meth:`~sympy.solvers.ode.classify_ode`. In general, hints earlier in the 

#: list should produce simpler solutions than those later in the list (for 

#: ODEs that fit both).  For now, the order of this list is based on empirical 

#: observations by the developers of SymPy. 

#: 

#: The hint used by :py:meth:`~sympy.solvers.ode.dsolve` for a specific ODE 

#: can be overridden (see the docstring). 

#: 

#: In general, ``_Integral`` hints are grouped at the end of the list, unless 

#: there is a method that returns an unevaluable integral most of the time 

#: (which go near the end of the list anyway).  ``default``, ``all``, 

#: ``best``, and ``all_Integral`` meta-hints should not be included in this 

#: list, but ``_best`` and ``_Integral`` hints should be included. 

allhints = ( 

    "separable", 

    "1st_exact", 

    "1st_linear", 

    "Bernoulli", 

    "Riccati_special_minus2", 

    "1st_homogeneous_coeff_best", 

    "1st_homogeneous_coeff_subs_indep_div_dep", 

    "1st_homogeneous_coeff_subs_dep_div_indep", 

    "almost_linear", 

    "linear_coefficients", 

    "separable_reduced", 

    "1st_power_series", 

    "lie_group", 

    "nth_linear_constant_coeff_homogeneous", 

    "nth_linear_euler_eq_homogeneous", 

    "nth_linear_constant_coeff_undetermined_coefficients", 

    "nth_linear_euler_eq_nonhomogeneous_undetermined_coefficients", 

    "nth_linear_constant_coeff_variation_of_parameters", 

    "nth_linear_euler_eq_nonhomogeneous_variation_of_parameters", 

    "Liouville", 

    "2nd_power_series_ordinary", 

    "2nd_power_series_regular", 

    "separable_Integral", 

    "1st_exact_Integral", 

    "1st_linear_Integral", 

    "Bernoulli_Integral", 

    "1st_homogeneous_coeff_subs_indep_div_dep_Integral", 

    "1st_homogeneous_coeff_subs_dep_div_indep_Integral", 

    "almost_linear_Integral", 

    "linear_coefficients_Integral", 

    "separable_reduced_Integral", 

    "nth_linear_constant_coeff_variation_of_parameters_Integral", 

    "nth_linear_euler_eq_nonhomogeneous_variation_of_parameters_Integral", 

    "Liouville_Integral", 

    ) 

 

lie_heuristics = ( 

    "abaco1_simple", 

    "abaco1_product", 

    "abaco2_similar", 

    "abaco2_unique_unknown", 

    "abaco2_unique_general", 

    "linear", 

    "function_sum", 

    "bivariate", 

    "chi" 

    ) 

 

 

def sub_func_doit(eq, func, new): 

    r""" 

    When replacing the func with something else, we usually want the 

    derivative evaluated, so this function helps in making that happen. 

 

    To keep subs from having to look through all derivatives, we mask them off 

    with dummy variables, do the func sub, and then replace masked-off 

    derivatives with their doit values. 

 

    Examples 

    ======== 

 

    >>> from sympy import Derivative, symbols, Function 

    >>> from sympy.solvers.ode import sub_func_doit 

    >>> x, z = symbols('x, z') 

    >>> y = Function('y') 

 

    >>> sub_func_doit(3*Derivative(y(x), x) - 1, y(x), x) 

    2 

 

    >>> sub_func_doit(x*Derivative(y(x), x) - y(x)**2 + y(x), y(x), 

    ... 1/(x*(z + 1/x))) 

    x*(-1/(x**2*(z + 1/x)) + 1/(x**3*(z + 1/x)**2)) + 1/(x*(z + 1/x)) 

    ...- 1/(x**2*(z + 1/x)**2) 

    """ 

    reps = {} 

    repu = {} 

    for d in eq.atoms(Derivative): 

        u = Dummy('u') 

        repu[u] = d.subs(func, new).doit() 

        reps[d] = u 

 

    return eq.subs(reps).subs(func, new).subs(repu) 

 

 

def get_numbered_constants(eq, num=1, start=1, prefix='C'): 

    """ 

    Returns a list of constants that do not occur 

    in eq already. 

    """ 

 

    if isinstance(eq, Expr): 

        eq = [eq] 

    elif not iterable(eq): 

        raise ValueError("Expected Expr or iterable but got %s" % eq) 

 

    atom_set = set().union(*[i.free_symbols for i in eq]) 

    ncs = numbered_symbols(start=start, prefix=prefix, exclude=atom_set) 

    Cs = [next(ncs) for i in range(num)] 

    return (Cs[0] if num == 1 else tuple(Cs)) 

 

 

def dsolve(eq, func=None, hint="default", simplify=True, 

    ics= None, xi=None, eta=None, x0=0, n=6, **kwargs): 

    r""" 

    Solves any (supported) kind of ordinary differential equation and 

    system of ordinary differential equations. 

 

    For single ordinary differential equation 

    ========================================= 

 

    It is classified under this when number of equation in ``eq`` is one. 

    **Usage** 

 

        ``dsolve(eq, f(x), hint)`` -> Solve ordinary differential equation 

        ``eq`` for function ``f(x)``, using method ``hint``. 

 

    **Details** 

 

        ``eq`` can be any supported ordinary differential equation (see the 

            :py:mod:`~sympy.solvers.ode` docstring for supported methods). 

            This can either be an :py:class:`~sympy.core.relational.Equality`, 

            or an expression, which is assumed to be equal to ``0``. 

 

        ``f(x)`` is a function of one variable whose derivatives in that 

            variable make up the ordinary differential equation ``eq``.  In 

            many cases it is not necessary to provide this; it will be 

            autodetected (and an error raised if it couldn't be detected). 

 

        ``hint`` is the solving method that you want dsolve to use.  Use 

            ``classify_ode(eq, f(x))`` to get all of the possible hints for an 

            ODE.  The default hint, ``default``, will use whatever hint is 

            returned first by :py:meth:`~sympy.solvers.ode.classify_ode`.  See 

            Hints below for more options that you can use for hint. 

 

        ``simplify`` enables simplification by 

            :py:meth:`~sympy.solvers.ode.odesimp`.  See its docstring for more 

            information.  Turn this off, for example, to disable solving of 

            solutions for ``func`` or simplification of arbitrary constants. 

            It will still integrate with this hint. Note that the solution may 

            contain more arbitrary constants than the order of the ODE with 

            this option enabled. 

 

        ``xi`` and ``eta`` are the infinitesimal functions of an ordinary 

            differential equation. They are the infinitesimals of the Lie group 

            of point transformations for which the differential equation is 

            invariant. The user can specify values for the infinitesimals. If 

            nothing is specified, ``xi`` and ``eta`` are calculated using 

            :py:meth:`~sympy.solvers.ode.infinitesimals` with the help of various 

            heuristics. 

 

        ``ics`` is the set of boundary conditions for the differential equation. 

          It should be given in the form of ``{f(x0): x1, f(x).diff(x).subs(x, x2): 

          x3}`` and so on. For now initial conditions are implemented only for 

          power series solutions of first-order differential equations which should 

          be given in the form of ``{f(x0): x1}`` (See issue 4720). If nothing is 

          specified for this case ``f(0)`` is assumed to be ``C0`` and the power 

          series solution is calculated about 0. 

 

        ``x0`` is the point about which the power series solution of a differential 

          equation is to be evaluated. 

 

        ``n`` gives the exponent of the dependent variable up to which the power series 

          solution of a differential equation is to be evaluated. 

 

    **Hints** 

 

        Aside from the various solving methods, there are also some meta-hints 

        that you can pass to :py:meth:`~sympy.solvers.ode.dsolve`: 

 

        ``default``: 

                This uses whatever hint is returned first by 

                :py:meth:`~sympy.solvers.ode.classify_ode`. This is the 

                default argument to :py:meth:`~sympy.solvers.ode.dsolve`. 

 

        ``all``: 

                To make :py:meth:`~sympy.solvers.ode.dsolve` apply all 

                relevant classification hints, use ``dsolve(ODE, func, 

                hint="all")``.  This will return a dictionary of 

                ``hint:solution`` terms.  If a hint causes dsolve to raise the 

                ``NotImplementedError``, value of that hint's key will be the 

                exception object raised.  The dictionary will also include 

                some special keys: 

 

                - ``order``: The order of the ODE.  See also 

                  :py:meth:`~sympy.solvers.deutils.ode_order` in 

                  ``deutils.py``. 

                - ``best``: The simplest hint; what would be returned by 

                  ``best`` below. 

                - ``best_hint``: The hint that would produce the solution 

                  given by ``best``.  If more than one hint produces the best 

                  solution, the first one in the tuple returned by 

                  :py:meth:`~sympy.solvers.ode.classify_ode` is chosen. 

                - ``default``: The solution that would be returned by default. 

                  This is the one produced by the hint that appears first in 

                  the tuple returned by 

                  :py:meth:`~sympy.solvers.ode.classify_ode`. 

 

        ``all_Integral``: 

                This is the same as ``all``, except if a hint also has a 

                corresponding ``_Integral`` hint, it only returns the 

                ``_Integral`` hint.  This is useful if ``all`` causes 

                :py:meth:`~sympy.solvers.ode.dsolve` to hang because of a 

                difficult or impossible integral.  This meta-hint will also be 

                much faster than ``all``, because 

                :py:meth:`~sympy.core.expr.Expr.integrate` is an expensive 

                routine. 

 

        ``best``: 

                To have :py:meth:`~sympy.solvers.ode.dsolve` try all methods 

                and return the simplest one.  This takes into account whether 

                the solution is solvable in the function, whether it contains 

                any Integral classes (i.e.  unevaluatable integrals), and 

                which one is the shortest in size. 

 

        See also the :py:meth:`~sympy.solvers.ode.classify_ode` docstring for 

        more info on hints, and the :py:mod:`~sympy.solvers.ode` docstring for 

        a list of all supported hints. 

 

    **Tips** 

 

        - You can declare the derivative of an unknown function this way: 

 

            >>> from sympy import Function, Derivative 

            >>> from sympy.abc import x # x is the independent variable 

            >>> f = Function("f")(x) # f is a function of x 

            >>> # f_ will be the derivative of f with respect to x 

            >>> f_ = Derivative(f, x) 

 

        - See ``test_ode.py`` for many tests, which serves also as a set of 

          examples for how to use :py:meth:`~sympy.solvers.ode.dsolve`. 

        - :py:meth:`~sympy.solvers.ode.dsolve` always returns an 

          :py:class:`~sympy.core.relational.Equality` class (except for the 

          case when the hint is ``all`` or ``all_Integral``).  If possible, it 

          solves the solution explicitly for the function being solved for. 

          Otherwise, it returns an implicit solution. 

        - Arbitrary constants are symbols named ``C1``, ``C2``, and so on. 

        - Because all solutions should be mathematically equivalent, some 

          hints may return the exact same result for an ODE. Often, though, 

          two different hints will return the same solution formatted 

          differently.  The two should be equivalent. Also note that sometimes 

          the values of the arbitrary constants in two different solutions may 

          not be the same, because one constant may have "absorbed" other 

          constants into it. 

        - Do ``help(ode.ode_<hintname>)`` to get help more information on a 

          specific hint, where ``<hintname>`` is the name of a hint without 

          ``_Integral``. 

 

    For system of ordinary differential equations 

    ============================================= 

 

   **Usage** 

        ``dsolve(eq, func)`` -> Solve a system of ordinary differential 

        equations ``eq`` for ``func`` being list of functions including 

        `x(t)`, `y(t)`, `z(t)` where number of functions in the list depends 

        upon the number of equations provided in ``eq``. 

 

    **Details** 

 

        ``eq`` can be any supported system of ordinary differential equations 

        This can either be an :py:class:`~sympy.core.relational.Equality`, 

        or an expression, which is assumed to be equal to ``0``. 

 

        ``func`` holds ``x(t)`` and ``y(t)`` being functions of one variable which 

        together with some of their derivatives make up the system of ordinary 

        differential equation ``eq``. It is not necessary to provide this; it 

        will be autodetected (and an error raised if it couldn't be detected). 

 

    **Hints** 

 

        The hints are formed by parameters returned by classify_sysode, combining 

        them give hints name used later for forming method name. 

 

    Examples 

    ======== 

 

    >>> from sympy import Function, dsolve, Eq, Derivative, sin, cos, symbols 

    >>> from sympy.abc import x 

    >>> f = Function('f') 

    >>> dsolve(Derivative(f(x), x, x) + 9*f(x), f(x)) 

    Eq(f(x), C1*sin(3*x) + C2*cos(3*x)) 

 

    >>> eq = sin(x)*cos(f(x)) + cos(x)*sin(f(x))*f(x).diff(x) 

    >>> dsolve(eq, hint='1st_exact') 

    [Eq(f(x), -acos(C1/cos(x)) + 2*pi), Eq(f(x), acos(C1/cos(x)))] 

    >>> dsolve(eq, hint='almost_linear') 

    [Eq(f(x), -acos(C1/sqrt(-cos(x)**2)) + 2*pi), Eq(f(x), acos(C1/sqrt(-cos(x)**2)))] 

    >>> t = symbols('t') 

    >>> x, y = symbols('x, y', function=True) 

    >>> eq = (Eq(Derivative(x(t),t), 12*t*x(t) + 8*y(t)), Eq(Derivative(y(t),t), 21*x(t) + 7*t*y(t))) 

    >>> dsolve(eq) 

    [Eq(x(t), C1*x0 + C2*x0*Integral(8*exp(Integral(7*t, t))*exp(Integral(12*t, t))/x0**2, t)), 

    Eq(y(t), C1*y0 + C2(y0*Integral(8*exp(Integral(7*t, t))*exp(Integral(12*t, t))/x0**2, t) + 

    exp(Integral(7*t, t))*exp(Integral(12*t, t))/x0))] 

    >>> eq = (Eq(Derivative(x(t),t),x(t)*y(t)*sin(t)), Eq(Derivative(y(t),t),y(t)**2*sin(t))) 

    >>> dsolve(eq) 

    set([Eq(x(t), -exp(C1)/(C2*exp(C1) - cos(t))), Eq(y(t), -1/(C1 - cos(t)))]) 

    """ 

    if iterable(eq): 

        match = classify_sysode(eq, func) 

        eq = match['eq'] 

        order = match['order'] 

        func = match['func'] 

        t = list(list(eq[0].atoms(Derivative))[0].atoms(Symbol))[0] 

 

        # keep highest order term coefficient positive 

        for i in range(len(eq)): 

            for func_ in func: 

                if isinstance(func_, list): 

                    pass 

                else: 

                    if eq[i].coeff(diff(func[i],t,ode_order(eq[i], func[i]))).is_negative: 

                        eq[i] = -eq[i] 

        match['eq'] = eq 

        if len(set(order.values()))!=1: 

            raise ValueError("It solves only those systems of equations whose orders are equal") 

        match['order'] = list(order.values())[0] 

        def recur_len(l): 

            return sum(recur_len(item) if isinstance(item,list) else 1 for item in l) 

        if recur_len(func) != len(eq): 

            raise ValueError("dsolve() and classify_sysode() work with " 

            "number of functions being equal to number of equations") 

        if match['type_of_equation'] is None: 

            raise NotImplementedError 

        else: 

            if match['is_linear'] == True: 

                if match['no_of_equation'] > 3: 

                    solvefunc = globals()['sysode_linear_neq_order%(order)s' % match] 

                else: 

                    solvefunc = globals()['sysode_linear_%(no_of_equation)seq_order%(order)s' % match] 

            else: 

                solvefunc = globals()['sysode_nonlinear_%(no_of_equation)seq_order%(order)s' % match] 

            sols = solvefunc(match) 

            return sols 

    else: 

        given_hint = hint  # hint given by the user 

 

        # See the docstring of _desolve for more details. 

        hints = _desolve(eq, func=func, 

            hint=hint, simplify=True, xi=xi, eta=eta, type='ode', ics=ics, 

            x0=x0, n=n, **kwargs) 

 

        eq = hints.pop('eq', eq) 

        all_ = hints.pop('all', False) 

        if all_: 

            retdict = {} 

            failed_hints = {} 

            gethints = classify_ode(eq, dict=True) 

            orderedhints = gethints['ordered_hints'] 

            for hint in hints: 

                try: 

                    rv = _helper_simplify(eq, hint, hints[hint], simplify) 

                except NotImplementedError as detail: 

                    failed_hints[hint] = detail 

                else: 

                    retdict[hint] = rv 

            func = hints[hint]['func'] 

 

            retdict['best'] = min(list(retdict.values()), key=lambda x: 

                ode_sol_simplicity(x, func, trysolving=not simplify)) 

            if given_hint == 'best': 

                return retdict['best'] 

            for i in orderedhints: 

                if retdict['best'] == retdict.get(i, None): 

                    retdict['best_hint'] = i 

                    break 

            retdict['default'] = gethints['default'] 

            retdict['order'] = gethints['order'] 

            retdict.update(failed_hints) 

            return retdict 

 

        else: 

            # The key 'hint' stores the hint needed to be solved for. 

            hint = hints['hint'] 

            return _helper_simplify(eq, hint, hints, simplify) 

 

def _helper_simplify(eq, hint, match, simplify=True, **kwargs): 

    r""" 

    Helper function of dsolve that calls the respective 

    :py:mod:`~sympy.solvers.ode` functions to solve for the ordinary 

    differential equations. This minimises the computation in calling 

    :py:meth:`~sympy.solvers.deutils._desolve` multiple times. 

    """ 

    r = match 

    if hint.endswith('_Integral'): 

        solvefunc = globals()['ode_' + hint[:-len('_Integral')]] 

    else: 

        solvefunc = globals()['ode_' + hint] 

    func = r['func'] 

    order = r['order'] 

    match = r[hint] 

 

    if simplify: 

        # odesimp() will attempt to integrate, if necessary, apply constantsimp(), 

        # attempt to solve for func, and apply any other hint specific 

        # simplifications 

        sols = solvefunc(eq, func, order, match) 

        free = eq.free_symbols 

        cons = lambda s: s.free_symbols.difference(free) 

        if isinstance(sols, Expr): 

            return odesimp(sols, func, order, cons(sols), hint) 

        return [odesimp(s, func, order, cons(s), hint) for s in sols] 

    else: 

        # We still want to integrate (you can disable it separately with the hint) 

        match['simplify'] = False  # Some hints can take advantage of this option 

        rv = _handle_Integral(solvefunc(eq, func, order, match), 

            func, order, hint) 

        return rv 

 

def classify_ode(eq, func=None, dict=False, ics=None, **kwargs): 

    r""" 

    Returns a tuple of possible :py:meth:`~sympy.solvers.ode.dsolve` 

    classifications for an ODE. 

 

    The tuple is ordered so that first item is the classification that 

    :py:meth:`~sympy.solvers.ode.dsolve` uses to solve the ODE by default.  In 

    general, classifications at the near the beginning of the list will 

    produce better solutions faster than those near the end, thought there are 

    always exceptions.  To make :py:meth:`~sympy.solvers.ode.dsolve` use a 

    different classification, use ``dsolve(ODE, func, 

    hint=<classification>)``.  See also the 

    :py:meth:`~sympy.solvers.ode.dsolve` docstring for different meta-hints 

    you can use. 

 

    If ``dict`` is true, :py:meth:`~sympy.solvers.ode.classify_ode` will 

    return a dictionary of ``hint:match`` expression terms. This is intended 

    for internal use by :py:meth:`~sympy.solvers.ode.dsolve`.  Note that 

    because dictionaries are ordered arbitrarily, this will most likely not be 

    in the same order as the tuple. 

 

    You can get help on different hints by executing 

    ``help(ode.ode_hintname)``, where ``hintname`` is the name of the hint 

    without ``_Integral``. 

 

    See :py:data:`~sympy.solvers.ode.allhints` or the 

    :py:mod:`~sympy.solvers.ode` docstring for a list of all supported hints 

    that can be returned from :py:meth:`~sympy.solvers.ode.classify_ode`. 

 

    Notes 

    ===== 

 

    These are remarks on hint names. 

 

    ``_Integral`` 

 

        If a classification has ``_Integral`` at the end, it will return the 

        expression with an unevaluated :py:class:`~sympy.integrals.Integral` 

        class in it.  Note that a hint may do this anyway if 

        :py:meth:`~sympy.core.expr.Expr.integrate` cannot do the integral, 

        though just using an ``_Integral`` will do so much faster.  Indeed, an 

        ``_Integral`` hint will always be faster than its corresponding hint 

        without ``_Integral`` because 

        :py:meth:`~sympy.core.expr.Expr.integrate` is an expensive routine. 

        If :py:meth:`~sympy.solvers.ode.dsolve` hangs, it is probably because 

        :py:meth:`~sympy.core.expr.Expr.integrate` is hanging on a tough or 

        impossible integral.  Try using an ``_Integral`` hint or 

        ``all_Integral`` to get it return something. 

 

        Note that some hints do not have ``_Integral`` counterparts.  This is 

        because :py:meth:`~sympy.solvers.ode.integrate` is not used in solving 

        the ODE for those method. For example, `n`\th order linear homogeneous 

        ODEs with constant coefficients do not require integration to solve, 

        so there is no ``nth_linear_homogeneous_constant_coeff_Integrate`` 

        hint. You can easily evaluate any unevaluated 

        :py:class:`~sympy.integrals.Integral`\s in an expression by doing 

        ``expr.doit()``. 

 

    Ordinals 

 

        Some hints contain an ordinal such as ``1st_linear``.  This is to help 

        differentiate them from other hints, as well as from other methods 

        that may not be implemented yet. If a hint has ``nth`` in it, such as 

        the ``nth_linear`` hints, this means that the method used to applies 

        to ODEs of any order. 

 

    ``indep`` and ``dep`` 

 

        Some hints contain the words ``indep`` or ``dep``.  These reference 

        the independent variable and the dependent function, respectively. For 

        example, if an ODE is in terms of `f(x)`, then ``indep`` will refer to 

        `x` and ``dep`` will refer to `f`. 

 

    ``subs`` 

 

        If a hints has the word ``subs`` in it, it means the the ODE is solved 

        by substituting the expression given after the word ``subs`` for a 

        single dummy variable.  This is usually in terms of ``indep`` and 

        ``dep`` as above.  The substituted expression will be written only in 

        characters allowed for names of Python objects, meaning operators will 

        be spelled out.  For example, ``indep``/``dep`` will be written as 

        ``indep_div_dep``. 

 

    ``coeff`` 

 

        The word ``coeff`` in a hint refers to the coefficients of something 

        in the ODE, usually of the derivative terms.  See the docstring for 

        the individual methods for more info (``help(ode)``).  This is 

        contrast to ``coefficients``, as in ``undetermined_coefficients``, 

        which refers to the common name of a method. 

 

    ``_best`` 

 

        Methods that have more than one fundamental way to solve will have a 

        hint for each sub-method and a ``_best`` meta-classification. This 

        will evaluate all hints and return the best, using the same 

        considerations as the normal ``best`` meta-hint. 

 

 

    Examples 

    ======== 

 

    >>> from sympy import Function, classify_ode, Eq 

    >>> from sympy.abc import x 

    >>> f = Function('f') 

    >>> classify_ode(Eq(f(x).diff(x), 0), f(x)) 

    ('separable', '1st_linear', '1st_homogeneous_coeff_best', 

    '1st_homogeneous_coeff_subs_indep_div_dep', 

    '1st_homogeneous_coeff_subs_dep_div_indep', 

    '1st_power_series', 'lie_group', 

    'nth_linear_constant_coeff_homogeneous', 

    'separable_Integral', '1st_linear_Integral', 

    '1st_homogeneous_coeff_subs_indep_div_dep_Integral', 

    '1st_homogeneous_coeff_subs_dep_div_indep_Integral') 

    >>> classify_ode(f(x).diff(x, 2) + 3*f(x).diff(x) + 2*f(x) - 4) 

    ('nth_linear_constant_coeff_undetermined_coefficients', 

    'nth_linear_constant_coeff_variation_of_parameters', 

    'nth_linear_constant_coeff_variation_of_parameters_Integral') 

 

    """ 

    prep = kwargs.pop('prep', True) 

 

    if func and len(func.args) != 1: 

        raise ValueError("dsolve() and classify_ode() only " 

        "work with functions of one variable, not %s" % func) 

    if prep or func is None: 

        eq, func_ = _preprocess(eq, func) 

        if func is None: 

            func = func_ 

    x = func.args[0] 

    f = func.func 

    y = Dummy('y') 

    xi = kwargs.get('xi') 

    eta = kwargs.get('eta') 

    terms = kwargs.get('n') 

 

    if isinstance(eq, Equality): 

        if eq.rhs != 0: 

            return classify_ode(eq.lhs - eq.rhs, func, ics=ics, xi=xi, 

                n=terms, eta=eta, prep=False) 

        eq = eq.lhs 

    order = ode_order(eq, f(x)) 

    # hint:matchdict or hint:(tuple of matchdicts) 

    # Also will contain "default":<default hint> and "order":order items. 

    matching_hints = {"order": order} 

 

    if not order: 

        if dict: 

            matching_hints["default"] = None 

            return matching_hints 

        else: 

            return () 

 

    df = f(x).diff(x) 

    a = Wild('a', exclude=[f(x)]) 

    b = Wild('b', exclude=[f(x)]) 

    c = Wild('c', exclude=[f(x)]) 

    d = Wild('d', exclude=[df, f(x).diff(x, 2)]) 

    e = Wild('e', exclude=[df]) 

    k = Wild('k', exclude=[df]) 

    n = Wild('n', exclude=[f(x)]) 

    c1 = Wild('c1', exclude=[x]) 

    a2 = Wild('a2', exclude=[x, f(x), df]) 

    b2 = Wild('b2', exclude=[x, f(x), df]) 

    c2 = Wild('c2', exclude=[x, f(x), df]) 

    d2 = Wild('d2', exclude=[x, f(x), df]) 

    a3 = Wild('a3', exclude=[f(x), df, f(x).diff(x, 2)]) 

    b3 = Wild('b3', exclude=[f(x), df, f(x).diff(x, 2)]) 

    c3 = Wild('c3', exclude=[f(x), df, f(x).diff(x, 2)]) 

    r3 = {'xi': xi, 'eta': eta}  # Used for the lie_group hint 

    boundary = {}  # Used to extract initial conditions 

    C1 = Symbol("C1") 

    eq = expand(eq) 

 

    # Preprocessing to get the initial conditions out 

    if ics is not None: 

        for funcarg in ics: 

            # Separating derivatives 

            if isinstance(funcarg, Subs): 

                deriv = funcarg.expr 

                old = funcarg.variables[0] 

                new = funcarg.point[0] 

                if isinstance(deriv, Derivative) and isinstance(deriv.args[0], 

                    AppliedUndef) and deriv.args[0].func == f and old == x and not new.has(x): 

                    dorder = ode_order(deriv, x) 

                    temp = 'f' + str(dorder) 

                    boundary.update({temp: new, temp + 'val': ics[funcarg]}) 

                else: 

                    raise ValueError("Enter valid boundary conditions for Derivatives") 

 

 

            # Separating functions 

            elif isinstance(funcarg, AppliedUndef): 

                if funcarg.func == f and len(funcarg.args) == 1 and \ 

                    not funcarg.args[0].has(x): 

                    boundary.update({'f0': funcarg.args[0], 'f0val': ics[funcarg]}) 

                else: 

                    raise ValueError("Enter valid boundary conditions for Function") 

 

            else: 

                raise ValueError("Enter boundary conditions of the form ics " 

                    " = {f(point}: value, f(point).diff(point, order).subs(arg, point) " 

                    ":value") 

 

    # Precondition to try remove f(x) from highest order derivative 

    reduced_eq = None 

    if eq.is_Add: 

        deriv_coef = eq.coeff(f(x).diff(x, order)) 

        if deriv_coef not in (1, 0): 

            r = deriv_coef.match(a*f(x)**c1) 

            if r and r[c1]: 

                den = f(x)**r[c1] 

                reduced_eq = Add(*[arg/den for arg in eq.args]) 

    if not reduced_eq: 

        reduced_eq = eq 

 

    if order == 1: 

 

        ## Linear case: a(x)*y'+b(x)*y+c(x) == 0 

        if eq.is_Add: 

            ind, dep = reduced_eq.as_independent(f) 

        else: 

            u = Dummy('u') 

            ind, dep = (reduced_eq + u).as_independent(f) 

            ind, dep = [tmp.subs(u, 0) for tmp in [ind, dep]] 

        r = {a: dep.coeff(df), 

             b: dep.coeff(f(x)), 

             c: ind} 

        # double check f[a] since the preconditioning may have failed 

        if not r[a].has(f) and not r[b].has(f) and ( 

                r[a]*df + r[b]*f(x) + r[c]).expand() - reduced_eq == 0: 

            r['a'] = a 

            r['b'] = b 

            r['c'] = c 

            matching_hints["1st_linear"] = r 

            matching_hints["1st_linear_Integral"] = r 

 

        ## Bernoulli case: a(x)*y'+b(x)*y+c(x)*y**n == 0 

        r = collect( 

            reduced_eq, f(x), exact=True).match(a*df + b*f(x) + c*f(x)**n) 

        if r and r[c] != 0 and r[n] != 1:  # See issue 4676 

            r['a'] = a 

            r['b'] = b 

            r['c'] = c 

            r['n'] = n 

            matching_hints["Bernoulli"] = r 

            matching_hints["Bernoulli_Integral"] = r 

 

        ## Riccati special n == -2 case: a2*y'+b2*y**2+c2*y/x+d2/x**2 == 0 

        r = collect(reduced_eq, 

            f(x), exact=True).match(a2*df + b2*f(x)**2 + c2*f(x)/x + d2/x**2) 

        if r and r[b2] != 0 and (r[c2] != 0 or r[d2] != 0): 

            r['a2'] = a2 

            r['b2'] = b2 

            r['c2'] = c2 

            r['d2'] = d2 

            matching_hints["Riccati_special_minus2"] = r 

 

        # NON-REDUCED FORM OF EQUATION matches 

        r = collect(eq, df, exact=True).match(d + e * df) 

        if r: 

            r['d'] = d 

            r['e'] = e 

            r['y'] = y 

            r[d] = r[d].subs(f(x), y) 

            r[e] = r[e].subs(f(x), y) 

 

            # FIRST ORDER POWER SERIES WHICH NEEDS INITIAL CONDITIONS 

            # TODO: Hint first order series should match only if d/e is analytic. 

            # For now, only d/e and (d/e).diff(arg) is checked for existence at 

            # at a given point. 

            # This is currently done internally in ode_1st_power_series. 

            point = boundary.get('f0', 0) 

            value = boundary.get('f0val', C1) 

            check = cancel(r[d]/r[e]) 

            check1 = check.subs({x: point, y: value}) 

            if not check1.has(oo) and not check1.has(zoo) and \ 

                not check1.has(NaN) and not check1.has(-oo): 

                check2 = (check1.diff(x)).subs({x: point, y: value}) 

                if not check2.has(oo) and not check2.has(zoo) and \ 

                    not check2.has(NaN) and not check2.has(-oo): 

                    rseries = r.copy() 

                    rseries.update({'terms': terms, 'f0': point, 'f0val': value}) 

                    matching_hints["1st_power_series"] = rseries 

 

            r3.update(r) 

            ## Exact Differential Equation: P(x, y) + Q(x, y)*y' = 0 where 

            # dP/dy == dQ/dx 

            try: 

                if r[d] != 0: 

                    numerator = simplify(r[d].diff(y) - r[e].diff(x)) 

                    # The following few conditions try to convert a non-exact 

                    # differential equation into an exact one. 

                    # References : Differential equations with applications 

                    # and historical notes - George E. Simmons 

 

                    if numerator: 

                        # If (dP/dy - dQ/dx) / Q = f(x) 

                        # then exp(integral(f(x))*equation becomes exact 

                        factor = simplify(numerator/r[e]) 

                        variables = factor.free_symbols 

                        if len(variables) == 1 and x == variables.pop(): 

                            factor = exp(Integral(factor).doit()) 

                            r[d] *= factor 

                            r[e] *= factor 

                            matching_hints["1st_exact"] = r 

                            matching_hints["1st_exact_Integral"] = r 

                        else: 

                            # If (dP/dy - dQ/dx) / -P = f(y) 

                            # then exp(integral(f(y))*equation becomes exact 

                            factor = simplify(-numerator/r[d]) 

                            variables = factor.free_symbols 

                            if len(variables) == 1 and y == variables.pop(): 

                                factor = exp(Integral(factor).doit()) 

                                r[d] *= factor 

                                r[e] *= factor 

                                matching_hints["1st_exact"] = r 

                                matching_hints["1st_exact_Integral"] = r 

                    else: 

                        matching_hints["1st_exact"] = r 

                        matching_hints["1st_exact_Integral"] = r 

 

            except NotImplementedError: 

                # Differentiating the coefficients might fail because of things 

                # like f(2*x).diff(x).  See issue 4624 and issue 4719. 

                pass 

 

        # Any first order ODE can be ideally solved by the Lie Group 

        # method 

        matching_hints["lie_group"] = r3 

 

        # This match is used for several cases below; we now collect on 

        # f(x) so the matching works. 

        r = collect(reduced_eq, df, exact=True).match(d + e*df) 

        if r: 

            # Using r[d] and r[e] without any modification for hints 

            # linear-coefficients and separable-reduced. 

            num, den = r[d], r[e]  # ODE = d/e + df 

            r['d'] = d 

            r['e'] = e 

            r['y'] = y 

            r[d] = num.subs(f(x), y) 

            r[e] = den.subs(f(x), y) 

 

            ## Separable Case: y' == P(y)*Q(x) 

            r[d] = separatevars(r[d]) 

            r[e] = separatevars(r[e]) 

            # m1[coeff]*m1[x]*m1[y] + m2[coeff]*m2[x]*m2[y]*y' 

            m1 = separatevars(r[d], dict=True, symbols=(x, y)) 

            m2 = separatevars(r[e], dict=True, symbols=(x, y)) 

            if m1 and m2: 

                r1 = {'m1': m1, 'm2': m2, 'y': y} 

                matching_hints["separable"] = r1 

                matching_hints["separable_Integral"] = r1 

 

            ## First order equation with homogeneous coefficients: 

            # dy/dx == F(y/x) or dy/dx == F(x/y) 

            ordera = homogeneous_order(r[d], x, y) 

            if ordera is not None: 

                orderb = homogeneous_order(r[e], x, y) 

                if ordera == orderb: 

                    # u1=y/x and u2=x/y 

                    u1 = Dummy('u1') 

                    u2 = Dummy('u2') 

                    s = "1st_homogeneous_coeff_subs" 

                    s1 = s + "_dep_div_indep" 

                    s2 = s + "_indep_div_dep" 

                    if simplify((r[d] + u1*r[e]).subs({x: 1, y: u1})) != 0: 

                        matching_hints[s1] = r 

                        matching_hints[s1 + "_Integral"] = r 

                    if simplify((r[e] + u2*r[d]).subs({x: u2, y: 1})) != 0: 

                        matching_hints[s2] = r 

                        matching_hints[s2 + "_Integral"] = r 

                    if s1 in matching_hints and s2 in matching_hints: 

                        matching_hints["1st_homogeneous_coeff_best"] = r 

 

            ## Linear coefficients of the form 

            # y'+ F((a*x + b*y + c)/(a'*x + b'y + c')) = 0 

            # that can be reduced to homogeneous form. 

            F = num/den 

            params = _linear_coeff_match(F, func) 

            if params: 

                xarg, yarg = params 

                u = Dummy('u') 

                t = Dummy('t') 

                # Dummy substitution for df and f(x). 

                dummy_eq = reduced_eq.subs(((df, t), (f(x), u))) 

                reps = ((x, x + xarg), (u, u + yarg), (t, df), (u, f(x))) 

                dummy_eq = simplify(dummy_eq.subs(reps)) 

                # get the re-cast values for e and d 

                r2 = collect(expand(dummy_eq), [df, f(x)]).match(e*df + d) 

                if r2: 

                    orderd = homogeneous_order(r2[d], x, f(x)) 

                    if orderd is not None: 

                        ordere = homogeneous_order(r2[e], x, f(x)) 

                        if orderd == ordere: 

                            # Match arguments are passed in such a way that it 

                            # is coherent with the already existing homogeneous 

                            # functions. 

                            r2[d] = r2[d].subs(f(x), y) 

                            r2[e] = r2[e].subs(f(x), y) 

                            r2.update({'xarg': xarg, 'yarg': yarg, 

                                'd': d, 'e': e, 'y': y}) 

                            matching_hints["linear_coefficients"] = r2 

                            matching_hints["linear_coefficients_Integral"] = r2 

 

            ## Equation of the form y' + (y/x)*H(x^n*y) = 0 

            # that can be reduced to separable form 

 

            factor = simplify(x/f(x)*num/den) 

 

            # Try representing factor in terms of x^n*y 

            # where n is lowest power of x in factor; 

            # first remove terms like sqrt(2)*3 from factor.atoms(Mul) 

            u = None 

            for mul in ordered(factor.atoms(Mul)): 

                if mul.has(x): 

                    _, u = mul.as_independent(x, f(x)) 

                    break 

            if u and u.has(f(x)): 

                h = x**(degree(Poly(u.subs(f(x), y), gen=x)))*f(x) 

                p = Wild('p') 

                if (u/h == 1) or ((u/h).simplify().match(x**p)): 

                    t = Dummy('t') 

                    r2 = {'t': t} 

                    xpart, ypart = u.as_independent(f(x)) 

                    test = factor.subs(((u, t), (1/u, 1/t))) 

                    free = test.free_symbols 

                    if len(free) == 1 and free.pop() == t: 

                        r2.update({'power': xpart.as_base_exp()[1], 'u': test}) 

                        matching_hints["separable_reduced"] = r2 

                        matching_hints["separable_reduced_Integral"] = r2 

 

        ## Almost-linear equation of the form f(x)*g(y)*y' + k(x)*l(y) + m(x) = 0 

        r = collect(eq, [df, f(x)]).match(e*df + d) 

        if r: 

            r2 = r.copy() 

            r2[c] = S.Zero 

            if r2[d].is_Add: 

                # Separate the terms having f(x) to r[d] and 

                # remaining to r[c] 

                no_f, r2[d] = r2[d].as_independent(f(x)) 

                r2[c] += no_f 

            factor = simplify(r2[d].diff(f(x))/r[e]) 

            if factor and not factor.has(f(x)): 

                r2[d] = factor_terms(r2[d]) 

                u = r2[d].as_independent(f(x), as_Add=False)[1] 

                r2.update({'a': e, 'b': d, 'c': c, 'u': u}) 

                r2[d] /= u 

                r2[e] /= u.diff(f(x)) 

                matching_hints["almost_linear"] = r2 

                matching_hints["almost_linear_Integral"] = r2 

 

 

    elif order == 2: 

        # Liouville ODE in the form 

        # f(x).diff(x, 2) + g(f(x))*(f(x).diff(x))**2 + h(x)*f(x).diff(x) 

        # See Goldstein and Braun, "Advanced Methods for the Solution of 

        # Differential Equations", pg. 98 

 

        s = d*f(x).diff(x, 2) + e*df**2 + k*df 

        r = reduced_eq.match(s) 

        if r and r[d] != 0: 

            y = Dummy('y') 

            g = simplify(r[e]/r[d]).subs(f(x), y) 

            h = simplify(r[k]/r[d]) 

            if h.has(f(x)) or g.has(x): 

                pass 

            else: 

                r = {'g': g, 'h': h, 'y': y} 

                matching_hints["Liouville"] = r 

                matching_hints["Liouville_Integral"] = r 

 

        # Homogeneous second order differential equation of the form 

        # a3*f(x).diff(x, 2) + b3*f(x).diff(x) + c3, where 

        # for simplicity, a3, b3 and c3 are assumed to be polynomials. 

        # It has a definite power series solution at point x0 if, b3/a3 and c3/a3 

        # are analytic at x0. 

        deq = a3*(f(x).diff(x, 2)) + b3*df + c3*f(x) 

        r = collect(reduced_eq, 

            [f(x).diff(x, 2), f(x).diff(x), f(x)]).match(deq) 

        ordinary = False 

        if r and r[a3] != 0: 

            if all([r[key].is_polynomial() for key in r]): 

                p = cancel(r[b3]/r[a3])  # Used below 

                q = cancel(r[c3]/r[a3])  # Used below 

                point = kwargs.get('x0', 0) 

                check = p.subs(x, point) 

                if not check.has(oo) and not check.has(NaN) and \ 

                    not check.has(zoo) and not check.has(-oo): 

                    check = q.subs(x, point) 

                    if not check.has(oo) and not check.has(NaN) and \ 

                        not check.has(zoo) and not check.has(-oo): 

                        ordinary = True 

                        r.update({'a3': a3, 'b3': b3, 'c3': c3, 'x0': point, 'terms': terms}) 

                        matching_hints["2nd_power_series_ordinary"] = r 

 

                # Checking if the differential equation has a regular singular point 

                # at x0. It has a regular singular point at x0, if (b3/a3)*(x - x0) 

                # and (c3/a3)*((x - x0)**2) are analytic at x0. 

                if not ordinary: 

                    p = cancel((x - point)*p) 

                    check = p.subs(x, point) 

                    if not check.has(oo) and not check.has(NaN) and \ 

                        not check.has(zoo) and not check.has(-oo): 

                        q = cancel(((x - point)**2)*q) 

                        check = q.subs(x, point) 

                        if not check.has(oo) and not check.has(NaN) and \ 

                            not check.has(zoo) and not check.has(-oo): 

                            coeff_dict = {'p': p, 'q': q, 'x0': point, 'terms': terms} 

                            matching_hints["2nd_power_series_regular"] = coeff_dict 

 

 

    if order > 0: 

        # nth order linear ODE 

        # a_n(x)y^(n) + ... + a_1(x)y' + a_0(x)y = F(x) = b 

 

        r = _nth_linear_match(reduced_eq, func, order) 

 

        # Constant coefficient case (a_i is constant for all i) 

        if r and not any(r[i].has(x) for i in r if i >= 0): 

            # Inhomogeneous case: F(x) is not identically 0 

            if r[-1]: 

                undetcoeff = _undetermined_coefficients_match(r[-1], x) 

                s = "nth_linear_constant_coeff_variation_of_parameters" 

                matching_hints[s] = r 

                matching_hints[s + "_Integral"] = r 

                if undetcoeff['test']: 

                    r['trialset'] = undetcoeff['trialset'] 

                    matching_hints[ 

                        "nth_linear_constant_coeff_undetermined_coefficients" 

                            ] = r 

 

            # Homogeneous case: F(x) is identically 0 

            else: 

                matching_hints["nth_linear_constant_coeff_homogeneous"] = r 

 

        # nth order Euler equation a_n*x**n*y^(n) + ... + a_1*x*y' + a_0*y = F(x) 

        #In case of Homogeneous euler equation F(x) = 0 

        def _test_term(coeff, order): 

            r""" 

            Linear Euler ODEs have the form  K*x**order*diff(y(x),x,order) = F(x), 

            where K is independent of x and y(x), order>= 0. 

            So we need to check that for each term, coeff == K*x**order from 

            some K.  We have a few cases, since coeff may have several 

            different types. 

            """ 

            if order < 0: 

                raise ValueError("order should be greater than 0") 

            if coeff == 0: 

                return True 

            if order == 0: 

                if x in coeff.free_symbols: 

                    return False 

                return True 

            if coeff.is_Mul: 

                if coeff.has(f(x)): 

                    return False 

                return x**order in coeff.args 

            elif coeff.is_Pow: 

                return coeff.as_base_exp() == (x, order) 

            elif order == 1: 

                return x == coeff 

            return False 

        if r and not any(not _test_term(r[i], i) for i in r if i >= 0): 

            if not r[-1]: 

                matching_hints["nth_linear_euler_eq_homogeneous"] = r 

            else: 

                matching_hints["nth_linear_euler_eq_nonhomogeneous_variation_of_parameters"] = r 

                matching_hints["nth_linear_euler_eq_nonhomogeneous_variation_of_parameters_Integral"] = r 

                e, re = posify(r[-1].subs(x, exp(x))) 

                undetcoeff = _undetermined_coefficients_match(e.subs(re), x) 

                if undetcoeff['test']: 

                    r['trialset'] = undetcoeff['trialset'] 

                    matching_hints["nth_linear_euler_eq_nonhomogeneous_undetermined_coefficients"] = r 

 

 

    # Order keys based on allhints. 

    retlist = [i for i in allhints if i in matching_hints] 

 

    if dict: 

        # Dictionaries are ordered arbitrarily, so make note of which 

        # hint would come first for dsolve().  Use an ordered dict in Py 3. 

        matching_hints["default"] = retlist[0] if retlist else None 

        matching_hints["ordered_hints"] = tuple(retlist) 

        return matching_hints 

    else: 

        return tuple(retlist) 

 

def classify_sysode(eq, funcs=None, **kwargs): 

    r""" 

    Returns a dictionary of parameter names and values that define the system 

    of ordinary differential equations in ``eq``. 

    The parameters are further used in 

    :py:meth:`~sympy.solvers.ode.dsolve` for solving that system. 

 

    The parameter names and values are: 

 

    'is_linear' (boolean), which tells whether the given system is linear. 

    Note that "linear" here refers to the operator: terms such as ``x*diff(x,t)`` are 

    nonlinear, whereas terms like ``sin(t)*diff(x,t)`` are still linear operators. 

 

    'func' (list) contains the :py:class:`~sympy.core.function.Function`s that 

    appear with a derivative in the ODE, i.e. those that we are trying to solve 

    the ODE for. 

 

    'order' (dict) with the maximum derivative for each element of the 'func' 

    parameter. 

 

    'func_coeff' (dict) with the coefficient for each triple ``(equation number, 

    function, order)```. The coefficients are those subexpressions that do not 

    appear in 'func', and hence can be considered constant for purposes of ODE 

    solving. 

 

    'eq' (list) with the equations from ``eq``, sympified and transformed into 

    expressions (we are solving for these expressions to be zero). 

 

    'no_of_equations' (int) is the number of equations (same as ``len(eq)``). 

 

    'type_of_equation' (string) is an internal classification of the type of 

    ODE. 

 

    References 

    ========== 

    -http://eqworld.ipmnet.ru/en/solutions/sysode/sode-toc1.htm 

    -A. D. Polyanin and A. V. Manzhirov, Handbook of Mathematics for Engineers and Scientists 

 

    Examples 

    ======== 

 

    >>> from sympy import Function, Eq, symbols, diff 

    >>> from sympy.solvers.ode import classify_sysode 

    >>> from sympy.abc import t 

    >>> f, x, y = symbols('f, x, y', function=True) 

    >>> k, l, m, n = symbols('k, l, m, n', Integer=True) 

    >>> x1 = diff(x(t), t) ; y1 = diff(y(t), t) 

    >>> x2 = diff(x(t), t, t) ; y2 = diff(y(t), t, t) 

    >>> eq = (Eq(5*x1, 12*x(t) - 6*y(t)), Eq(2*y1, 11*x(t) + 3*y(t))) 

    >>> classify_sysode(eq) 

    {'eq': [-12*x(t) + 6*y(t) + 5*Derivative(x(t), t), -11*x(t) - 3*y(t) + 2*Derivative(y(t), t)], 

    'func': [x(t), y(t)], 'func_coeff': {(0, x(t), 0): -12, (0, x(t), 1): 5, (0, y(t), 0): 6, 

    (0, y(t), 1): 0, (1, x(t), 0): -11, (1, x(t), 1): 0, (1, y(t), 0): -3, (1, y(t), 1): 2}, 

    'is_linear': True, 'no_of_equation': 2, 'order': {x(t): 1, y(t): 1}, 'type_of_equation': 'type1'} 

    >>> eq = (Eq(diff(x(t),t), 5*t*x(t) + t**2*y(t)), Eq(diff(y(t),t), -t**2*x(t) + 5*t*y(t))) 

    >>> classify_sysode(eq) 

    {'eq': [-t**2*y(t) - 5*t*x(t) + Derivative(x(t), t), t**2*x(t) - 5*t*y(t) + Derivative(y(t), t)], 

    'func': [x(t), y(t)], 'func_coeff': {(0, x(t), 0): -5*t, (0, x(t), 1): 1, (0, y(t), 0): -t**2, 

    (0, y(t), 1): 0, (1, x(t), 0): t**2, (1, x(t), 1): 0, (1, y(t), 0): -5*t, (1, y(t), 1): 1}, 

    'is_linear': True, 'no_of_equation': 2, 'order': {x(t): 1, y(t): 1}, 'type_of_equation': 'type4'} 

 

    """ 

 

    # Sympify equations and convert iterables of equations into 

    # a list of equations 

    def _sympify(eq): 

        return list(map(sympify, eq if iterable(eq) else [eq])) 

 

    eq, funcs = (_sympify(w) for w in [eq, funcs]) 

    for i, fi in enumerate(eq): 

        if isinstance(fi, Equality): 

            eq[i] = fi.lhs - fi.rhs 

    matching_hints = {"no_of_equation":i+1} 

    matching_hints['eq'] = eq 

    if i==0: 

        raise ValueError("classify_sysode() works for systems of ODEs. " 

        "For scalar ODEs, classify_ode should be used") 

    t = list(list(eq[0].atoms(Derivative))[0].atoms(Symbol))[0] 

 

    # find all the functions if not given 

    order = dict() 

    if funcs==[None]: 

        funcs = [] 

        for eqs in eq: 

            derivs = eqs.atoms(Derivative) 

            func = set().union(*[d.atoms(AppliedUndef) for d in derivs]) 

            for func_ in  func: 

                order[func_] = 0 

                funcs.append(func_) 

    funcs = list(set(funcs)) 

    if len(funcs) < len(eq): 

        raise ValueError("Number of functions given is less than number of equations %s" % funcs) 

    func_dict = dict() 

    for func in funcs: 

        if not order[func]: 

            max_order = 0 

            for i, eqs_ in enumerate(eq): 

                order_ = ode_order(eqs_,func) 

                if max_order < order_: 

                    max_order = order_ 

                    eq_no = i 

        if eq_no in func_dict: 

            list_func = [] 

            list_func.append(func_dict[eq_no]) 

            list_func.append(func) 

            func_dict[eq_no] = list_func 

        else: 

            func_dict[eq_no] = func 

        order[func] = max_order 

    funcs = [func_dict[i] for i in range(len(func_dict))] 

    matching_hints['func'] = funcs 

    for func in funcs: 

        if isinstance(func, list): 

            for func_elem in func: 

                if len(func_elem.args) != 1: 

                    raise ValueError("dsolve() and classify_sysode() work with " 

                    "functions of one variable only, not %s" % func) 

        else: 

            if func and len(func.args) != 1: 

                raise ValueError("dsolve() and classify_sysode() work with " 

                "functions of one variable only, not %s" % func) 

 

    # find the order of all equation in system of odes 

    matching_hints["order"] = order 

 

    # find coefficients of terms f(t), diff(f(t),t) and higher derivatives 

    # and similarly for other functions g(t), diff(g(t),t) in all equations. 

    # Here j denotes the equation number, funcs[l] denotes the function about 

    # which we are talking about and k denotes the order of function funcs[l] 

    # whose coefficient we are calculating. 

    def linearity_check(eqs, j, func, is_linear_): 

        for k in range(order[func]+1): 

            func_coef[j,func,k] = collect(eqs.expand(),[diff(func,t,k)]).coeff(diff(func,t,k)) 

            if is_linear_ == True: 

                if func_coef[j,func,k]==0: 

                    if k==0: 

                        coef = eqs.as_independent(func)[1] 

                        for xr in range(1, ode_order(eqs,func)+1): 

                            coef -= eqs.as_independent(diff(func,t,xr))[1] 

                        if coef != 0: 

                            is_linear_ = False 

                    else: 

                        if eqs.as_independent(diff(func,t,k))[1]: 

                            is_linear_ = False 

                else: 

                    for func_ in funcs: 

                        if isinstance(func_, list): 

                            for elem_func_ in func_: 

                                dep = func_coef[j,func,k].as_independent(elem_func_)[1] 

                                if dep!=1 and dep!=0: 

                                    is_linear_ = False 

                        else: 

                            dep = func_coef[j,func,k].as_independent(func_)[1] 

                            if dep!=1 and dep!=0: 

                                is_linear_ = False 

        return is_linear_ 

 

    func_coef = {} 

    is_linear = True 

    for j, eqs in enumerate(eq): 

        for func in funcs: 

            if isinstance(func, list): 

                for func_elem in func: 

                    is_linear = linearity_check(eqs, j, func_elem, is_linear) 

            else: 

                is_linear = linearity_check(eqs, j, func, is_linear) 

    matching_hints['func_coeff'] = func_coef 

    matching_hints['is_linear'] = is_linear 

 

    if len(set(order.values()))==1: 

        order_eq = list(matching_hints['order'].values())[0] 

        if matching_hints['is_linear'] == True: 

            if matching_hints['no_of_equation'] == 2: 

                if order_eq == 1: 

                    type_of_equation = check_linear_2eq_order1(eq, funcs, func_coef) 

                elif order_eq == 2: 

                    type_of_equation = check_linear_2eq_order2(eq, funcs, func_coef) 

                else: 

                    type_of_equation = None 

 

            elif matching_hints['no_of_equation'] == 3: 

                if order_eq == 1: 

                    type_of_equation = check_linear_3eq_order1(eq, funcs, func_coef) 

                    if type_of_equation==None: 

                        type_of_equation = check_linear_neq_order1(eq, funcs, func_coef) 

                else: 

                    type_of_equation = None 

            else: 

                if order_eq == 1: 

                    type_of_equation = check_linear_neq_order1(eq, funcs, func_coef) 

                else: 

                    type_of_equation = None 

        else: 

            if matching_hints['no_of_equation'] == 2: 

                if order_eq == 1: 

                    type_of_equation = check_nonlinear_2eq_order1(eq, funcs, func_coef) 

                else: 

                    type_of_equation = None 

            elif matching_hints['no_of_equation'] == 3: 

                if order_eq == 1: 

                    type_of_equation = check_nonlinear_3eq_order1(eq, funcs, func_coef) 

                else: 

                    type_of_equation = None 

            else: 

                type_of_equation = None 

    else: 

        type_of_equation = None 

 

    matching_hints['type_of_equation'] = type_of_equation 

 

    return matching_hints 

 

 

def check_linear_2eq_order1(eq, func, func_coef): 

    x = func[0].func 

    y = func[1].func 

    fc = func_coef 

    t = list(list(eq[0].atoms(Derivative))[0].atoms(Symbol))[0] 

    r = dict() 

    # for equations Eq(a1*diff(x(t),t), b1*x(t) + c1*y(t) + d1) 

    # and Eq(a2*diff(y(t),t), b2*x(t) + c2*y(t) + d2) 

    r['a1'] = fc[0,x(t),1] ; r['a2'] = fc[1,y(t),1] 

    r['b1'] = -fc[0,x(t),0]/fc[0,x(t),1] ; r['b2'] = -fc[1,x(t),0]/fc[1,y(t),1] 

    r['c1'] = -fc[0,y(t),0]/fc[0,x(t),1] ; r['c2'] = -fc[1,y(t),0]/fc[1,y(t),1] 

    forcing = [S(0),S(0)] 

    for i in range(2): 

        for j in Add.make_args(eq[i]): 

            if not j.has(x(t), y(t)): 

                forcing[i] += j 

    if not (forcing[0].has(t) or forcing[1].has(t)): 

        # We can handle homogeneous case and simple constant forcings 

        r['d1'] = forcing[0] 

        r['d2'] = forcing[1] 

    else: 

        # Issue #9244: nonhomogeneous linear systems are not supported 

        return None 

 

    # Conditions to check for type 6 whose equations are Eq(diff(x(t),t), f(t)*x(t) + g(t)*y(t)) and 

    # Eq(diff(y(t),t), a*[f(t) + a*h(t)]x(t) + a*[g(t) - h(t)]*y(t)) 

    p = 0 

    q = 0 

    p1 = cancel(r['b2']/(cancel(r['b2']/r['c2']).as_numer_denom()[0])) 

    p2 = cancel(r['b1']/(cancel(r['b1']/r['c1']).as_numer_denom()[0])) 

    for n, i in enumerate([p1, p2]): 

        for j in Mul.make_args(collect_const(i)): 

            if not j.has(t): 

                q = j 

            if q and n==0: 

                if ((r['b2']/j - r['b1'])/(r['c1'] - r['c2']/j)) == j: 

                    p = 1 

            elif q and n==1: 

                if ((r['b1']/j - r['b2'])/(r['c2'] - r['c1']/j)) == j: 

                    p = 2 

    # End of condition for type 6 

 

    if r['d1']!=0 or r['d2']!=0: 

        if not r['d1'].has(t) and not r['d2'].has(t): 

            if all(not r[k].has(t) for k in 'a1 a2 b1 b2 c1 c2'.split()): 

                # Equations for type 2 are Eq(a1*diff(x(t),t),b1*x(t)+c1*y(t)+d1) and Eq(a2*diff(y(t),t),b2*x(t)+c2*y(t)+d2) 

                return "type2" 

        else: 

            return None 

    else: 

        if all(not r[k].has(t) for k in 'a1 a2 b1 b2 c1 c2'.split()): 

             # Equations for type 1 are Eq(a1*diff(x(t),t),b1*x(t)+c1*y(t)) and Eq(a2*diff(y(t),t),b2*x(t)+c2*y(t)) 

            return "type1" 

        else: 

            r['b1'] = r['b1']/r['a1'] ; r['b2'] = r['b2']/r['a2'] 

            r['c1'] = r['c1']/r['a1'] ; r['c2'] = r['c2']/r['a2'] 

            if (r['b1'] == r['c2']) and (r['c1'] == r['b2']): 

                # Equation for type 3 are Eq(diff(x(t),t), f(t)*x(t) + g(t)*y(t)) and Eq(diff(y(t),t), g(t)*x(t) + f(t)*y(t)) 

                return "type3" 

            elif (r['b1'] == r['c2']) and (r['c1'] == -r['b2']) or (r['b1'] == -r['c2']) and (r['c1'] == r['b2']): 

                # Equation for type 4 are Eq(diff(x(t),t), f(t)*x(t) + g(t)*y(t)) and Eq(diff(y(t),t), -g(t)*x(t) + f(t)*y(t)) 

                return "type4" 

            elif (not cancel(r['b2']/r['c1']).has(t) and not cancel((r['c2']-r['b1'])/r['c1']).has(t)) \ 

            or (not cancel(r['b1']/r['c2']).has(t) and not cancel((r['c1']-r['b2'])/r['c2']).has(t)): 

                # Equations for type 5 are Eq(diff(x(t),t), f(t)*x(t) + g(t)*y(t)) and Eq(diff(y(t),t), a*g(t)*x(t) + [f(t) + b*g(t)]*y(t) 

                return "type5" 

            elif p: 

                return "type6" 

            else: 

                # Equations for type 7 are Eq(diff(x(t),t), f(t)*x(t) + g(t)*y(t)) and Eq(diff(y(t),t), h(t)*x(t) + p(t)*y(t)) 

                return "type7" 

 

def check_linear_2eq_order2(eq, func, func_coef): 

    x = func[0].func 

    y = func[1].func 

    fc = func_coef 

    t = list(list(eq[0].atoms(Derivative))[0].atoms(Symbol))[0] 

    r = dict() 

    a = Wild('a', exclude=[1/t]) 

    b = Wild('b', exclude=[1/t**2]) 

    u = Wild('u', exclude=[t, t**2]) 

    v = Wild('v', exclude=[t, t**2]) 

    w = Wild('w', exclude=[t, t**2]) 

    p = Wild('p', exclude=[t, t**2]) 

    r['a1'] = fc[0,x(t),2] ; r['a2'] = fc[1,y(t),2] 

    r['b1'] = fc[0,x(t),1] ; r['b2'] = fc[1,x(t),1] 

    r['c1'] = fc[0,y(t),1] ; r['c2'] = fc[1,y(t),1] 

    r['d1'] = fc[0,x(t),0] ; r['d2'] = fc[1,x(t),0] 

    r['e1'] = fc[0,y(t),0] ; r['e2'] = fc[1,y(t),0] 

    const = [S(0), S(0)] 

    for i in range(2): 

        for j in Add.make_args(eq[i]): 

            if not (j.has(x(t)) or j.has(y(t))): 

                const[i] += j 

    r['f1'] = const[0] 

    r['f2'] = const[1] 

    if r['f1']!=0 or r['f2']!=0: 

        if all(not r[k].has(t) for k in 'a1 a2 d1 d2 e1 e2 f1 f2'.split()) \ 

        and r['b1']==r['c1']==r['b2']==r['c2']==0: 

            return "type2" 

 

        elif all(not r[k].has(t) for k in 'a1 a2 b1 b2 c1 c2 d1 d2 e1 e1'.split()): 

            p = [S(0), S(0)] ; q = [S(0), S(0)] 

            for n, e in enumerate([r['f1'], r['f2']]): 

                if e.has(t): 

                    tpart = e.as_independent(t, Mul)[1] 

                    for i in Mul.make_args(tpart): 

                        if i.has(exp): 

                            b, e = i.as_base_exp() 

                            co = e.coeff(t) 

                            if co and not co.has(t) and co.has(I): 

                                p[n] = 1 

                            else: 

                                q[n] = 1 

                        else: 

                            q[n] = 1 

                else: 

                    q[n] = 1 

 

            if p[0]==1 and p[1]==1 and q[0]==0 and q[1]==0: 

                    return "type4" 

            else: 

                return None 

        else: 

            return None 

    else: 

        if r['b1']==r['b2']==r['c1']==r['c2']==0 and all(not r[k].has(t) \ 

        for k in 'a1 a2 d1 d2 e1 e2'.split()): 

            return "type1" 

 

        elif r['b1']==r['e1']==r['c2']==r['d2']==0 and all(not r[k].has(t) \ 

        for k in 'a1 a2 b2 c1 d1 e2'.split()) and r['c1'] == -r['b2'] and \ 

        r['d1'] == r['e2']: 

            return "type3" 

 

        elif cancel(-r['b2']/r['d2'])==t and cancel(-r['c1']/r['e1'])==t and not \ 

        (r['d2']/r['a2']).has(t) and not (r['e1']/r['a1']).has(t) and \ 

        r['b1']==r['d1']==r['c2']==r['e2']==0: 

            return "type5" 

 

        elif ((r['a1']/r['d1']).expand()).match((p*(u*t**2+v*t+w)**2).expand()) and not \ 

        (cancel(r['a1']*r['d2']/(r['a2']*r['d1']))).has(t) and not (r['d1']/r['e1']).has(t) and not \ 

        (r['d2']/r['e2']).has(t) and r['b1'] == r['b2'] == r['c1'] == r['c2'] == 0: 

            return "type10" 

 

        elif not cancel(r['d1']/r['e1']).has(t) and not cancel(r['d2']/r['e2']).has(t) and not \ 

        cancel(r['d1']*r['a2']/(r['d2']*r['a1'])).has(t) and r['b1']==r['b2']==r['c1']==r['c2']==0: 

            return "type6" 

 

        elif not cancel(r['b1']/r['c1']).has(t) and not cancel(r['b2']/r['c2']).has(t) and not \ 

        cancel(r['b1']*r['a2']/(r['b2']*r['a1'])).has(t) and r['d1']==r['d2']==r['e1']==r['e2']==0: 

            return "type7" 

 

        elif cancel(-r['b2']/r['d2'])==t and cancel(-r['c1']/r['e1'])==t and not \ 

        cancel(r['e1']*r['a2']/(r['d2']*r['a1'])).has(t) and r['e1'].has(t) \ 

        and r['b1']==r['d1']==r['c2']==r['e2']==0: 

            return "type8" 

 

        elif (r['b1']/r['a1']).match(a/t) and (r['b2']/r['a2']).match(a/t) and not \ 

        (r['b1']/r['c1']).has(t) and not (r['b2']/r['c2']).has(t) and \ 

        (r['d1']/r['a1']).match(b/t**2) and (r['d2']/r['a2']).match(b/t**2) \ 

        and not (r['d1']/r['e1']).has(t) and not (r['d2']/r['e2']).has(t): 

            return "type9" 

 

        elif -r['b1']/r['d1']==-r['c1']/r['e1']==-r['b2']/r['d2']==-r['c2']/r['e2']==t: 

            return "type11" 

 

        else: 

            return None 

 

def check_linear_3eq_order1(eq, func, func_coef): 

    x = func[0].func 

    y = func[1].func 

    z = func[2].func 

    fc = func_coef 

    t = list(list(eq[0].atoms(Derivative))[0].atoms(Symbol))[0] 

    r = dict() 

    r['a1'] = fc[0,x(t),1]; r['a2'] = fc[1,y(t),1]; r['a3'] = fc[2,z(t),1] 

    r['b1'] = fc[0,x(t),0]; r['b2'] = fc[1,x(t),0]; r['b3'] = fc[2,x(t),0] 

    r['c1'] = fc[0,y(t),0]; r['c2'] = fc[1,y(t),0]; r['c3'] = fc[2,y(t),0] 

    r['d1'] = fc[0,z(t),0]; r['d2'] = fc[1,z(t),0]; r['d3'] = fc[2,z(t),0] 

    forcing = [S(0), S(0), S(0)] 

    for i in range(3): 

        for j in Add.make_args(eq[i]): 

            if not j.has(x(t), y(t), z(t)): 

                forcing[i] += j 

    if forcing[0].has(t) or forcing[1].has(t) or forcing[2].has(t): 

        # We can handle homogeneous case and simple constant forcings. 

        # Issue #9244: nonhomogeneous linear systems are not supported 

        return None 

 

    if all(not r[k].has(t) for k in 'a1 a2 a3 b1 b2 b3 c1 c2 c3 d1 d2 d3'.split()): 

        if r['c1']==r['d1']==r['d2']==0: 

            return 'type1' 

        elif r['c1'] == -r['b2'] and r['d1'] == -r['b3'] and r['d2'] == -r['c3'] \ 

        and r['b1'] == r['c2'] == r['d3'] == 0: 

            return 'type2' 

        elif r['b1'] == r['c2'] == r['d3'] == 0 and r['c1']/r['a1'] == -r['d1']/r['a1'] \ 

        and r['d2']/r['a2'] == -r['b2']/r['a2'] and r['b3']/r['a3'] == -r['c3']/r['a3']: 

            return 'type3' 

        else: 

            return None 

    else: 

        for k1 in 'c1 d1 b2 d2 b3 c3'.split(): 

            if r[k1] == 0: 

                continue 

            else: 

                if all(not cancel(r[k1]/r[k]).has(t) for k in 'd1 b2 d2 b3 c3'.split() if r[k]!=0) \ 

                and all(not cancel(r[k1]/(r['b1'] - r[k])).has(t) for k in 'b1 c2 d3'.split() if r['b1']!=r[k]): 

                    return 'type4' 

                else: 

                    break 

    return None 

 

def check_linear_neq_order1(eq, func, func_coef): 

    x = func[0].func 

    y = func[1].func 

    z = func[2].func 

    fc = func_coef 

    t = list(list(eq[0].atoms(Derivative))[0].atoms(Symbol))[0] 

    r = dict() 

    n = len(eq) 

    for i in range(n): 

        for j in range(n): 

            if (fc[i,func[j],0]/fc[i,func[i],1]).has(t): 

                return None 

    if len(eq)==3: 

        return 'type6' 

    return 'type1' 

 

def check_nonlinear_2eq_order1(eq, func, func_coef): 

    t = list(list(eq[0].atoms(Derivative))[0].atoms(Symbol))[0] 

    f = Wild('f') 

    g = Wild('g') 

    u, v = symbols('u, v', cls=Dummy) 

    def check_type(x, y): 

        r1 = eq[0].match(t*diff(x(t),t) - x(t) + f) 

        r2 = eq[1].match(t*diff(y(t),t) - y(t) + g) 

        if not (r1 and r2): 

            r1 = eq[0].match(diff(x(t),t) - x(t)/t + f/t) 

            r2 = eq[1].match(diff(y(t),t) - y(t)/t + g/t) 

        if not (r1 and r2): 

            r1 = (-eq[0]).match(t*diff(x(t),t) - x(t) + f) 

            r2 = (-eq[1]).match(t*diff(y(t),t) - y(t) + g) 

        if not (r1 and r2): 

            r1 = (-eq[0]).match(diff(x(t),t) - x(t)/t + f/t) 

            r2 = (-eq[1]).match(diff(y(t),t) - y(t)/t + g/t) 

        if r1 and r2 and not (r1[f].subs(diff(x(t),t),u).subs(diff(y(t),t),v).has(t) \ 

        or r2[g].subs(diff(x(t),t),u).subs(diff(y(t),t),v).has(t)): 

            return 'type5' 

        else: 

            return None 

    for func_ in func: 

        if isinstance(func_, list): 

            x = func[0][0].func 

            y = func[0][1].func 

            eq_type = check_type(x, y) 

            if not eq_type: 

                eq_type = check_type(y, x) 

            return eq_type 

    x = func[0].func 

    y = func[1].func 

    fc = func_coef 

    n = Wild('n', exclude=[x(t),y(t)]) 

    f1 = Wild('f1', exclude=[v,t]) 

    f2 = Wild('f2', exclude=[v,t]) 

    g1 = Wild('g1', exclude=[u,t]) 

    g2 = Wild('g2', exclude=[u,t]) 

    for i in range(2): 

        eqs = 0 

        for terms in Add.make_args(eq[i]): 

            eqs += terms/fc[i,func[i],1] 

        eq[i] = eqs 

    r = eq[0].match(diff(x(t),t) - x(t)**n*f) 

    if r: 

        g = (diff(y(t),t) - eq[1])/r[f] 

    if r and not (g.has(x(t)) or g.subs(y(t),v).has(t) or r[f].subs(x(t),u).subs(y(t),v).has(t)): 

        return 'type1' 

    r = eq[0].match(diff(x(t),t) - exp(n*x(t))*f) 

    if r: 

        g = (diff(y(t),t) - eq[1])/r[f] 

    if r and not (g.has(x(t)) or g.subs(y(t),v).has(t) or r[f].subs(x(t),u).subs(y(t),v).has(t)): 

        return 'type2' 

    g = Wild('g') 

    r1 = eq[0].match(diff(x(t),t) - f) 

    r2 = eq[1].match(diff(y(t),t) - g) 

    if r1 and r2 and not (r1[f].subs(x(t),u).subs(y(t),v).has(t) or \ 

    r2[g].subs(x(t),u).subs(y(t),v).has(t)): 

        return 'type3' 

    r1 = eq[0].match(diff(x(t),t) - f) 

    r2 = eq[1].match(diff(y(t),t) - g) 

    num, denum = ((r1[f].subs(x(t),u).subs(y(t),v))/(r2[g].subs(x(t),u).subs(y(t),v))).as_numer_denom() 

    R1 = num.match(f1*g1) 

    R2 = denum.match(f2*g2) 

    phi = (r1[f].subs(x(t),u).subs(y(t),v))/num 

    if R1 and R2: 

        return 'type4' 

    return None 

 

 

def check_nonlinear_2eq_order2(eq, func, func_coef): 

    return None 

 

def check_nonlinear_3eq_order1(eq, func, func_coef): 

    x = func[0].func 

    y = func[1].func 

    z = func[2].func 

    fc = func_coef 

    t = list(list(eq[0].atoms(Derivative))[0].atoms(Symbol))[0] 

    u, v, w = symbols('u, v, w', cls=Dummy) 

    a = Wild('a', exclude=[x(t), y(t), z(t), t]) 

    b = Wild('b', exclude=[x(t), y(t), z(t), t]) 

    c = Wild('c', exclude=[x(t), y(t), z(t), t]) 

    f = Wild('f') 

    F1 = Wild('F1') 

    F2 = Wild('F2') 

    F3 = Wild('F3') 

    for i in range(3): 

        eqs = 0 

        for terms in Add.make_args(eq[i]): 

            eqs += terms/fc[i,func[i],1] 

        eq[i] = eqs 

    r1 = eq[0].match(diff(x(t),t) - a*y(t)*z(t)) 

    r2 = eq[1].match(diff(y(t),t) - b*z(t)*x(t)) 

    r3 = eq[2].match(diff(z(t),t) - c*x(t)*y(t)) 

    if r1 and r2 and r3: 

        num1, den1 = r1[a].as_numer_denom() 

        num2, den2 = r2[b].as_numer_denom() 

        num3, den3 = r3[c].as_numer_denom() 

        if solve([num1*u-den1*(v-w), num2*v-den2*(w-u), num3*w-den3*(u-v)],[u, v]): 

            return 'type1' 

    r = eq[0].match(diff(x(t),t) - y(t)*z(t)*f) 

    if r: 

        r1 = collect_const(r[f]).match(a*f) 

        r2 = ((diff(y(t),t) - eq[1])/r1[f]).match(b*z(t)*x(t)) 

        r3 = ((diff(z(t),t) - eq[2])/r1[f]).match(c*x(t)*y(t)) 

    if r1 and r2 and r3: 

        num1, den1 = r1[a].as_numer_denom() 

        num2, den2 = r2[b].as_numer_denom() 

        num3, den3 = r3[c].as_numer_denom() 

        if solve([num1*u-den1*(v-w), num2*v-den2*(w-u), num3*w-den3*(u-v)],[u, v]): 

            return 'type2' 

    r = eq[0].match(diff(x(t),t) - (F2-F3)) 

    if r: 

        r1 = collect_const(r[F2]).match(c*F2) 

        r1.update(collect_const(r[F3]).match(b*F3)) 

        if r1: 

            if eq[1].has(r1[F2]) and not eq[1].has(r1[F3]): 

                r1[F2], r1[F3] = r1[F3], r1[F2] 

                r1[c], r1[b] = -r1[b], -r1[c] 

            r2 = eq[1].match(diff(y(t),t) - a*r1[F3] + r1[c]*F1) 

        if r2: 

            r3 = (eq[2] == diff(z(t),t) - r1[b]*r2[F1] + r2[a]*r1[F2]) 

        if r1 and r2 and r3: 

            return 'type3' 

    r = eq[0].match(diff(x(t),t) - z(t)*F2 + y(t)*F3) 

    if r: 

        r1 = collect_const(r[F2]).match(c*F2) 

        r1.update(collect_const(r[F3]).match(b*F3)) 

        if r1: 

            if eq[1].has(r1[F2]) and not eq[1].has(r1[F3]): 

                r1[F2], r1[F3] = r1[F3], r1[F2] 

                r1[c], r1[b] = -r1[b], -r1[c] 

            r2 = (diff(y(t),t) - eq[1]).match(a*x(t)*r1[F3] - r1[c]*z(t)*F1) 

        if r2: 

            r3 = (diff(z(t),t) - eq[2] == r1[b]*y(t)*r2[F1] - r2[a]*x(t)*r1[F2]) 

        if r1 and r2 and r3: 

            return 'type4' 

    r = (diff(x(t),t) - eq[0]).match(x(t)*(F2 - F3)) 

    if r: 

        r1 = collect_const(r[F2]).match(c*F2) 

        r1.update(collect_const(r[F3]).match(b*F3)) 

        if r1: 

            if eq[1].has(r1[F2]) and not eq[1].has(r1[F3]): 

                r1[F2], r1[F3] = r1[F3], r1[F2] 

                r1[c], r1[b] = -r1[b], -r1[c] 

            r2 = (diff(y(t),t) - eq[1]).match(y(t)*(a*r1[F3] - r1[c]*F1)) 

        if r2: 

            r3 = (diff(z(t),t) - eq[2] == z(t)*(r1[b]*r2[F1] - r2[a]*r1[F2])) 

        if r1 and r2 and r3: 

            return 'type5' 

    return None 

 

def check_nonlinear_3eq_order2(eq, func, func_coef): 

    return None 

 

 

def checksysodesol(eqs, sols, func=None): 

    r""" 

    Substitutes corresponding ``sols`` for each functions into each ``eqs`` and 

    checks that the result of substitutions for each equation is ``0``. The 

    equations and solutions passed can be any iterable. 

 

    This only works when each ``sols`` have one function only, like `x(t)` or `y(t)`. 

    For each function, ``sols`` can have a single solution or a list of solutions. 

    In most cases it will not be necessary to explicitly identify the function, 

    but if the function cannot be inferred from the original equation it 

    can be supplied through the ``func`` argument. 

 

    When a sequence of equations is passed, the same sequence is used to return 

    the result for each equation with each function substitued with corresponding 

    solutions. 

 

    It tries the following method to find zero equivalence for each equation: 

 

    Substitute the solutions for functions, like `x(t)` and `y(t)` into the 

    original equations containing those functions. 

    This function returns a tuple.  The first item in the tuple is ``True`` if 

    the substitution results for each equation is ``0``, and ``False`` otherwise. 

    The second item in the tuple is what the substitution results in.  Each element 

    of the ``list`` should always be ``0`` corresponding to each equation if the 

    first item is ``True``. Note that sometimes this function may return ``False``, 

    but with an expression that is identically equal to ``0``, instead of returning 

    ``True``.  This is because :py:meth:`~sympy.simplify.simplify.simplify` cannot 

    reduce the expression to ``0``.  If an expression returned by each function 

    vanishes identically, then ``sols`` really is a solution to ``eqs``. 

 

    If this function seems to hang, it is probably because of a difficult simplification. 

 

    Examples 

    ======== 

 

    >>> from sympy import Eq, diff, symbols, sin, cos, exp, sqrt, S 

    >>> from sympy.solvers.ode import checksysodesol 

    >>> C1, C2 = symbols('C1:3') 

    >>> t = symbols('t') 

    >>> x, y = symbols('x, y', function=True) 

    >>> eq = (Eq(diff(x(t),t), x(t) + y(t) + 17), Eq(diff(y(t),t), -2*x(t) + y(t) + 12)) 

    >>> sol = [Eq(x(t), (C1*sin(sqrt(2)*t) + C2*cos(sqrt(2)*t))*exp(t) - S(5)/3), 

    ... Eq(y(t), (sqrt(2)*C1*cos(sqrt(2)*t) - sqrt(2)*C2*sin(sqrt(2)*t))*exp(t) - S(46)/3)] 

    >>> checksysodesol(eq, sol) 

    (True, [0, 0]) 

    >>> eq = (Eq(diff(x(t),t),x(t)*y(t)**4), Eq(diff(y(t),t),y(t)**3)) 

    >>> sol = [Eq(x(t), C1*exp(-1/(4*(C2 + t)))), Eq(y(t), -sqrt(2)*sqrt(-1/(C2 + t))/2), 

    ... Eq(x(t), C1*exp(-1/(4*(C2 + t)))), Eq(y(t), sqrt(2)*sqrt(-1/(C2 + t))/2)] 

    >>> checksysodesol(eq, sol) 

    (True, [0, 0]) 

 

    """ 

    def _sympify(eq): 

        return list(map(sympify, eq if iterable(eq) else [eq])) 

    eqs = _sympify(eqs) 

    for i in range(len(eqs)): 

        if isinstance(eqs[i], Equality): 

            eqs[i] = eqs[i].lhs - eqs[i].rhs 

    if func is None: 

        funcs = [] 

        for eq in eqs: 

            derivs = eq.atoms(Derivative) 

            func = set().union(*[d.atoms(AppliedUndef) for d in derivs]) 

            for func_ in  func: 

                funcs.append(func_) 

        funcs = list(set(funcs)) 

    if not all(isinstance(func, AppliedUndef) and len(func.args) == 1 for func in funcs)\ 

    and len(set([func.args for func in funcs]))!=1: 

        raise ValueError("func must be a function of one variable, not %s" % func) 

    for sol in sols: 

        if len(sol.atoms(AppliedUndef)) != 1: 

            raise ValueError("solutions should have one function only") 

    if len(funcs) != len(set([sol.lhs for sol in sols])): 

        raise ValueError("number of solutions provided does not match the number of equations") 

    t = funcs[0].args[0] 

    dictsol = dict() 

    for sol in sols: 

        sol_func = list(sol.atoms(AppliedUndef))[0] 

        if not (sol.lhs == sol_func and not sol.rhs.has(sol_func)) and not (\ 

        sol.rhs == sol_func and not sol.lhs.has(sol_func)): 

            solved = solve(sol, sol_func) 

            if not solved: 

                raise NotImplementedError 

            dictsol[sol_func] = solved 

        if sol.lhs == sol_func: 

            dictsol[sol_func] = sol.rhs 

        if sol.rhs == sol_func: 

            dictsol[sol_func] = sol.lhs 

    checkeq = [] 

    for eq in eqs: 

        for func in funcs: 

            eq = sub_func_doit(eq, func, dictsol[func]) 

        ss = simplify(eq) 

        if ss != 0: 

            eq = ss.expand(force=True) 

        else: 

            eq = 0 

        checkeq.append(eq) 

    if len(set(checkeq)) == 1 and list(set(checkeq))[0] == 0: 

        return (True, checkeq) 

    else: 

        return (False, checkeq) 

 

 

@vectorize(0) 

def odesimp(eq, func, order, constants, hint): 

    r""" 

    Simplifies ODEs, including trying to solve for ``func`` and running 

    :py:meth:`~sympy.solvers.ode.constantsimp`. 

 

    It may use knowledge of the type of solution that the hint returns to 

    apply additional simplifications. 

 

    It also attempts to integrate any :py:class:`~sympy.integrals.Integral`\s 

    in the expression, if the hint is not an ``_Integral`` hint. 

 

    This function should have no effect on expressions returned by 

    :py:meth:`~sympy.solvers.ode.dsolve`, as 

    :py:meth:`~sympy.solvers.ode.dsolve` already calls 

    :py:meth:`~sympy.solvers.ode.odesimp`, but the individual hint functions 

    do not call :py:meth:`~sympy.solvers.ode.odesimp` (because the 

    :py:meth:`~sympy.solvers.ode.dsolve` wrapper does).  Therefore, this 

    function is designed for mainly internal use. 

 

    Examples 

    ======== 

 

    >>> from sympy import sin, symbols, dsolve, pprint, Function 

    >>> from sympy.solvers.ode import odesimp 

    >>> x , u2, C1= symbols('x,u2,C1') 

    >>> f = Function('f') 

 

    >>> eq = dsolve(x*f(x).diff(x) - f(x) - x*sin(f(x)/x), f(x), 

    ... hint='1st_homogeneous_coeff_subs_indep_div_dep_Integral', 

    ... simplify=False) 

    >>> pprint(eq, wrap_line=False) 

                            x 

                           ---- 

                           f(x) 

                             / 

                            | 

                            |   /        1   \ 

                            |  -|u2 + -------| 

                            |   |        /1 \| 

                            |   |     sin|--|| 

                            |   \        \u2// 

    log(f(x)) = log(C1) +   |  ---------------- d(u2) 

                            |          2 

                            |        u2 

                            | 

                           / 

 

    >>> pprint(odesimp(eq, f(x), 1, set([C1]), 

    ... hint='1st_homogeneous_coeff_subs_indep_div_dep' 

    ... )) #doctest: +SKIP 

        x 

    --------- = C1 

       /f(x)\ 

    tan|----| 

       \2*x / 

 

    """ 

    x = func.args[0] 

    f = func.func 

    C1 = get_numbered_constants(eq, num=1) 

 

    # First, integrate if the hint allows it. 

    eq = _handle_Integral(eq, func, order, hint) 

    if hint.startswith("nth_linear_euler_eq_nonhomogeneous"): 

        eq = simplify(eq) 

    if not isinstance(eq, Equality): 

        raise TypeError("eq should be an instance of Equality") 

 

    # Second, clean up the arbitrary constants. 

    # Right now, nth linear hints can put as many as 2*order constants in an 

    # expression.  If that number grows with another hint, the third argument 

    # here should be raised accordingly, or constantsimp() rewritten to handle 

    # an arbitrary number of constants. 

    eq = constantsimp(eq, constants) 

 

    # Lastly, now that we have cleaned up the expression, try solving for func. 

    # When RootOf is implemented in solve(), we will want to return a RootOf 

    # everytime instead of an Equality. 

 

    # Get the f(x) on the left if possible. 

    if eq.rhs == func and not eq.lhs.has(func): 

        eq = [Eq(eq.rhs, eq.lhs)] 

 

    # make sure we are working with lists of solutions in simplified form. 

    if eq.lhs == func and not eq.rhs.has(func): 

        # The solution is already solved 

        eq = [eq] 

 

        # special simplification of the rhs 

        if hint.startswith("nth_linear_constant_coeff"): 

            # Collect terms to make the solution look nice. 

            # This is also necessary for constantsimp to remove unnecessary 

            # terms from the particular solution from variation of parameters 

            # 

            # Collect is not behaving reliably here.  The results for 

            # some linear constant-coefficient equations with repeated 

            # roots do not properly simplify all constants sometimes. 

            # 'collectterms' gives different orders sometimes, and results 

            # differ in collect based on that order.  The 

            # sort-reverse trick fixes things, but may fail in the 

            # future. In addition, collect is splitting exponentials with 

            # rational powers for no reason.  We have to do a match 

            # to fix this using Wilds. 

            global collectterms 

            try: 

                collectterms.sort(key=default_sort_key) 

                collectterms.reverse() 

            except Exception: 

                pass 

            assert len(eq) == 1 and eq[0].lhs == f(x) 

            sol = eq[0].rhs 

            sol = expand_mul(sol) 

            for i, reroot, imroot in collectterms: 

                sol = collect(sol, x**i*exp(reroot*x)*sin(abs(imroot)*x)) 

                sol = collect(sol, x**i*exp(reroot*x)*cos(imroot*x)) 

            for i, reroot, imroot in collectterms: 

                sol = collect(sol, x**i*exp(reroot*x)) 

            del collectterms 

 

            # Collect is splitting exponentials with rational powers for 

            # no reason.  We call powsimp to fix. 

            sol = powsimp(sol) 

 

            eq[0] = Eq(f(x), sol) 

 

    else: 

        # The solution is not solved, so try to solve it 

        try: 

            eqsol = solve(eq, func, force=True) 

            if not eqsol: 

                raise NotImplementedError 

        except (NotImplementedError, PolynomialError): 

            eq = [eq] 

        else: 

            def _expand(expr): 

                numer, denom = expr.as_numer_denom() 

 

                if denom.is_Add: 

                    return expr 

                else: 

                    return powsimp(expr.expand(), combine='exp', deep=True) 

 

            # XXX: the rest of odesimp() expects each ``t`` to be in a 

            # specific normal form: rational expression with numerator 

            # expanded, but with combined exponential functions (at 

            # least in this setup all tests pass). 

            eq = [Eq(f(x), _expand(t)) for t in eqsol] 

 

        # special simplification of the lhs. 

        if hint.startswith("1st_homogeneous_coeff"): 

            for j, eqi in enumerate(eq): 

                newi = logcombine(eqi, force=True) 

                if newi.lhs.func is log and newi.rhs == 0: 

                    newi = Eq(newi.lhs.args[0]/C1, C1) 

                eq[j] = newi 

 

    # We cleaned up the constants before solving to help the solve engine with 

    # a simpler expression, but the solved expression could have introduced 

    # things like -C1, so rerun constantsimp() one last time before returning. 

    for i, eqi in enumerate(eq): 

        eq[i] = constantsimp(eqi, constants) 

        eq[i] = constant_renumber(eq[i], 'C', 1, 2*order) 

 

    # If there is only 1 solution, return it; 

    # otherwise return the list of solutions. 

    if len(eq) == 1: 

        eq = eq[0] 

    return eq 

 

def checkodesol(ode, sol, func=None, order='auto', solve_for_func=True): 

    r""" 

    Substitutes ``sol`` into ``ode`` and checks that the result is ``0``. 

 

    This only works when ``func`` is one function, like `f(x)`.  ``sol`` can 

    be a single solution or a list of solutions.  Each solution may be an 

    :py:class:`~sympy.core.relational.Equality` that the solution satisfies, 

    e.g. ``Eq(f(x), C1), Eq(f(x) + C1, 0)``; or simply an 

    :py:class:`~sympy.core.expr.Expr`, e.g. ``f(x) - C1``. In most cases it 

    will not be necessary to explicitly identify the function, but if the 

    function cannot be inferred from the original equation it can be supplied 

    through the ``func`` argument. 

 

    If a sequence of solutions is passed, the same sort of container will be 

    used to return the result for each solution. 

 

    It tries the following methods, in order, until it finds zero equivalence: 

 

    1. Substitute the solution for `f` in the original equation.  This only 

       works if ``ode`` is solved for `f`.  It will attempt to solve it first 

       unless ``solve_for_func == False``. 

    2. Take `n` derivatives of the solution, where `n` is the order of 

       ``ode``, and check to see if that is equal to the solution.  This only 

       works on exact ODEs. 

    3. Take the 1st, 2nd, ..., `n`\th derivatives of the solution, each time 

       solving for the derivative of `f` of that order (this will always be 

       possible because `f` is a linear operator). Then back substitute each 

       derivative into ``ode`` in reverse order. 

 

    This function returns a tuple.  The first item in the tuple is ``True`` if 

    the substitution results in ``0``, and ``False`` otherwise. The second 

    item in the tuple is what the substitution results in.  It should always 

    be ``0`` if the first item is ``True``. Note that sometimes this function 

    will ``False``, but with an expression that is identically equal to ``0``, 

    instead of returning ``True``.  This is because 

    :py:meth:`~sympy.simplify.simplify.simplify` cannot reduce the expression 

    to ``0``.  If an expression returned by this function vanishes 

    identically, then ``sol`` really is a solution to ``ode``. 

 

    If this function seems to hang, it is probably because of a hard 

    simplification. 

 

    To use this function to test, test the first item of the tuple. 

 

    Examples 

    ======== 

 

    >>> from sympy import Eq, Function, checkodesol, symbols 

    >>> x, C1 = symbols('x,C1') 

    >>> f = Function('f') 

    >>> checkodesol(f(x).diff(x), Eq(f(x), C1)) 

    (True, 0) 

    >>> assert checkodesol(f(x).diff(x), C1)[0] 

    >>> assert not checkodesol(f(x).diff(x), x)[0] 

    >>> checkodesol(f(x).diff(x, 2), x**2) 

    (False, 2) 

 

    """ 

    if not isinstance(ode, Equality): 

        ode = Eq(ode, 0) 

    if func is None: 

        try: 

            _, func = _preprocess(ode.lhs) 

        except ValueError: 

            funcs = [s.atoms(AppliedUndef) for s in ( 

                sol if is_sequence(sol, set) else [sol])] 

            funcs = set().union(*funcs) 

            if len(funcs) != 1: 

                raise ValueError( 

                    'must pass func arg to checkodesol for this case.') 

            func = funcs.pop() 

    if not isinstance(func, AppliedUndef) or len(func.args) != 1: 

        raise ValueError( 

            "func must be a function of one variable, not %s" % func) 

    if is_sequence(sol, set): 

        return type(sol)([checkodesol(ode, i, order=order, solve_for_func=solve_for_func) for i in sol]) 

 

    if not isinstance(sol, Equality): 

        sol = Eq(func, sol) 

    x = func.args[0] 

    s = True 

    testnum = 0 

    if order == 'auto': 

        order = ode_order(ode, func) 

    if solve_for_func and not ( 

            sol.lhs == func and not sol.rhs.has(func)) and not ( 

            sol.rhs == func and not sol.lhs.has(func)): 

        try: 

            solved = solve(sol, func) 

            if not solved: 

                raise NotImplementedError 

        except NotImplementedError: 

            pass 

        else: 

            if len(solved) == 1: 

                result = checkodesol(ode, Eq(func, solved[0]), 

                    order=order, solve_for_func=False) 

            else: 

                result = checkodesol(ode, [Eq(func, t) for t in solved], 

                order=order, solve_for_func=False) 

 

            return result 

 

    while s: 

        if testnum == 0: 

            # First pass, try substituting a solved solution directly into the 

            # ODE. This has the highest chance of succeeding. 

            ode_diff = ode.lhs - ode.rhs 

 

            if sol.lhs == func: 

                s = sub_func_doit(ode_diff, func, sol.rhs) 

            elif sol.rhs == func: 

                s = sub_func_doit(ode_diff, func, sol.lhs) 

            else: 

                testnum += 1 

                continue 

            ss = simplify(s) 

            if ss: 

                # with the new numer_denom in power.py, if we do a simple 

                # expansion then testnum == 0 verifies all solutions. 

                s = ss.expand(force=True) 

            else: 

                s = 0 

            testnum += 1 

        elif testnum == 1: 

            # Second pass. If we cannot substitute f, try seeing if the nth 

            # derivative is equal, this will only work for odes that are exact, 

            # by definition. 

            s = simplify( 

                trigsimp(diff(sol.lhs, x, order) - diff(sol.rhs, x, order)) - 

                trigsimp(ode.lhs) + trigsimp(ode.rhs)) 

            # s2 = simplify( 

            #     diff(sol.lhs, x, order) - diff(sol.rhs, x, order) - \ 

            #     ode.lhs + ode.rhs) 

            testnum += 1 

        elif testnum == 2: 

            # Third pass. Try solving for df/dx and substituting that into the 

            # ODE. Thanks to Chris Smith for suggesting this method.  Many of 

            # the comments below are his too. 

            # The method: 

            # - Take each of 1..n derivatives of the solution. 

            # - Solve each nth derivative for d^(n)f/dx^(n) 

            #   (the differential of that order) 

            # - Back substitute into the ODE in decreasing order 

            #   (i.e., n, n-1, ...) 

            # - Check the result for zero equivalence 

            if sol.lhs == func and not sol.rhs.has(func): 

                diffsols = {0: sol.rhs} 

            elif sol.rhs == func and not sol.lhs.has(func): 

                diffsols = {0: sol.lhs} 

            else: 

                diffsols = {} 

            sol = sol.lhs - sol.rhs 

            for i in range(1, order + 1): 

                # Differentiation is a linear operator, so there should always 

                # be 1 solution. Nonetheless, we test just to make sure. 

                # We only need to solve once.  After that, we automatically 

                # have the solution to the differential in the order we want. 

                if i == 1: 

                    ds = sol.diff(x) 

                    try: 

                        sdf = solve(ds, func.diff(x, i)) 

                        if not sdf: 

                            raise NotImplementedError 

                    except NotImplementedError: 

                        testnum += 1 

                        break 

                    else: 

                        diffsols[i] = sdf[0] 

                else: 

                    # This is what the solution says df/dx should be. 

                    diffsols[i] = diffsols[i - 1].diff(x) 

 

            # Make sure the above didn't fail. 

            if testnum > 2: 

                continue 

            else: 

                # Substitute it into ODE to check for self consistency. 

                lhs, rhs = ode.lhs, ode.rhs 

                for i in range(order, -1, -1): 

                    if i == 0 and 0 not in diffsols: 

                        # We can only substitute f(x) if the solution was 

                        # solved for f(x). 

                        break 

                    lhs = sub_func_doit(lhs, func.diff(x, i), diffsols[i]) 

                    rhs = sub_func_doit(rhs, func.diff(x, i), diffsols[i]) 

                    ode_or_bool = Eq(lhs, rhs) 

                    ode_or_bool = simplify(ode_or_bool) 

 

                    if isinstance(ode_or_bool, (bool, BooleanAtom)): 

                        if ode_or_bool: 

                            lhs = rhs = S.Zero 

                    else: 

                        lhs = ode_or_bool.lhs 

                        rhs = ode_or_bool.rhs 

                # No sense in overworking simplify -- just prove that the 

                # numerator goes to zero 

                num = trigsimp((lhs - rhs).as_numer_denom()[0]) 

                # since solutions are obtained using force=True we test 

                # using the same level of assumptions 

                ## replace function with dummy so assumptions will work 

                _func = Dummy('func') 

                num = num.subs(func, _func) 

                ## posify the expression 

                num, reps = posify(num) 

                s = simplify(num).xreplace(reps).xreplace({_func: func}) 

                testnum += 1 

        else: 

            break 

 

    if not s: 

        return (True, s) 

    elif s is True:  # The code above never was able to change s 

        raise NotImplementedError("Unable to test if " + str(sol) + 

            " is a solution to " + str(ode) + ".") 

    else: 

        return (False, s) 

 

 

def ode_sol_simplicity(sol, func, trysolving=True): 

    r""" 

    Returns an extended integer representing how simple a solution to an ODE 

    is. 

 

    The following things are considered, in order from most simple to least: 

 

    - ``sol`` is solved for ``func``. 

    - ``sol`` is not solved for ``func``, but can be if passed to solve (e.g., 

      a solution returned by ``dsolve(ode, func, simplify=False``). 

    - If ``sol`` is not solved for ``func``, then base the result on the 

      length of ``sol``, as computed by ``len(str(sol))``. 

    - If ``sol`` has any unevaluated :py:class:`~sympy.integrals.Integral`\s, 

      this will automatically be considered less simple than any of the above. 

 

    This function returns an integer such that if solution A is simpler than 

    solution B by above metric, then ``ode_sol_simplicity(sola, func) < 

    ode_sol_simplicity(solb, func)``. 

 

    Currently, the following are the numbers returned, but if the heuristic is 

    ever improved, this may change.  Only the ordering is guaranteed. 

 

    +----------------------------------------------+-------------------+ 

    | Simplicity                                   | Return            | 

    +==============================================+===================+ 

    | ``sol`` solved for ``func``                  | ``-2``            | 

    +----------------------------------------------+-------------------+ 

    | ``sol`` not solved for ``func`` but can be   | ``-1``            | 

    +----------------------------------------------+-------------------+ 

    | ``sol`` is not solved nor solvable for       | ``len(str(sol))`` | 

    | ``func``                                     |                   | 

    +----------------------------------------------+-------------------+ 

    | ``sol`` contains an                          | ``oo``            | 

    | :py:class:`~sympy.integrals.Integral`        |                   | 

    +----------------------------------------------+-------------------+ 

 

    ``oo`` here means the SymPy infinity, which should compare greater than 

    any integer. 

 

    If you already know :py:meth:`~sympy.solvers.solvers.solve` cannot solve 

    ``sol``, you can use ``trysolving=False`` to skip that step, which is the 

    only potentially slow step.  For example, 

    :py:meth:`~sympy.solvers.ode.dsolve` with the ``simplify=False`` flag 

    should do this. 

 

    If ``sol`` is a list of solutions, if the worst solution in the list 

    returns ``oo`` it returns that, otherwise it returns ``len(str(sol))``, 

    that is, the length of the string representation of the whole list. 

 

    Examples 

    ======== 

 

    This function is designed to be passed to ``min`` as the key argument, 

    such as ``min(listofsolutions, key=lambda i: ode_sol_simplicity(i, 

    f(x)))``. 

 

    >>> from sympy import symbols, Function, Eq, tan, cos, sqrt, Integral 

    >>> from sympy.solvers.ode import ode_sol_simplicity 

    >>> x, C1, C2 = symbols('x, C1, C2') 

    >>> f = Function('f') 

 

    >>> ode_sol_simplicity(Eq(f(x), C1*x**2), f(x)) 

    -2 

    >>> ode_sol_simplicity(Eq(x**2 + f(x), C1), f(x)) 

    -1 

    >>> ode_sol_simplicity(Eq(f(x), C1*Integral(2*x, x)), f(x)) 

    oo 

    >>> eq1 = Eq(f(x)/tan(f(x)/(2*x)), C1) 

    >>> eq2 = Eq(f(x)/tan(f(x)/(2*x) + f(x)), C2) 

    >>> [ode_sol_simplicity(eq, f(x)) for eq in [eq1, eq2]] 

    [28, 35] 

    >>> min([eq1, eq2], key=lambda i: ode_sol_simplicity(i, f(x))) 

    Eq(f(x)/tan(f(x)/(2*x)), C1) 

 

    """ 

    # TODO: if two solutions are solved for f(x), we still want to be 

    # able to get the simpler of the two 

 

    # See the docstring for the coercion rules.  We check easier (faster) 

    # things here first, to save time. 

 

    if iterable(sol): 

        # See if there are Integrals 

        for i in sol: 

            if ode_sol_simplicity(i, func, trysolving=trysolving) == oo: 

                return oo 

 

        return len(str(sol)) 

 

    if sol.has(Integral): 

        return oo 

 

    # Next, try to solve for func.  This code will change slightly when RootOf 

    # is implemented in solve().  Probably a RootOf solution should fall 

    # somewhere between a normal solution and an unsolvable expression. 

 

    # First, see if they are already solved 

    if sol.lhs == func and not sol.rhs.has(func) or \ 

            sol.rhs == func and not sol.lhs.has(func): 

        return -2 

    # We are not so lucky, try solving manually 

    if trysolving: 

        try: 

            sols = solve(sol, func) 

            if not sols: 

                raise NotImplementedError 

        except NotImplementedError: 

            pass 

        else: 

            return -1 

 

    # Finally, a naive computation based on the length of the string version 

    # of the expression.  This may favor combined fractions because they 

    # will not have duplicate denominators, and may slightly favor expressions 

    # with fewer additions and subtractions, as those are separated by spaces 

    # by the printer. 

 

    # Additional ideas for simplicity heuristics are welcome, like maybe 

    # checking if a equation has a larger domain, or if constantsimp has 

    # introduced arbitrary constants numbered higher than the order of a 

    # given ODE that sol is a solution of. 

    return len(str(sol)) 

 

 

def _get_constant_subexpressions(expr, Cs): 

    Cs = set(Cs) 

    Ces = [] 

    def _recursive_walk(expr): 

        expr_syms = expr.free_symbols 

        if len(expr_syms) > 0 and expr_syms.issubset(Cs): 

            Ces.append(expr) 

        else: 

            if expr.func == exp: 

                expr = expr.expand(mul=True) 

            if expr.func in (Add, Mul): 

                d = sift(expr.args, lambda i : i.free_symbols.issubset(Cs)) 

                if len(d[True]) > 1: 

                    x = expr.func(*d[True]) 

                    if not x.is_number: 

                        Ces.append(x) 

            elif isinstance(expr, Integral): 

                if expr.free_symbols.issubset(Cs) and \ 

                            all(len(x) == 3 for x in expr.limits): 

                    Ces.append(expr) 

            for i in expr.args: 

                _recursive_walk(i) 

        return 

    _recursive_walk(expr) 

    return Ces 

 

def __remove_linear_redundancies(expr, Cs): 

    cnts = dict([(i, expr.count(i)) for i in Cs]) 

    Cs = [i for i in Cs if cnts[i] > 0] 

 

    def _linear(expr): 

        if expr.func is Add: 

            xs = [i for i in Cs if expr.count(i)==cnts[i] \ 

                and 0 == expr.diff(i, 2)] 

            d = {} 

            for x in xs: 

                y = expr.diff(x) 

                if y not in d: 

                    d[y]=[] 

                d[y].append(x) 

            for y in d: 

                if len(d[y]) > 1: 

                    d[y].sort(key=str) 

                    for x in d[y][1:]: 

                        expr = expr.subs(x, 0) 

        return expr 

 

    def _recursive_walk(expr): 

        if len(expr.args) != 0: 

            expr = expr.func(*[_recursive_walk(i) for i in expr.args]) 

        expr = _linear(expr) 

        return expr 

 

    if expr.func is Equality: 

        lhs, rhs = [_recursive_walk(i) for i in expr.args] 

        f = lambda i: isinstance(i, Number) or i in Cs 

        if lhs.func is Symbol and lhs in Cs: 

            rhs, lhs = lhs, rhs 

        if lhs.func in (Add, Symbol) and rhs.func in (Add, Symbol): 

            dlhs = sift([lhs] if isinstance(lhs, AtomicExpr) else lhs.args, f) 

            drhs = sift([rhs] if isinstance(rhs, AtomicExpr) else rhs.args, f) 

            for i in [True, False]: 

                for hs in [dlhs, drhs]: 

                    if i not in hs: 

                        hs[i] = [0] 

            # this calculation can be simplified 

            lhs = Add(*dlhs[False]) - Add(*drhs[False]) 

            rhs = Add(*drhs[True]) - Add(*dlhs[True]) 

        elif lhs.func in (Mul, Symbol) and rhs.func in (Mul, Symbol): 

            dlhs = sift([lhs] if isinstance(lhs, AtomicExpr) else lhs.args, f) 

            if True in dlhs: 

                if False not in dlhs: 

                    dlhs[False] = [1] 

                lhs = Mul(*dlhs[False]) 

                rhs = rhs/Mul(*dlhs[True]) 

        return Eq(lhs, rhs) 

    else: 

        return _recursive_walk(expr) 

 

@vectorize(0) 

def constantsimp(expr, constants): 

    r""" 

    Simplifies an expression with arbitrary constants in it. 

 

    This function is written specifically to work with 

    :py:meth:`~sympy.solvers.ode.dsolve`, and is not intended for general use. 

 

    Simplification is done by "absorbing" the arbitrary constants into other 

    arbitrary constants, numbers, and symbols that they are not independent 

    of. 

 

    The symbols must all have the same name with numbers after it, for 

    example, ``C1``, ``C2``, ``C3``.  The ``symbolname`` here would be 

    '``C``', the ``startnumber`` would be 1, and the ``endnumber`` would be 3. 

    If the arbitrary constants are independent of the variable ``x``, then the 

    independent symbol would be ``x``.  There is no need to specify the 

    dependent function, such as ``f(x)``, because it already has the 

    independent symbol, ``x``, in it. 

 

    Because terms are "absorbed" into arbitrary constants and because 

    constants are renumbered after simplifying, the arbitrary constants in 

    expr are not necessarily equal to the ones of the same name in the 

    returned result. 

 

    If two or more arbitrary constants are added, multiplied, or raised to the 

    power of each other, they are first absorbed together into a single 

    arbitrary constant.  Then the new constant is combined into other terms if 

    necessary. 

 

    Absorption of constants is done with limited assistance: 

 

    1. terms of :py:class:`~sympy.core.add.Add`\s are collected to try join 

       constants so `e^x (C_1 \cos(x) + C_2 \cos(x))` will simplify to `e^x 

       C_1 \cos(x)`; 

 

    2. powers with exponents that are :py:class:`~sympy.core.add.Add`\s are 

       expanded so `e^{C_1 + x}` will be simplified to `C_1 e^x`. 

 

    Use :py:meth:`~sympy.solvers.ode.constant_renumber` to renumber constants 

    after simplification or else arbitrary numbers on constants may appear, 

    e.g. `C_1 + C_3 x`. 

 

    In rare cases, a single constant can be "simplified" into two constants. 

    Every differential equation solution should have as many arbitrary 

    constants as the order of the differential equation.  The result here will 

    be technically correct, but it may, for example, have `C_1` and `C_2` in 

    an expression, when `C_1` is actually equal to `C_2`.  Use your discretion 

    in such situations, and also take advantage of the ability to use hints in 

    :py:meth:`~sympy.solvers.ode.dsolve`. 

 

    Examples 

    ======== 

 

    >>> from sympy import symbols 

    >>> from sympy.solvers.ode import constantsimp 

    >>> C1, C2, C3, x, y = symbols('C1, C2, C3, x, y') 

    >>> constantsimp(2*C1*x, set([C1, C2, C3])) 

    C1*x 

    >>> constantsimp(C1 + 2 + x, set([C1, C2, C3])) 

    C1 + x 

    >>> constantsimp(C1*C2 + 2 + C2 + C3*x, set([C1, C2, C3])) 

    C1 + C3*x 

 

    """ 

    # This function works recursively.  The idea is that, for Mul, 

    # Add, Pow, and Function, if the class has a constant in it, then 

    # we can simplify it, which we do by recursing down and 

    # simplifying up.  Otherwise, we can skip that part of the 

    # expression. 

 

    Cs = constants 

 

    orig_expr = expr 

 

    constant_subexprs = _get_constant_subexpressions(expr, Cs) 

    for xe in constant_subexprs: 

        xes = list(xe.free_symbols) 

        if not xes: 

            continue 

        if all([expr.count(c) == xe.count(c) for c in xes]): 

            xes.sort(key=str) 

            expr = expr.subs(xe, xes[0]) 

 

    # try to perform common sub-expression elimination of constant terms 

    try: 

        commons, rexpr = cse(expr) 

        commons.reverse() 

        rexpr = rexpr[0] 

        for s in commons: 

            cs = list(s[1].atoms(Symbol)) 

            if len(cs) == 1 and cs[0] in Cs: 

                rexpr = rexpr.subs(s[0], cs[0]) 

            else: 

                rexpr = rexpr.subs(*s) 

        expr = rexpr 

    except Exception: 

        pass 

    expr = __remove_linear_redundancies(expr, Cs) 

 

    def _conditional_term_factoring(expr): 

        new_expr = terms_gcd(expr, clear=False, deep=True, expand=False) 

 

        # we do not want to factor exponentials, so handle this separately 

        if new_expr.is_Mul: 

            infac = False 

            asfac = False 

            for m in new_expr.args: 

                if m.func is exp: 

                    asfac = True 

                elif m.is_Add: 

                    infac = any(fi.func is exp for t in m.args 

                        for fi in Mul.make_args(t)) 

                if asfac and infac: 

                    new_expr = expr 

                    break 

        return new_expr 

 

    expr = _conditional_term_factoring(expr) 

 

    # call recursively if more simplification is possible 

    if orig_expr != expr: 

        return constantsimp(expr, Cs) 

    return expr 

 

 

def constant_renumber(expr, symbolname, startnumber, endnumber): 

    r""" 

    Renumber arbitrary constants in ``expr`` to have numbers 1 through `N` 

    where `N` is ``endnumber - startnumber + 1`` at most. 

    In the process, this reorders expression terms in a standard way. 

 

    This is a simple function that goes through and renumbers any 

    :py:class:`~sympy.core.symbol.Symbol` with a name in the form ``symbolname 

    + num`` where ``num`` is in the range from ``startnumber`` to 

    ``endnumber``. 

 

    Symbols are renumbered based on ``.sort_key()``, so they should be 

    numbered roughly in the order that they appear in the final, printed 

    expression.  Note that this ordering is based in part on hashes, so it can 

    produce different results on different machines. 

 

    The structure of this function is very similar to that of 

    :py:meth:`~sympy.solvers.ode.constantsimp`. 

 

    Examples 

    ======== 

 

    >>> from sympy import symbols, Eq, pprint 

    >>> from sympy.solvers.ode import constant_renumber 

    >>> x, C0, C1, C2, C3, C4 = symbols('x,C:5') 

 

    Only constants in the given range (inclusive) are renumbered; 

    the renumbering always starts from 1: 

 

    >>> constant_renumber(C1 + C3 + C4, 'C', 1, 3) 

    C1 + C2 + C4 

    >>> constant_renumber(C0 + C1 + C3 + C4, 'C', 2, 4) 

    C0 + 2*C1 + C2 

    >>> constant_renumber(C0 + 2*C1 + C2, 'C', 0, 1) 

    C1 + 3*C2 

    >>> pprint(C2 + C1*x + C3*x**2) 

                    2 

    C1*x + C2 + C3*x 

    >>> pprint(constant_renumber(C2 + C1*x + C3*x**2, 'C', 1, 3)) 

                    2 

    C1 + C2*x + C3*x 

 

    """ 

    if type(expr) in (set, list, tuple): 

        return type(expr)( 

            [constant_renumber(i, symbolname=symbolname, startnumber=startnumber, endnumber=endnumber) 

                for i in expr] 

        ) 

    global newstartnumber 

    newstartnumber = 1 

    constants_found = [None]*(endnumber + 2) 

    constantsymbols = [Symbol( 

        symbolname + "%d" % t) for t in range(startnumber, 

        endnumber + 1)] 

 

    # make a mapping to send all constantsymbols to S.One and use 

    # that to make sure that term ordering is not dependent on 

    # the indexed value of C 

    C_1 = [(ci, S.One) for ci in constantsymbols] 

    sort_key=lambda arg: default_sort_key(arg.subs(C_1)) 

 

    def _constant_renumber(expr): 

        r""" 

        We need to have an internal recursive function so that 

        newstartnumber maintains its values throughout recursive calls. 

 

        """ 

        global newstartnumber 

 

        if isinstance(expr, Equality): 

            return Eq( 

                _constant_renumber(expr.lhs), 

                _constant_renumber(expr.rhs)) 

 

        if type(expr) not in (Mul, Add, Pow) and not expr.is_Function and \ 

                not expr.has(*constantsymbols): 

            # Base case, as above.  Hope there aren't constants inside 

            # of some other class, because they won't be renumbered. 

            return expr 

        elif expr.is_Piecewise: 

            return expr 

        elif expr in constantsymbols: 

            if expr not in constants_found: 

                constants_found[newstartnumber] = expr 

                newstartnumber += 1 

            return expr 

        elif expr.is_Function or expr.is_Pow or isinstance(expr, Tuple): 

            return expr.func( 

                *[_constant_renumber(x) for x in expr.args]) 

        else: 

            sortedargs = list(expr.args) 

            sortedargs.sort(key=sort_key) 

            return expr.func(*[_constant_renumber(x) for x in sortedargs]) 

    expr = _constant_renumber(expr) 

    # Renumbering happens here 

    newconsts = symbols('C1:%d' % newstartnumber) 

    expr = expr.subs(zip(constants_found[1:], newconsts), simultaneous=True) 

    return expr 

 

 

def _handle_Integral(expr, func, order, hint): 

    r""" 

    Converts a solution with Integrals in it into an actual solution. 

 

    For most hints, this simply runs ``expr.doit()``. 

 

    """ 

    global y 

    x = func.args[0] 

    f = func.func 

    if hint == "1st_exact": 

        sol = (expr.doit()).subs(y, f(x)) 

        del y 

    elif hint == "1st_exact_Integral": 

        sol = expr.subs(y, f(x)) 

        del y 

    elif hint == "nth_linear_constant_coeff_homogeneous": 

        sol = expr 

    elif not hint.endswith("_Integral"): 

        sol = expr.doit() 

    else: 

        sol = expr 

    return sol 

 

 

# FIXME: replace the general solution in the docstring with 

# dsolve(equation, hint='1st_exact_Integral').  You will need to be able 

# to have assumptions on P and Q that dP/dy = dQ/dx. 

def ode_1st_exact(eq, func, order, match): 

    r""" 

    Solves 1st order exact ordinary differential equations. 

 

    A 1st order differential equation is called exact if it is the total 

    differential of a function. That is, the differential equation 

 

    .. math:: P(x, y) \,\partial{}x + Q(x, y) \,\partial{}y = 0 

 

    is exact if there is some function `F(x, y)` such that `P(x, y) = 

    \partial{}F/\partial{}x` and `Q(x, y) = \partial{}F/\partial{}y`.  It can 

    be shown that a necessary and sufficient condition for a first order ODE 

    to be exact is that `\partial{}P/\partial{}y = \partial{}Q/\partial{}x`. 

    Then, the solution will be as given below:: 

 

        >>> from sympy import Function, Eq, Integral, symbols, pprint 

        >>> x, y, t, x0, y0, C1= symbols('x,y,t,x0,y0,C1') 

        >>> P, Q, F= map(Function, ['P', 'Q', 'F']) 

        >>> pprint(Eq(Eq(F(x, y), Integral(P(t, y), (t, x0, x)) + 

        ... Integral(Q(x0, t), (t, y0, y))), C1)) 

                    x                y 

                    /                / 

                   |                | 

        F(x, y) =  |  P(t, y) dt +  |  Q(x0, t) dt = C1 

                   |                | 

                  /                / 

                  x0               y0 

 

    Where the first partials of `P` and `Q` exist and are continuous in a 

    simply connected region. 

 

    A note: SymPy currently has no way to represent inert substitution on an 

    expression, so the hint ``1st_exact_Integral`` will return an integral 

    with `dy`.  This is supposed to represent the function that you are 

    solving for. 

 

    Examples 

    ======== 

 

    >>> from sympy import Function, dsolve, cos, sin 

    >>> from sympy.abc import x 

    >>> f = Function('f') 

    >>> dsolve(cos(f(x)) - (x*sin(f(x)) - f(x)**2)*f(x).diff(x), 

    ... f(x), hint='1st_exact') 

    Eq(x*cos(f(x)) + f(x)**3/3, C1) 

 

    References 

    ========== 

 

    - http://en.wikipedia.org/wiki/Exact_differential_equation 

    - M. Tenenbaum & H. Pollard, "Ordinary Differential Equations", 

      Dover 1963, pp. 73 

 

    # indirect doctest 

 

    """ 

    x = func.args[0] 

    f = func.func 

    r = match  # d+e*diff(f(x),x) 

    e = r[r['e']] 

    d = r[r['d']] 

    global y  # This is the only way to pass dummy y to _handle_Integral 

    y = r['y'] 

    C1 = get_numbered_constants(eq, num=1) 

    # Refer Joel Moses, "Symbolic Integration - The Stormy Decade", 

    # Communications of the ACM, Volume 14, Number 8, August 1971, pp. 558 

    # which gives the method to solve an exact differential equation. 

    sol = Integral(d, x) + Integral((e - (Integral(d, x).diff(y))), y) 

    return Eq(sol, C1) 

 

 

def ode_1st_homogeneous_coeff_best(eq, func, order, match): 

    r""" 

    Returns the best solution to an ODE from the two hints 

    ``1st_homogeneous_coeff_subs_dep_div_indep`` and 

    ``1st_homogeneous_coeff_subs_indep_div_dep``. 

 

    This is as determined by :py:meth:`~sympy.solvers.ode.ode_sol_simplicity`. 

 

    See the 

    :py:meth:`~sympy.solvers.ode.ode_1st_homogeneous_coeff_subs_indep_div_dep` 

    and 

    :py:meth:`~sympy.solvers.ode.ode_1st_homogeneous_coeff_subs_dep_div_indep` 

    docstrings for more information on these hints.  Note that there is no 

    ``ode_1st_homogeneous_coeff_best_Integral`` hint. 

 

    Examples 

    ======== 

 

    >>> from sympy import Function, dsolve, pprint 

    >>> from sympy.abc import x 

    >>> f = Function('f') 

    >>> pprint(dsolve(2*x*f(x) + (x**2 + f(x)**2)*f(x).diff(x), f(x), 

    ... hint='1st_homogeneous_coeff_best', simplify=False)) 

                             /    2    \ 

                             | 3*x     | 

                          log|----- + 1| 

                             | 2       | 

                             \f (x)    / 

    log(f(x)) = log(C1) - -------------- 

                                3 

 

    References 

    ========== 

 

    - http://en.wikipedia.org/wiki/Homogeneous_differential_equation 

    - M. Tenenbaum & H. Pollard, "Ordinary Differential Equations", 

      Dover 1963, pp. 59 

 

    # indirect doctest 

 

    """ 

    # There are two substitutions that solve the equation, u1=y/x and u2=x/y 

    # They produce different integrals, so try them both and see which 

    # one is easier. 

    sol1 = ode_1st_homogeneous_coeff_subs_indep_div_dep(eq, 

    func, order, match) 

    sol2 = ode_1st_homogeneous_coeff_subs_dep_div_indep(eq, 

    func, order, match) 

    simplify = match.get('simplify', True) 

    if simplify: 

        # why is odesimp called here?  Should it be at the usual spot? 

        constants = sol1.free_symbols.difference(eq.free_symbols) 

        sol1 = odesimp( 

            sol1, func, order, constants, 

            "1st_homogeneous_coeff_subs_indep_div_dep") 

        constants = sol2.free_symbols.difference(eq.free_symbols) 

        sol2 = odesimp( 

            sol2, func, order, constants, 

            "1st_homogeneous_coeff_subs_dep_div_indep") 

    return min([sol1, sol2], key=lambda x: ode_sol_simplicity(x, func, 

        trysolving=not simplify)) 

 

 

def ode_1st_homogeneous_coeff_subs_dep_div_indep(eq, func, order, match): 

    r""" 

    Solves a 1st order differential equation with homogeneous coefficients 

    using the substitution `u_1 = \frac{\text{<dependent 

    variable>}}{\text{<independent variable>}}`. 

 

    This is a differential equation 

 

    .. math:: P(x, y) + Q(x, y) dy/dx = 0 

 

    such that `P` and `Q` are homogeneous and of the same order.  A function 

    `F(x, y)` is homogeneous of order `n` if `F(x t, y t) = t^n F(x, y)`. 

    Equivalently, `F(x, y)` can be rewritten as `G(y/x)` or `H(x/y)`.  See 

    also the docstring of :py:meth:`~sympy.solvers.ode.homogeneous_order`. 

 

    If the coefficients `P` and `Q` in the differential equation above are 

    homogeneous functions of the same order, then it can be shown that the 

    substitution `y = u_1 x` (i.e. `u_1 = y/x`) will turn the differential 

    equation into an equation separable in the variables `x` and `u`.  If 

    `h(u_1)` is the function that results from making the substitution `u_1 = 

    f(x)/x` on `P(x, f(x))` and `g(u_2)` is the function that results from the 

    substitution on `Q(x, f(x))` in the differential equation `P(x, f(x)) + 

    Q(x, f(x)) f'(x) = 0`, then the general solution is:: 

 

        >>> from sympy import Function, dsolve, pprint 

        >>> from sympy.abc import x 

        >>> f, g, h = map(Function, ['f', 'g', 'h']) 

        >>> genform = g(f(x)/x) + h(f(x)/x)*f(x).diff(x) 

        >>> pprint(genform) 

         /f(x)\    /f(x)\ d 

        g|----| + h|----|*--(f(x)) 

         \ x  /    \ x  / dx 

        >>> pprint(dsolve(genform, f(x), 

        ... hint='1st_homogeneous_coeff_subs_dep_div_indep_Integral')) 

                       f(x) 

                       ---- 

                        x 

                         / 

                        | 

                        |       -h(u1) 

        log(x) = C1 +   |  ---------------- d(u1) 

                        |  u1*h(u1) + g(u1) 

                        | 

                       / 

 

    Where `u_1 h(u_1) + g(u_1) \ne 0` and `x \ne 0`. 

 

    See also the docstrings of 

    :py:meth:`~sympy.solvers.ode.ode_1st_homogeneous_coeff_best` and 

    :py:meth:`~sympy.solvers.ode.ode_1st_homogeneous_coeff_subs_indep_div_dep`. 

 

    Examples 

    ======== 

 

    >>> from sympy import Function, dsolve 

    >>> from sympy.abc import x 

    >>> f = Function('f') 

    >>> pprint(dsolve(2*x*f(x) + (x**2 + f(x)**2)*f(x).diff(x), f(x), 

    ... hint='1st_homogeneous_coeff_subs_dep_div_indep', simplify=False)) 

                          /          3   \ 

                          |3*f(x)   f (x)| 

                       log|------ + -----| 

                          |  x         3 | 

                          \           x  / 

    log(x) = log(C1) - ------------------- 

                                3 

 

    References 

    ========== 

 

    - http://en.wikipedia.org/wiki/Homogeneous_differential_equation 

    - M. Tenenbaum & H. Pollard, "Ordinary Differential Equations", 

      Dover 1963, pp. 59 

 

    # indirect doctest 

 

    """ 

    x = func.args[0] 

    f = func.func 

    u = Dummy('u') 

    u1 = Dummy('u1')  # u1 == f(x)/x 

    r = match  # d+e*diff(f(x),x) 

    C1 = get_numbered_constants(eq, num=1) 

    xarg = match.get('xarg', 0) 

    yarg = match.get('yarg', 0) 

    int = Integral( 

        (-r[r['e']]/(r[r['d']] + u1*r[r['e']])).subs({x: 1, r['y']: u1}), 

        (u1, None, f(x)/x)) 

    sol = logcombine(Eq(log(x), int + log(C1)), force=True) 

    sol = sol.subs(f(x), u).subs(((u, u - yarg), (x, x - xarg), (u, f(x)))) 

    return sol 

 

 

def ode_1st_homogeneous_coeff_subs_indep_div_dep(eq, func, order, match): 

    r""" 

    Solves a 1st order differential equation with homogeneous coefficients 

    using the substitution `u_2 = \frac{\text{<independent 

    variable>}}{\text{<dependent variable>}}`. 

 

    This is a differential equation 

 

    .. math:: P(x, y) + Q(x, y) dy/dx = 0 

 

    such that `P` and `Q` are homogeneous and of the same order.  A function 

    `F(x, y)` is homogeneous of order `n` if `F(x t, y t) = t^n F(x, y)`. 

    Equivalently, `F(x, y)` can be rewritten as `G(y/x)` or `H(x/y)`.  See 

    also the docstring of :py:meth:`~sympy.solvers.ode.homogeneous_order`. 

 

    If the coefficients `P` and `Q` in the differential equation above are 

    homogeneous functions of the same order, then it can be shown that the 

    substitution `x = u_2 y` (i.e. `u_2 = x/y`) will turn the differential 

    equation into an equation separable in the variables `y` and `u_2`.  If 

    `h(u_2)` is the function that results from making the substitution `u_2 = 

    x/f(x)` on `P(x, f(x))` and `g(u_2)` is the function that results from the 

    substitution on `Q(x, f(x))` in the differential equation `P(x, f(x)) + 

    Q(x, f(x)) f'(x) = 0`, then the general solution is: 

 

    >>> from sympy import Function, dsolve, pprint 

    >>> from sympy.abc import x 

    >>> f, g, h = map(Function, ['f', 'g', 'h']) 

    >>> genform = g(x/f(x)) + h(x/f(x))*f(x).diff(x) 

    >>> pprint(genform) 

     / x  \    / x  \ d 

    g|----| + h|----|*--(f(x)) 

     \f(x)/    \f(x)/ dx 

    >>> pprint(dsolve(genform, f(x), 

    ... hint='1st_homogeneous_coeff_subs_indep_div_dep_Integral')) 

                 x 

                ---- 

                f(x) 

                  / 

                 | 

                 |       -g(u2) 

                 |  ---------------- d(u2) 

                 |  u2*g(u2) + h(u2) 

                 | 

                / 

    <BLANKLINE> 

    f(x) = C1*e 

 

    Where `u_2 g(u_2) + h(u_2) \ne 0` and `f(x) \ne 0`. 

 

    See also the docstrings of 

    :py:meth:`~sympy.solvers.ode.ode_1st_homogeneous_coeff_best` and 

    :py:meth:`~sympy.solvers.ode.ode_1st_homogeneous_coeff_subs_dep_div_indep`. 

 

    Examples 

    ======== 

 

    >>> from sympy import Function, pprint, dsolve 

    >>> from sympy.abc import x 

    >>> f = Function('f') 

    >>> pprint(dsolve(2*x*f(x) + (x**2 + f(x)**2)*f(x).diff(x), f(x), 

    ... hint='1st_homogeneous_coeff_subs_indep_div_dep', 

    ... simplify=False)) 

                             /    2    \ 

                             | 3*x     | 

                          log|----- + 1| 

                             | 2       | 

                             \f (x)    / 

    log(f(x)) = log(C1) - -------------- 

                                3 

 

    References 

    ========== 

 

    - http://en.wikipedia.org/wiki/Homogeneous_differential_equation 

    - M. Tenenbaum & H. Pollard, "Ordinary Differential Equations", 

      Dover 1963, pp. 59 

 

    # indirect doctest 

 

    """ 

    x = func.args[0] 

    f = func.func 

    u = Dummy('u') 

    u2 = Dummy('u2')  # u2 == x/f(x) 

    r = match  # d+e*diff(f(x),x) 

    C1 = get_numbered_constants(eq, num=1) 

    xarg = match.get('xarg', 0)  # If xarg present take xarg, else zero 

    yarg = match.get('yarg', 0)  # If yarg present take yarg, else zero 

    int = Integral( 

        simplify( 

            (-r[r['d']]/(r[r['e']] + u2*r[r['d']])).subs({x: u2, r['y']: 1})), 

        (u2, None, x/f(x))) 

    sol = logcombine(Eq(log(f(x)), int + log(C1)), force=True) 

    sol = sol.subs(f(x), u).subs(((u, u - yarg), (x, x - xarg), (u, f(x)))) 

    return sol 

 

# XXX: Should this function maybe go somewhere else? 

 

 

def homogeneous_order(eq, *symbols): 

    r""" 

    Returns the order `n` if `g` is homogeneous and ``None`` if it is not 

    homogeneous. 

 

    Determines if a function is homogeneous and if so of what order.  A 

    function `f(x, y, \cdots)` is homogeneous of order `n` if `f(t x, t y, 

    \cdots) = t^n f(x, y, \cdots)`. 

 

    If the function is of two variables, `F(x, y)`, then `f` being homogeneous 

    of any order is equivalent to being able to rewrite `F(x, y)` as `G(x/y)` 

    or `H(y/x)`.  This fact is used to solve 1st order ordinary differential 

    equations whose coefficients are homogeneous of the same order (see the 

    docstrings of 

    :py:meth:`~solvers.ode.ode_1st_homogeneous_coeff_subs_dep_div_indep` and 

    :py:meth:`~solvers.ode.ode_1st_homogeneous_coeff_subs_indep_div_dep`). 

 

    Symbols can be functions, but every argument of the function must be a 

    symbol, and the arguments of the function that appear in the expression 

    must match those given in the list of symbols.  If a declared function 

    appears with different arguments than given in the list of symbols, 

    ``None`` is returned. 

 

    Examples 

    ======== 

 

    >>> from sympy import Function, homogeneous_order, sqrt 

    >>> from sympy.abc import x, y 

    >>> f = Function('f') 

    >>> homogeneous_order(f(x), f(x)) is None 

    True 

    >>> homogeneous_order(f(x,y), f(y, x), x, y) is None 

    True 

    >>> homogeneous_order(f(x), f(x), x) 

    1 

    >>> homogeneous_order(x**2*f(x)/sqrt(x**2+f(x)**2), x, f(x)) 

    2 

    >>> homogeneous_order(x**2+f(x), x, f(x)) is None 

    True 

 

    """ 

 

    if not symbols: 

        raise ValueError("homogeneous_order: no symbols were given.") 

    symset = set(symbols) 

    eq = sympify(eq) 

 

    # The following are not supported 

    if eq.has(Order, Derivative): 

        return None 

 

    # These are all constants 

    if (eq.is_Number or 

        eq.is_NumberSymbol or 

        eq.is_number 

            ): 

        return S.Zero 

 

    # Replace all functions with dummy variables 

    dum = numbered_symbols(prefix='d', cls=Dummy) 

    newsyms = set() 

    for i in [j for j in symset if getattr(j, 'is_Function')]: 

        iargs = set(i.args) 

        if iargs.difference(symset): 

            return None 

        else: 

            dummyvar = next(dum) 

            eq = eq.subs(i, dummyvar) 

            symset.remove(i) 

            newsyms.add(dummyvar) 

    symset.update(newsyms) 

 

    if not eq.free_symbols & symset: 

        return None 

 

    # assuming order of a nested function can only be equal to zero 

    if isinstance(eq, Function): 

        return None if homogeneous_order( 

            eq.args[0], *tuple(symset)) != 0 else S.Zero 

 

    # make the replacement of x with x*t and see if t can be factored out 

    t = Dummy('t', positive=True)  # It is sufficient that t > 0 

    eqs = separatevars(eq.subs([(i, t*i) for i in symset]), [t], dict=True)[t] 

    if eqs is S.One: 

        return S.Zero  # there was no term with only t 

    i, d = eqs.as_independent(t, as_Add=False) 

    b, e = d.as_base_exp() 

    if b == t: 

        return e 

 

 

def ode_1st_linear(eq, func, order, match): 

    r""" 

    Solves 1st order linear differential equations. 

 

    These are differential equations of the form 

 

    .. math:: dy/dx + P(x) y = Q(x)\text{.} 

 

    These kinds of differential equations can be solved in a general way.  The 

    integrating factor `e^{\int P(x) \,dx}` will turn the equation into a 

    separable equation.  The general solution is:: 

 

        >>> from sympy import Function, dsolve, Eq, pprint, diff, sin 

        >>> from sympy.abc import x 

        >>> f, P, Q = map(Function, ['f', 'P', 'Q']) 

        >>> genform = Eq(f(x).diff(x) + P(x)*f(x), Q(x)) 

        >>> pprint(genform) 

                    d 

        P(x)*f(x) + --(f(x)) = Q(x) 

                    dx 

        >>> pprint(dsolve(genform, f(x), hint='1st_linear_Integral')) 

               /       /                   \ 

               |      |                    | 

               |      |         /          |     / 

               |      |        |           |    | 

               |      |        | P(x) dx   |  - | P(x) dx 

               |      |        |           |    | 

               |      |       /            |   / 

        f(x) = |C1 +  | Q(x)*e           dx|*e 

               |      |                    | 

               \     /                     / 

 

 

    Examples 

    ======== 

 

    >>> f = Function('f') 

    >>> pprint(dsolve(Eq(x*diff(f(x), x) - f(x), x**2*sin(x)), 

    ... f(x), '1st_linear')) 

    f(x) = x*(C1 - cos(x)) 

 

    References 

    ========== 

 

    - http://en.wikipedia.org/wiki/Linear_differential_equation#First_order_equation 

    - M. Tenenbaum & H. Pollard, "Ordinary Differential Equations", 

      Dover 1963, pp. 92 

 

    # indirect doctest 

 

    """ 

    x = func.args[0] 

    f = func.func 

    r = match  # a*diff(f(x),x) + b*f(x) + c 

    C1 = get_numbered_constants(eq, num=1) 

    t = exp(Integral(r[r['b']]/r[r['a']], x)) 

    tt = Integral(t*(-r[r['c']]/r[r['a']]), x) 

    f = match.get('u', f(x))  # take almost-linear u if present, else f(x) 

    return Eq(f, (tt + C1)/t) 

 

 

def ode_Bernoulli(eq, func, order, match): 

    r""" 

    Solves Bernoulli differential equations. 

 

    These are equations of the form 

 

    .. math:: dy/dx + P(x) y = Q(x) y^n\text{, }n \ne 1`\text{.} 

 

    The substitution `w = 1/y^{1-n}` will transform an equation of this form 

    into one that is linear (see the docstring of 

    :py:meth:`~sympy.solvers.ode.ode_1st_linear`).  The general solution is:: 

 

        >>> from sympy import Function, dsolve, Eq, pprint 

        >>> from sympy.abc import x, n 

        >>> f, P, Q = map(Function, ['f', 'P', 'Q']) 

        >>> genform = Eq(f(x).diff(x) + P(x)*f(x), Q(x)*f(x)**n) 

        >>> pprint(genform) 

                    d                n 

        P(x)*f(x) + --(f(x)) = Q(x)*f (x) 

                    dx 

        >>> pprint(dsolve(genform, f(x), hint='Bernoulli_Integral')) #doctest: +SKIP 

                                                                                       1 

                                                                                      ---- 

                                                                                     1 - n 

               //                /                            \                     \ 

               ||               |                             |                     | 

               ||               |                  /          |             /       | 

               ||               |                 |           |            |        | 

               ||               |        (1 - n)* | P(x) dx   |  (-1 + n)* | P(x) dx| 

               ||               |                 |           |            |        | 

               ||               |                /            |           /         | 

        f(x) = ||C1 + (-1 + n)* | -Q(x)*e                   dx|*e                   | 

               ||               |                             |                     | 

               \\               /                            /                     / 

 

 

    Note that the equation is separable when `n = 1` (see the docstring of 

    :py:meth:`~sympy.solvers.ode.ode_separable`). 

 

    >>> pprint(dsolve(Eq(f(x).diff(x) + P(x)*f(x), Q(x)*f(x)), f(x), 

    ... hint='separable_Integral')) 

     f(x) 

       / 

      |                / 

      |  1            | 

      |  - dy = C1 +  | (-P(x) + Q(x)) dx 

      |  y            | 

      |              / 

     / 

 

 

    Examples 

    ======== 

 

    >>> from sympy import Function, dsolve, Eq, pprint, log 

    >>> from sympy.abc import x 

    >>> f = Function('f') 

 

    >>> pprint(dsolve(Eq(x*f(x).diff(x) + f(x), log(x)*f(x)**2), 

    ... f(x), hint='Bernoulli')) 

                    1 

    f(x) = ------------------- 

             /     log(x)   1\ 

           x*|C1 + ------ + -| 

             \       x      x/ 

 

    References 

    ========== 

 

    - http://en.wikipedia.org/wiki/Bernoulli_differential_equation 

    - M. Tenenbaum & H. Pollard, "Ordinary Differential Equations", 

      Dover 1963, pp. 95 

 

    # indirect doctest 

 

    """ 

    x = func.args[0] 

    f = func.func 

    r = match  # a*diff(f(x),x) + b*f(x) + c*f(x)**n, n != 1 

    C1 = get_numbered_constants(eq, num=1) 

    t = exp((1 - r[r['n']])*Integral(r[r['b']]/r[r['a']], x)) 

    tt = (r[r['n']] - 1)*Integral(t*r[r['c']]/r[r['a']], x) 

    return Eq(f(x), ((tt + C1)/t)**(1/(1 - r[r['n']]))) 

 

 

def ode_Riccati_special_minus2(eq, func, order, match): 

    r""" 

    The general Riccati equation has the form 

 

    .. math:: dy/dx = f(x) y^2 + g(x) y + h(x)\text{.} 

 

    While it does not have a general solution [1], the "special" form, `dy/dx 

    = a y^2 - b x^c`, does have solutions in many cases [2].  This routine 

    returns a solution for `a(dy/dx) = b y^2 + c y/x + d/x^2` that is obtained 

    by using a suitable change of variables to reduce it to the special form 

    and is valid when neither `a` nor `b` are zero and either `c` or `d` is 

    zero. 

 

    >>> from sympy.abc import x, y, a, b, c, d 

    >>> from sympy.solvers.ode import dsolve, checkodesol 

    >>> from sympy import pprint, Function 

    >>> f = Function('f') 

    >>> y = f(x) 

    >>> genform = a*y.diff(x) - (b*y**2 + c*y/x + d/x**2) 

    >>> sol = dsolve(genform, y) 

    >>> pprint(sol, wrap_line=False) 

            /                                 /        __________________       \\ 

            |           __________________    |       /                2        || 

            |          /                2     |     \/  4*b*d - (a + c)  *log(x)|| 

           -|a + c - \/  4*b*d - (a + c)  *tan|C1 + ----------------------------|| 

            \                                 \                 2*a             // 

    f(x) = ------------------------------------------------------------------------ 

                                            2*b*x 

 

    >>> checkodesol(genform, sol, order=1)[0] 

    True 

 

    References 

    ========== 

 

    1. http://www.maplesoft.com/support/help/Maple/view.aspx?path=odeadvisor/Riccati 

    2. http://eqworld.ipmnet.ru/en/solutions/ode/ode0106.pdf - 

       http://eqworld.ipmnet.ru/en/solutions/ode/ode0123.pdf 

    """ 

 

    x = func.args[0] 

    f = func.func 

    r = match  # a2*diff(f(x),x) + b2*f(x) + c2*f(x)/x + d2/x**2 

    a2, b2, c2, d2 = [r[r[s]] for s in 'a2 b2 c2 d2'.split()] 

    C1 = get_numbered_constants(eq, num=1) 

    mu = sqrt(4*d2*b2 - (a2 - c2)**2) 

    return Eq(f(x), (a2 - c2 - mu*tan(mu/(2*a2)*log(x) + C1))/(2*b2*x)) 

 

 

def ode_Liouville(eq, func, order, match): 

    r""" 

    Solves 2nd order Liouville differential equations. 

 

    The general form of a Liouville ODE is 

 

    .. math:: \frac{d^2 y}{dx^2} + g(y) \left(\! 

                \frac{dy}{dx}\!\right)^2 + h(x) 

                \frac{dy}{dx}\text{.} 

 

    The general solution is: 

 

        >>> from sympy import Function, dsolve, Eq, pprint, diff 

        >>> from sympy.abc import x 

        >>> f, g, h = map(Function, ['f', 'g', 'h']) 

        >>> genform = Eq(diff(f(x),x,x) + g(f(x))*diff(f(x),x)**2 + 

        ... h(x)*diff(f(x),x), 0) 

        >>> pprint(genform) 

                          2                    2 

                /d       \         d          d 

        g(f(x))*|--(f(x))|  + h(x)*--(f(x)) + ---(f(x)) = 0 

                \dx      /         dx           2 

                                              dx 

        >>> pprint(dsolve(genform, f(x), hint='Liouville_Integral')) 

                                          f(x) 

                  /                     / 

                 |                     | 

                 |     /               |     / 

                 |    |                |    | 

                 |  - | h(x) dx        |    | g(y) dy 

                 |    |                |    | 

                 |   /                 |   / 

        C1 + C2* | e            dx +   |  e           dy = 0 

                 |                     | 

                /                     / 

 

    Examples 

    ======== 

 

    >>> from sympy import Function, dsolve, Eq, pprint 

    >>> from sympy.abc import x 

    >>> f = Function('f') 

    >>> pprint(dsolve(diff(f(x), x, x) + diff(f(x), x)**2/f(x) + 

    ... diff(f(x), x)/x, f(x), hint='Liouville')) 

               ________________           ________________ 

    [f(x) = -\/ C1 + C2*log(x) , f(x) = \/ C1 + C2*log(x) ] 

 

    References 

    ========== 

 

    - Goldstein and Braun, "Advanced Methods for the Solution of Differential 

      Equations", pp. 98 

    - http://www.maplesoft.com/support/help/Maple/view.aspx?path=odeadvisor/Liouville 

 

    # indirect doctest 

 

    """ 

    # Liouville ODE: 

    #  f(x).diff(x, 2) + g(f(x))*(f(x).diff(x, 2))**2 + h(x)*f(x).diff(x) 

    # See Goldstein and Braun, "Advanced Methods for the Solution of 

    # Differential Equations", pg. 98, as well as 

    # http://www.maplesoft.com/support/help/view.aspx?path=odeadvisor/Liouville 

    x = func.args[0] 

    f = func.func 

    r = match  # f(x).diff(x, 2) + g*f(x).diff(x)**2 + h*f(x).diff(x) 

    y = r['y'] 

    C1, C2 = get_numbered_constants(eq, num=2) 

    int = Integral(exp(Integral(r['g'], y)), (y, None, f(x))) 

    sol = Eq(int + C1*Integral(exp(-Integral(r['h'], x)), x) + C2, 0) 

    return sol 

 

 

def ode_2nd_power_series_ordinary(eq, func, order, match): 

    r""" 

    Gives a power series solution to a second order homogeneous differential 

    equation with polynomial coefficients at an ordinary point. A homogenous 

    differential equation is of the form 

 

    .. math :: P(x)\frac{d^2y}{dx^2} + Q(x)\frac{dy}{dx} + R(x) = 0 

 

    For simplicity it is assumed that `P(x)`, `Q(x)` and `R(x)` are polynomials, 

    it is sufficient that `\frac{Q(x)}{P(x)}` and `\frac{R(x)}{P(x)}` exists at 

    `x_{0}`. A recurrence relation is obtained by substituting `y` as `\sum_{n=0}^\infty a_{n}x^{n}`, 

    in the differential equation, and equating the nth term. Using this relation 

    various terms can be generated. 

 

 

    Examples 

    ======== 

 

    >>> from sympy import dsolve, Function, pprint 

    >>> from sympy.abc import x, y 

    >>> f = Function("f") 

    >>> eq = f(x).diff(x, 2) + f(x) 

    >>> pprint(dsolve(eq, hint='2nd_power_series_ordinary')) 

              / 4    2    \        /   2    \ 

              |x    x     |        |  x     |    / 6\ 

    f(x) = C2*|-- - -- + 1| + C1*x*|- -- + 1| + O\x / 

              \24   2     /        \  6     / 

 

 

    References 

    ========== 

    - http://tutorial.math.lamar.edu/Classes/DE/SeriesSolutions.aspx 

    - George E. Simmons, "Differential Equations with Applications and 

      Historical Notes", p.p 176 - 184 

 

    """ 

    x = func.args[0] 

    f = func.func 

    C0, C1 = get_numbered_constants(eq, num=2) 

    n = Dummy("n", integer=True) 

    s = Wild("s") 

    k = Wild("k", exclude=[x]) 

    x0 = match.get('x0') 

    terms = match.get('terms', 5) 

    p = match[match['a3']] 

    q = match[match['b3']] 

    r = match[match['c3']] 

    seriesdict = {} 

    recurr = Function("r") 

 

    # Generating the recurrence relation which works this way 

    # a] For the second order term the summation begins at n = 2. The coefficients 

    # p is multiplied with an*(n - 1)*(n - 2)*x**n-2 and a substitution is made such that 

    # the exponent of x becomes n. 

    # For example, if p is x, then the second degree recurrence term is 

    # an*(n - 1)*(n - 2)*x**n-1, substituting (n - 1) as n, it transforms to 

    # an+1*n*(n - 1)*x**n. 

    # A similar process is done with the first order and zeroth order term. 

 

    coefflist = [(recurr(n), r), (n*recurr(n), q), (n*(n - 1)*recurr(n), p)] 

    for index, coeff in enumerate(coefflist): 

        if coeff[1]: 

            f2 = powsimp(expand((coeff[1]*(x - x0)**(n - index)).subs(x, x + x0))) 

            if f2.is_Add: 

                addargs = f2.args 

            else: 

                addargs = [f2] 

            for arg in addargs: 

                powm = arg.match(s*x**k) 

                term = coeff[0]*powm[s] 

                if not powm[k].is_Symbol: 

                    term = term.subs(n, n - powm[k].as_independent(n)[0]) 

                startind = powm[k].subs(n, index) 

                # Seeing if the startterm can be reduced further. 

                # If it vanishes for n lesser than startind, it is 

                # equal to summation from n. 

                if startind: 

                    for i in reversed(range(startind)): 

                        if not term.subs(n, i): 

                            seriesdict[term] = i 

                        else: 

                            seriesdict[term] = i + 1 

                            break 

                else: 

                    seriesdict[term] = S(0) 

 

    # Stripping of terms so that the sum starts with the same number. 

    teq = S(0) 

    suminit = seriesdict.values() 

    rkeys = seriesdict.keys() 

    req = Add(*rkeys) 

    if any(suminit): 

        maxval = max(suminit) 

        for term in seriesdict: 

            val = seriesdict[term] 

            if val != maxval: 

                for i in range(val, maxval): 

                    teq += term.subs(n, val) 

 

    finaldict = {} 

    if teq: 

        fargs = teq.atoms(AppliedUndef) 

        if len(fargs) == 1: 

            finaldict[fargs.pop()] = 0 

        else: 

            maxf = max(fargs, key = lambda x: x.args[0]) 

            sol = solve(teq, maxf) 

            if isinstance(sol, list): 

                sol = sol[0] 

            finaldict[maxf] = sol 

 

    # Finding the recurrence relation in terms of the largest term. 

    fargs = req.atoms(AppliedUndef) 

    maxf = max(fargs, key = lambda x: x.args[0]) 

    minf = min(fargs, key = lambda x: x.args[0]) 

    if minf.args[0].is_Symbol: 

        startiter = 0 

    else: 

        startiter = -minf.args[0].as_independent(n)[0] 

    lhs = maxf 

    rhs =  solve(req, maxf) 

    if isinstance(rhs, list): 

        rhs = rhs[0] 

 

    # Checking how many values are already present 

    tcounter = len([t for t in finaldict.values() if t]) 

 

    for count in range(tcounter, terms - 3):  # Assuming c0 and c1 to be arbitrary 

    #while tcounter < terms - 2:  # Assuming c0 and c1 to be arbitrary 

        check = rhs.subs(n, startiter) 

        nlhs = lhs.subs(n, startiter) 

        nrhs = check.subs(finaldict) 

        finaldict[nlhs] = nrhs 

        startiter += 1 

 

    # Post processing 

    series = C0 + C1*(x - x0) 

    for term in finaldict: 

        if finaldict[term]: 

            fact = term.args[0] 

            series += (finaldict[term].subs([(recurr(0), C0), (recurr(1), C1)])*( 

                x - x0)**fact) 

    series = collect(expand_mul(series), [C0, C1]) + Order(x**terms) 

    return Eq(f(x), series) 

 

 

 

def ode_2nd_power_series_regular(eq, func, order, match): 

    r""" 

    Gives a power series solution to a second order homogeneous differential 

    equation with polynomial coefficients at a regular point. A second order 

    homogenous differential equation is of the form 

 

    .. math :: P(x)\frac{d^2y}{dx^2} + Q(x)\frac{dy}{dx} + R(x) = 0 

 

    A point is said to regular singular at `x0` if `x - x0\frac{Q(x)}{P(x)}` 

    and `(x - x0)^{2}\frac{R(x)}{P(x)}` are analytic at `x0`. For simplicity 

    `P(x)`, `Q(x)` and `R(x)` are assumed to be polynomials. The algorithm for 

    finding the power series solutions is: 

 

    1.  Try expressing `(x - x0)P(x)` and `((x - x0)^{2})Q(x)` as power series 

        solutions about x0. Find `p0` and `q0` which are the constants of the 

        power series expansions. 

    2.  Solve the indicial equation `f(m) = m(m - 1) + m*p0 + q0`, to obtain the 

        roots `m1` and `m2` of the indicial equation. 

    3.  If `m1 - m2` is a non integer there exists two series solutions. If 

        `m1 = m2`, there exists only one solution. If `m1 - m2` is an integer, 

        then the existence of one solution is confirmed. The other solution may 

        or may not exist. 

 

    The power series solution is of the form `x^{m}\sum_{n=0}^\infty a_{n}x^{n}`. The 

    coefficients are determined by the following recurrence relation. 

    `a_{n} = -\frac{\sum_{k=0}^{n-1} q_{n-k} + (m + k)p_{n-k}}{f(m + n)}`. For the case 

    in which `m1 - m2` is an integer, it can be seen from the recurrence relation 

    that for the lower root `m`, when `n` equals the difference of both the 

    roots, the denominator becomes zero. So if the numerator is not equal to zero, 

    a second series solution exists. 

 

 

    Examples 

    ======== 

 

    >>> from sympy import dsolve, Function, pprint 

    >>> from sympy.abc import x, y 

    >>> f = Function("f") 

    >>> eq = x*(f(x).diff(x, 2)) + 2*(f(x).diff(x)) + x*f(x) 

    >>> pprint(dsolve(eq)) 

                                  /    6    4    2    \ 

                                  |   x    x    x     | 

              /  4    2    \   C1*|- --- + -- - -- + 1| 

              | x    x     |      \  720   24   2     /    / 6\ 

    f(x) = C2*|--- - -- + 1| + ------------------------ + O\x / 

              \120   6     /              x 

 

 

    References 

    ========== 

    - George E. Simmons, "Differential Equations with Applications and 

      Historical Notes", p.p 176 - 184 

 

    """ 

    x = func.args[0] 

    f = func.func 

    C0, C1 = get_numbered_constants(eq, num=2) 

    n = Dummy("n") 

    m = Dummy("m")  # for solving the indicial equation 

    s = Wild("s") 

    k = Wild("k", exclude=[x]) 

    x0 = match.get('x0') 

    terms = match.get('terms', 5) 

    p = match['p'] 

    q = match['q'] 

 

    # Generating the indicial equation 

    indicial = [] 

    for term in [p, q]: 

        if not term.has(x): 

            indicial.append(term) 

        else: 

            term = series(term, n=1, x0=x0) 

            if isinstance(term, Order): 

                indicial.append(S(0)) 

            else: 

                for arg in term.args: 

                    if not arg.has(x): 

                        indicial.append(arg) 

                        break 

 

    p0, q0 = indicial 

    sollist = solve(m*(m - 1) + m*p0 + q0, m) 

    if sollist and isinstance(sollist, list) and all( 

        [sol.is_real for sol in sollist]): 

        serdict1 = {} 

        serdict2 = {} 

        if len(sollist) == 1: 

            # Only one series solution exists in this case. 

            m1 = m2 = sollist.pop() 

            if terms-m1-1 <= 0: 

              return Eq(f(x), Order(terms)) 

            serdict1 = _frobenius(terms-m1-1, m1, p0, q0, p, q, x0, x, C0) 

 

        else: 

            m1 = sollist[0] 

            m2 = sollist[1] 

            if m1 < m2: 

                m1, m2 = m2, m1 

            # Irrespective of whether m1 - m2 is an integer or not, one 

            # Frobenius series solution exists. 

            serdict1 = _frobenius(terms-m1-1, m1, p0, q0, p, q, x0, x, C0) 

            if not (m1 - m2).is_integer: 

                # Second frobenius series solution exists. 

                serdict2 = _frobenius(terms-m2-1, m2, p0, q0, p, q, x0, x, C1) 

            else: 

                # Check if second frobenius series solution exists. 

                serdict2 = _frobenius(terms-m2-1, m2, p0, q0, p, q, x0, x, C1, check=m1) 

 

        if serdict1: 

            finalseries1 = C0 

            for key in serdict1: 

                power = int(key.name[1:]) 

                finalseries1 += serdict1[key]*(x - x0)**power 

            finalseries1 = (x - x0)**m1*finalseries1 

            finalseries2 = S(0) 

            if serdict2: 

                for key in serdict2: 

                    power = int(key.name[1:]) 

                    finalseries2 += serdict2[key]*(x - x0)**power 

                finalseries2 += C1 

                finalseries2 = (x - x0)**m2*finalseries2 

            return Eq(f(x), collect(finalseries1 + finalseries2, 

                [C0, C1]) + Order(x**terms)) 

 

def _frobenius(n, m, p0, q0, p, q, x0, x, c, check=None): 

    r""" 

    Returns a dict with keys as coefficients and values as their values in terms of C0 

    """ 

    n = int(n) 

    # In cases where m1 - m2 is not an integer 

    m2 = check 

 

    d = Dummy("d") 

    numsyms = numbered_symbols("C", start=0) 

    numsyms = [next(numsyms) for i in range(n + 1)] 

    C0 = Symbol("C0") 

    serlist = [] 

    for ser in [p, q]: 

        # Order term not present 

        if ser.is_polynomial(x) and Poly(ser, x).degree() <= n: 

            if x0: 

                ser = ser.subs(x, x + x0) 

            dict_ = Poly(ser, x).as_dict() 

        # Order term present 

        else: 

            tseries = series(ser, x=x0, n=n+1) 

            # Removing order 

            dict_ = Poly(list(ordered(tseries.args))[: -1], x).as_dict() 

        # Fill in with zeros, if coefficients are zero. 

        for i in range(n + 1): 

            if (i,) not in dict_: 

                dict_[(i,)] = S(0) 

        serlist.append(dict_) 

 

    pseries = serlist[0] 

    qseries = serlist[1] 

    indicial = d*(d - 1) + d*p0 + q0 

    frobdict = {} 

    for i in range(1, n + 1): 

        num = c*(m*pseries[(i,)] + qseries[(i,)]) 

        for j in range(1, i): 

            sym = Symbol("C" + str(j)) 

            num += frobdict[sym]*((m + j)*pseries[(i - j,)] + qseries[(i - j,)]) 

 

        # Checking for cases when m1 - m2 is an integer. If num equals zero 

        # then a second Frobenius series solution cannot be found. If num is not zero 

        # then set constant as zero and proceed. 

        if m2 is not None and i == m2 - m: 

            if num: 

                return False 

            else: 

                frobdict[numsyms[i]] = S(0) 

        else: 

            frobdict[numsyms[i]] = -num/(indicial.subs(d, m+i)) 

 

    return frobdict 

 

def _nth_linear_match(eq, func, order): 

    r""" 

    Matches a differential equation to the linear form: 

 

    .. math:: a_n(x) y^{(n)} + \cdots + a_1(x)y' + a_0(x) y + B(x) = 0 

 

    Returns a dict of order:coeff terms, where order is the order of the 

    derivative on each term, and coeff is the coefficient of that derivative. 

    The key ``-1`` holds the function `B(x)`. Returns ``None`` if the ODE is 

    not linear.  This function assumes that ``func`` has already been checked 

    to be good. 

 

    Examples 

    ======== 

 

    >>> from sympy import Function, cos, sin 

    >>> from sympy.abc import x 

    >>> from sympy.solvers.ode import _nth_linear_match 

    >>> f = Function('f') 

    >>> _nth_linear_match(f(x).diff(x, 3) + 2*f(x).diff(x) + 

    ... x*f(x).diff(x, 2) + cos(x)*f(x).diff(x) + x - f(x) - 

    ... sin(x), f(x), 3) 

    {-1: x - sin(x), 0: -1, 1: cos(x) + 2, 2: x, 3: 1} 

    >>> _nth_linear_match(f(x).diff(x, 3) + 2*f(x).diff(x) + 

    ... x*f(x).diff(x, 2) + cos(x)*f(x).diff(x) + x - f(x) - 

    ... sin(f(x)), f(x), 3) == None 

    True 

 

    """ 

    x = func.args[0] 

    one_x = set([x]) 

    terms = dict([(i, S.Zero) for i in range(-1, order + 1)]) 

    for i in Add.make_args(eq): 

        if not i.has(func): 

            terms[-1] += i 

        else: 

            c, f = i.as_independent(func) 

            if not ((isinstance(f, Derivative) and set(f.variables) == one_x) \ 

                    or f == func): 

                return None 

            else: 

                terms[len(f.args[1:])] += c 

    return terms 

 

 

def ode_nth_linear_euler_eq_homogeneous(eq, func, order, match, returns='sol'): 

    r""" 

    Solves an `n`\th order linear homogeneous variable-coefficient 

    Cauchy-Euler equidimensional ordinary differential equation. 

 

    This is an equation with form `0 = a_0 f(x) + a_1 x f'(x) + a_2 x^2 f''(x) 

    \cdots`. 

 

    These equations can be solved in a general manner, by substituting 

    solutions of the form `f(x) = x^r`, and deriving a characteristic equation 

    for `r`.  When there are repeated roots, we include extra terms of the 

    form `C_{r k} \ln^k(x) x^r`, where `C_{r k}` is an arbitrary integration 

    constant, `r` is a root of the characteristic equation, and `k` ranges 

    over the multiplicity of `r`.  In the cases where the roots are complex, 

    solutions of the form `C_1 x^a \sin(b \log(x)) + C_2 x^a \cos(b \log(x))` 

    are returned, based on expansions with Eulers formula.  The general 

    solution is the sum of the terms found.  If SymPy cannot find exact roots 

    to the characteristic equation, a 

    :py:class:`~sympy.polys.rootoftools.RootOf` instance will be returned 

    instead. 

 

    >>> from sympy import Function, dsolve, Eq 

    >>> from sympy.abc import x 

    >>> f = Function('f') 

    >>> dsolve(4*x**2*f(x).diff(x, 2) + f(x), f(x), 

    ... hint='nth_linear_euler_eq_homogeneous') 

    ... # doctest: +NORMALIZE_WHITESPACE 

    Eq(f(x), sqrt(x)*(C1 + C2*log(x))) 

 

    Note that because this method does not involve integration, there is no 

    ``nth_linear_euler_eq_homogeneous_Integral`` hint. 

 

    The following is for internal use: 

 

    - ``returns = 'sol'`` returns the solution to the ODE. 

    - ``returns = 'list'`` returns a list of linearly independent solutions, 

      corresponding to the fundamental solution set, for use with non 

      homogeneous solution methods like variation of parameters and 

      undetermined coefficients.  Note that, though the solutions should be 

      linearly independent, this function does not explicitly check that.  You 

      can do ``assert simplify(wronskian(sollist)) != 0`` to check for linear 

      independence.  Also, ``assert len(sollist) == order`` will need to pass. 

    - ``returns = 'both'``, return a dictionary ``{'sol': <solution to ODE>, 

      'list': <list of linearly independent solutions>}``. 

 

    Examples 

    ======== 

 

    >>> from sympy import Function, dsolve, pprint 

    >>> from sympy.abc import x 

    >>> f = Function('f') 

    >>> eq = f(x).diff(x, 2)*x**2 - 4*f(x).diff(x)*x + 6*f(x) 

    >>> pprint(dsolve(eq, f(x), 

    ... hint='nth_linear_euler_eq_homogeneous')) 

            2 

    f(x) = x *(C1 + C2*x) 

 

    References 

    ========== 

 

    - http://en.wikipedia.org/wiki/Cauchy%E2%80%93Euler_equation 

    - C. Bender & S. Orszag, "Advanced Mathematical Methods for Scientists and 

      Engineers", Springer 1999, pp. 12 

 

    # indirect doctest 

 

    """ 

    global collectterms 

    collectterms = [] 

 

    x = func.args[0] 

    f = func.func 

    r = match 

 

    # First, set up characteristic equation. 

    chareq, symbol = S.Zero, Dummy('x') 

 

    for i in r.keys(): 

        if not isinstance(i, str) and i >= 0: 

            chareq += (r[i]*diff(x**symbol, x, i)*x**-symbol).expand() 

 

    chareq = Poly(chareq, symbol) 

    chareqroots = [RootOf(chareq, k) for k in range(chareq.degree())] 

 

    # A generator of constants 

    constants = list(get_numbered_constants(eq, num=chareq.degree()*2)) 

    constants.reverse() 

 

    # Create a dict root: multiplicity or charroots 

    charroots = defaultdict(int) 

    for root in chareqroots: 

        charroots[root] += 1 

    gsol = S(0) 

    # We need keep track of terms so we can run collect() at the end. 

    # This is necessary for constantsimp to work properly. 

    ln = log 

    for root, multiplicity in charroots.items(): 

        for i in range(multiplicity): 

            if isinstance(root, RootOf): 

                gsol += (x**root) * constants.pop() 

                if multiplicity != 1: 

                    raise ValueError("Value should be 1") 

                collectterms = [(0, root, 0)] + collectterms 

            elif root.is_real: 

                gsol += ln(x)**i*(x**root) * constants.pop() 

                collectterms = [(i, root, 0)] + collectterms 

            else: 

                reroot = re(root) 

                imroot = im(root) 

                gsol += ln(x)**i * (x**reroot) * ( 

                    constants.pop() * sin(abs(imroot)*ln(x)) 

                    + constants.pop() * cos(imroot*ln(x))) 

                # Preserve ordering (multiplicity, real part, imaginary part) 

                # It will be assumed implicitly when constructing 

                # fundamental solution sets. 

                collectterms = [(i, reroot, imroot)] + collectterms 

    if returns == 'sol': 

        return Eq(f(x), gsol) 

    elif returns in ('list' 'both'): 

        # HOW TO TEST THIS CODE? (dsolve does not pass 'returns' through) 

        # Create a list of (hopefully) linearly independent solutions 

        gensols = [] 

        # Keep track of when to use sin or cos for nonzero imroot 

        for i, reroot, imroot in collectterms: 

            if imroot == 0: 

                gensols.append(ln(x)**i*x**reroot) 

            else: 

                sin_form = ln(x)**i*x**reroot*sin(abs(imroot)*ln(x)) 

                if sin_form in gensols: 

                    cos_form = ln(x)**i*x**reroot*cos(imroot*ln(x)) 

                    gensols.append(cos_form) 

                else: 

                    gensols.append(sin_form) 

        if returns == 'list': 

            return gensols 

        else: 

            return {'sol': Eq(f(x), gsol), 'list': gensols} 

    else: 

        raise ValueError('Unknown value for key "returns".') 

 

 

def ode_nth_linear_euler_eq_nonhomogeneous_undetermined_coefficients(eq, func, order, match, returns='sol'): 

    r""" 

    Solves an `n`\th order linear non homogeneous Cauchy-Euler equidimensional 

    ordinary differential equation using undetermined coefficients. 

 

    This is an equation with form `g(x) = a_0 f(x) + a_1 x f'(x) + a_2 x^2 f''(x) 

    \cdots`. 

 

    These equations can be solved in a general manner, by substituting 

    solutions of the form `x = exp(t)`, and deriving a characteristic equation 

    of form `g(exp(t)) = b_0 f(t) + b_1 f'(t) + b_2 f''(t) \cdots` which can 

    be then solved by nth_linear_constant_coeff_undetermined_coefficients if 

    g(exp(t)) has finite number of lineary independent derivatives. 

 

    Functions that fit this requirement are finite sums functions of the form 

    `a x^i e^{b x} \sin(c x + d)` or `a x^i e^{b x} \cos(c x + d)`, where `i` 

    is a non-negative integer and `a`, `b`, `c`, and `d` are constants.  For 

    example any polynomial in `x`, functions like `x^2 e^{2 x}`, `x \sin(x)`, 

    and `e^x \cos(x)` can all be used.  Products of `\sin`'s and `\cos`'s have 

    a finite number of derivatives, because they can be expanded into `\sin(a 

    x)` and `\cos(b x)` terms.  However, SymPy currently cannot do that 

    expansion, so you will need to manually rewrite the expression in terms of 

    the above to use this method.  So, for example, you will need to manually 

    convert `\sin^2(x)` into `(1 + \cos(2 x))/2` to properly apply the method 

    of undetermined coefficients on it. 

 

    After replacement of x by exp(t), this method works by creating a trial function 

    from the expression and all of its linear independent derivatives and 

    substituting them into the original ODE.  The coefficients for each term 

    will be a system of linear equations, which are be solved for and 

    substituted, giving the solution. If any of the trial functions are linearly 

    dependent on the solution to the homogeneous equation, they are multiplied 

    by sufficient `x` to make them linearly independent. 

 

    Examples 

    ======== 

 

    >>> from sympy import dsolve, Function, Derivative, log 

    >>> from sympy.abc import x 

    >>> f = Function('f') 

    >>> eq = x**2*Derivative(f(x), x, x) - 2*x*Derivative(f(x), x) + 2*f(x) - log(x) 

    >>> dsolve(eq, f(x), 

    ... hint='nth_linear_euler_eq_nonhomogeneous_undetermined_coefficients').expand() 

    Eq(f(x), C1*x + C2*x**2 + log(x)/2 + 3/4) 

 

    """ 

    x = func.args[0] 

    f = func.func 

    r = match 

 

    chareq, eq, symbol = S.Zero, S.Zero, Dummy('x') 

 

    for i in r.keys(): 

        if not isinstance(i, str) and i >= 0: 

            chareq += (r[i]*diff(x**symbol, x, i)*x**-symbol).expand() 

 

    for i in range(1,degree(Poly(chareq, symbol))+1): 

        eq += chareq.coeff(symbol**i)*diff(f(x), x, i) 

 

    if chareq.as_coeff_add(symbol)[0]: 

        eq += chareq.as_coeff_add(symbol)[0]*f(x) 

    e, re = posify(r[-1].subs(x, exp(x))) 

    eq += e.subs(re) 

 

    match = _nth_linear_match(eq, f(x), ode_order(eq, f(x))) 

    match['trialset'] = r['trialset'] 

    return ode_nth_linear_constant_coeff_undetermined_coefficients(eq, func, order, match).subs(x, log(x)).subs(f(log(x)), f(x)).expand() 

 

 

def ode_nth_linear_euler_eq_nonhomogeneous_variation_of_parameters(eq, func, order, match, returns='sol'): 

    r""" 

    Solves an `n`\th order linear non homogeneous Cauchy-Euler equidimensional 

    ordinary differential equation using variation of parameters. 

 

    This is an equation with form `g(x) = a_0 f(x) + a_1 x f'(x) + a_2 x^2 f''(x) 

    \cdots`. 

 

    This method works by assuming that the particular solution takes the form 

 

    .. math:: \sum_{x=1}^{n} c_i(x) y_i(x) {a_n} {x^n} \text{,} 

 

    where `y_i` is the `i`\th solution to the homogeneous equation.  The 

    solution is then solved using Wronskian's and Cramer's Rule.  The 

    particular solution is given by multiplying eq given below with `a_n x^{n}` 

 

    .. math:: \sum_{x=1}^n \left( \int \frac{W_i(x)}{W(x)} \,dx 

                \right) y_i(x) \text{,} 

 

    where `W(x)` is the Wronskian of the fundamental system (the system of `n` 

    linearly independent solutions to the homogeneous equation), and `W_i(x)` 

    is the Wronskian of the fundamental system with the `i`\th column replaced 

    with `[0, 0, \cdots, 0, \frac{x^{- n}}{a_n} g{\left (x \right )}]`. 

 

    This method is general enough to solve any `n`\th order inhomogeneous 

    linear differential equation, but sometimes SymPy cannot simplify the 

    Wronskian well enough to integrate it.  If this method hangs, try using the 

    ``nth_linear_constant_coeff_variation_of_parameters_Integral`` hint and 

    simplifying the integrals manually.  Also, prefer using 

    ``nth_linear_constant_coeff_undetermined_coefficients`` when it 

    applies, because it doesn't use integration, making it faster and more 

    reliable. 

 

    Warning, using simplify=False with 

    'nth_linear_constant_coeff_variation_of_parameters' in 

    :py:meth:`~sympy.solvers.ode.dsolve` may cause it to hang, because it will 

    not attempt to simplify the Wronskian before integrating.  It is 

    recommended that you only use simplify=False with 

    'nth_linear_constant_coeff_variation_of_parameters_Integral' for this 

    method, especially if the solution to the homogeneous equation has 

    trigonometric functions in it. 

 

    Examples 

    ======== 

 

    >>> from sympy import Function, dsolve, Derivative 

    >>> from sympy.abc import x 

    >>> f = Function('f') 

    >>> eq = x**2*Derivative(f(x), x, x) - 2*x*Derivative(f(x), x) + 2*f(x) - x**4 

    >>> dsolve(eq, f(x), 

    ... hint='nth_linear_euler_eq_nonhomogeneous_variation_of_parameters').expand() 

    Eq(f(x), C1*x + C2*x**2 + x**4/6) 

 

    """ 

    x = func.args[0] 

    f = func.func 

    r = match 

 

    gensol = ode_nth_linear_euler_eq_homogeneous(eq, func, order, match, returns='both') 

    match.update(gensol) 

    r[-1] = r[-1]/r[ode_order(eq, f(x))] 

    sol = _solve_variation_of_parameters(eq, func, order, match) 

    return Eq(f(x), r['sol'].rhs + (sol.rhs - r['sol'].rhs)*r[ode_order(eq, f(x))]) 

 

 

def ode_almost_linear(eq, func, order, match): 

    r""" 

    Solves an almost-linear differential equation. 

 

    The general form of an almost linear differential equation is 

 

    .. math:: f(x) g(y) y + k(x) l(y) + m(x) = 0 

                \text{where} l'(y) = g(y)\text{.} 

 

    This can be solved by substituting `l(y) = u(y)`.  Making the given 

    substitution reduces it to a linear differential equation of the form `u' 

    + P(x) u + Q(x) = 0`. 

 

    The general solution is 

 

        >>> from sympy import Function, dsolve, Eq, pprint 

        >>> from sympy.abc import x, y, n 

        >>> f, g, k, l = map(Function, ['f', 'g', 'k', 'l']) 

        >>> genform = Eq(f(x)*(l(y).diff(y)) + k(x)*l(y) + g(x)) 

        >>> pprint(genform) 

             d 

        f(x)*--(l(y)) + g(x) + k(x)*l(y) = 0 

             dy 

        >>> pprint(dsolve(genform, hint = 'almost_linear')) 

               /     //   -y*g(x)                  \\ 

               |     ||   --------     for k(x) = 0|| 

               |     ||     f(x)                   ||  -y*k(x) 

               |     ||                            ||  -------- 

               |     ||       y*k(x)               ||    f(x) 

        l(y) = |C1 + |<       ------               ||*e 

               |     ||        f(x)                || 

               |     ||-g(x)*e                     || 

               |     ||--------------   otherwise  || 

               |     ||     k(x)                   || 

               \     \\                            // 

 

 

    See Also 

    ======== 

    :meth:`sympy.solvers.ode.ode_1st_linear` 

 

    Examples 

    ======== 

 

    >>> from sympy import Function, Derivative, pprint 

    >>> from sympy.solvers.ode import dsolve, classify_ode 

    >>> from sympy.abc import x 

    >>> f = Function('f') 

    >>> d = f(x).diff(x) 

    >>> eq = x*d + x*f(x) + 1 

    >>> dsolve(eq, f(x), hint='almost_linear') 

    Eq(f(x), (C1 - Ei(x))*exp(-x)) 

    >>> pprint(dsolve(eq, f(x), hint='almost_linear')) 

                         -x 

    f(x) = (C1 - Ei(x))*e 

 

    References 

    ========== 

 

    - Joel Moses, "Symbolic Integration - The Stormy Decade", Communications 

      of the ACM, Volume 14, Number 8, August 1971, pp. 558 

    """ 

 

    # Since ode_1st_linear has already been implemented, and the 

    # coefficients have been modified to the required form in 

    # classify_ode, just passing eq, func, order and match to 

    # ode_1st_linear will give the required output. 

    return ode_1st_linear(eq, func, order, match) 

 

def _linear_coeff_match(expr, func): 

    r""" 

    Helper function to match hint ``linear_coefficients``. 

 

    Matches the expression to the form `(a_1 x + b_1 f(x) + c_1)/(a_2 x + b_2 

    f(x) + c_2)` where the following conditions hold: 

 

    1. `a_1`, `b_1`, `c_1`, `a_2`, `b_2`, `c_2` are Rationals; 

    2. `c_1` or `c_2` are not equal to zero; 

    3. `a_2 b_1 - a_1 b_2` is not equal to zero. 

 

    Return ``xarg``, ``yarg`` where 

 

    1. ``xarg`` = `(b_2 c_1 - b_1 c_2)/(a_2 b_1 - a_1 b_2)` 

    2. ``yarg`` = `(a_1 c_2 - a_2 c_1)/(a_2 b_1 - a_1 b_2)` 

 

 

    Examples 

    ======== 

 

    >>> from sympy import Function 

    >>> from sympy.abc import x 

    >>> from sympy.solvers.ode import _linear_coeff_match 

    >>> from sympy.functions.elementary.trigonometric import sin 

    >>> f = Function('f') 

    >>> _linear_coeff_match(( 

    ... (-25*f(x) - 8*x + 62)/(4*f(x) + 11*x - 11)), f(x)) 

    (1/9, 22/9) 

    >>> _linear_coeff_match( 

    ... sin((-5*f(x) - 8*x + 6)/(4*f(x) + x - 1)), f(x)) 

    (19/27, 2/27) 

    >>> _linear_coeff_match(sin(f(x)/x), f(x)) 

 

    """ 

    f = func.func 

    x = func.args[0] 

    def abc(eq): 

        r''' 

        Internal function of _linear_coeff_match 

        that returns Rationals a, b, c 

        if eq is a*x + b*f(x) + c, else None. 

        ''' 

        eq = _mexpand(eq) 

        c = eq.as_independent(x, f(x), as_Add = True)[0] 

        if not c.is_Rational: 

            return 

        a = eq.coeff(x) 

        if not a.is_Rational: 

            return 

        b = eq.coeff(f(x)) 

        if not b.is_Rational: 

            return 

        if eq == a*x + b*f(x) + c: 

            return a, b, c 

 

    def match(arg): 

        r''' 

        Internal function of _linear_coeff_match that returns Rationals a1, 

        b1, c1, a2, b2, c2 and a2*b1 - a1*b2 of the expression (a1*x + b1*f(x) 

        + c1)/(a2*x + b2*f(x) + c2) if one of c1 or c2 and a2*b1 - a1*b2 is 

        non-zero, else None. 

        ''' 

        n, d = arg.together().as_numer_denom() 

        m = abc(n) 

        if m is not None: 

            a1, b1, c1 = m 

            m = abc(d) 

            if m is not None: 

                a2, b2, c2 = m 

                d = a2*b1 - a1*b2 

                if (c1 or c2) and d: 

                    return a1, b1, c1, a2, b2, c2, d 

 

    m = [fi.args[0] for fi in expr.atoms(Function) if fi.func != f and 

         len(fi.args) == 1 and not fi.args[0].is_Function] or set([expr]) 

    m1 = match(m.pop()) 

    if m1 and all(match(mi) == m1 for mi in m): 

        a1, b1, c1, a2, b2, c2, denom = m1 

        return (b2*c1 - b1*c2)/denom, (a1*c2 - a2*c1)/denom 

 

def ode_linear_coefficients(eq, func, order, match): 

    r""" 

    Solves a differential equation with linear coefficients. 

 

    The general form of a differential equation with linear coefficients is 

 

    .. math:: y' + F\left(\!\frac{a_1 x + b_1 y + c_1}{a_2 x + b_2 y + 

                c_2}\!\right) = 0\text{,} 

 

    where `a_1`, `b_1`, `c_1`, `a_2`, `b_2`, `c_2` are constants and `a_1 b_2 

    - a_2 b_1 \ne 0`. 

 

    This can be solved by substituting: 

 

    .. math:: x = x' + \frac{b_2 c_1 - b_1 c_2}{a_2 b_1 - a_1 b_2} 

 

              y = y' + \frac{a_1 c_2 - a_2 c_1}{a_2 b_1 - a_1 

                  b_2}\text{.} 

 

    This substitution reduces the equation to a homogeneous differential 

    equation. 

 

    See Also 

    ======== 

    :meth:`sympy.solvers.ode.ode_1st_homogeneous_coeff_best` 

    :meth:`sympy.solvers.ode.ode_1st_homogeneous_coeff_subs_indep_div_dep` 

    :meth:`sympy.solvers.ode.ode_1st_homogeneous_coeff_subs_dep_div_indep` 

 

    Examples 

    ======== 

 

    >>> from sympy import Function, Derivative, pprint 

    >>> from sympy.solvers.ode import dsolve, classify_ode 

    >>> from sympy.abc import x 

    >>> f = Function('f') 

    >>> df = f(x).diff(x) 

    >>> eq = (x + f(x) + 1)*df + (f(x) - 6*x + 1) 

    >>> dsolve(eq, hint='linear_coefficients') 

    [Eq(f(x), -x - sqrt(C1 + 7*x**2) - 1), Eq(f(x), -x + sqrt(C1 + 7*x**2) - 1)] 

    >>> pprint(dsolve(eq, hint='linear_coefficients')) 

                      ___________                     ___________ 

                   /         2                     /         2 

    [f(x) = -x - \/  C1 + 7*x   - 1, f(x) = -x + \/  C1 + 7*x   - 1] 

 

 

    References 

    ========== 

 

    - Joel Moses, "Symbolic Integration - The Stormy Decade", Communications 

      of the ACM, Volume 14, Number 8, August 1971, pp. 558 

    """ 

 

    return ode_1st_homogeneous_coeff_best(eq, func, order, match) 

 

 

def ode_separable_reduced(eq, func, order, match): 

    r""" 

    Solves a differential equation that can be reduced to the separable form. 

 

    The general form of this equation is 

 

    .. math:: y' + (y/x) H(x^n y) = 0\text{}. 

 

    This can be solved by substituting `u(y) = x^n y`.  The equation then 

    reduces to the separable form `\frac{u'}{u (\mathrm{power} - H(u))} - 

    \frac{1}{x} = 0`. 

 

    The general solution is: 

 

        >>> from sympy import Function, dsolve, Eq, pprint 

        >>> from sympy.abc import x, n 

        >>> f, g = map(Function, ['f', 'g']) 

        >>> genform = f(x).diff(x) + (f(x)/x)*g(x**n*f(x)) 

        >>> pprint(genform) 

                         / n     \ 

        d          f(x)*g\x *f(x)/ 

        --(f(x)) + --------------- 

        dx                x 

        >>> pprint(dsolve(genform, hint='separable_reduced')) 

         n 

        x *f(x) 

          / 

         | 

         |         1 

         |    ------------ dy = C1 + log(x) 

         |    y*(n - g(y)) 

         | 

         / 

 

    See Also 

    ======== 

    :meth:`sympy.solvers.ode.ode_separable` 

 

    Examples 

    ======== 

 

    >>> from sympy import Function, Derivative, pprint 

    >>> from sympy.solvers.ode import dsolve, classify_ode 

    >>> from sympy.abc import x 

    >>> f = Function('f') 

    >>> d = f(x).diff(x) 

    >>> eq = (x - x**2*f(x))*d - f(x) 

    >>> dsolve(eq, hint='separable_reduced') 

    [Eq(f(x), (-sqrt(C1*x**2 + 1) + 1)/x), Eq(f(x), (sqrt(C1*x**2 + 1) + 1)/x)] 

    >>> pprint(dsolve(eq, hint='separable_reduced')) 

                 ___________                ___________ 

                /     2                    /     2 

            - \/  C1*x  + 1  + 1         \/  C1*x  + 1  + 1 

    [f(x) = --------------------, f(x) = ------------------] 

                     x                           x 

 

    References 

    ========== 

 

    - Joel Moses, "Symbolic Integration - The Stormy Decade", Communications 

      of the ACM, Volume 14, Number 8, August 1971, pp. 558 

    """ 

 

    # Arguments are passed in a way so that they are coherent with the 

    # ode_separable function 

    x = func.args[0] 

    f = func.func 

    y = Dummy('y') 

    u = match['u'].subs(match['t'], y) 

    ycoeff = 1/(y*(match['power'] - u)) 

    m1 = {y: 1, x: -1/x, 'coeff': 1} 

    m2 = {y: ycoeff, x: 1, 'coeff': 1} 

    r = {'m1': m1, 'm2': m2, 'y': y, 'hint': x**match['power']*f(x)} 

    return ode_separable(eq, func, order, r) 

 

 

def ode_1st_power_series(eq, func, order, match): 

    r""" 

    The power series solution is a method which gives the Taylor series expansion 

    to the solution of a differential equation. 

 

    For a first order differential equation `\frac{dy}{dx} = h(x, y)`, a power 

    series solution exists at a point `x = x_{0}` if `h(x, y)` is analytic at `x_{0}`. 

    The solution is given by 

 

    .. math:: y(x) = y(x_{0}) + \sum_{n = 1}^{\infty} \frac{F_{n}(x_{0},b)(x - x_{0})^n}{n!}, 

 

    where `y(x_{0}) = b` is the value of y at the initial value of `x_{0}`. 

    To compute the values of the `F_{n}(x_{0},b)` the following algorithm is 

    followed, until the required number of terms are generated. 

 

    1. `F_1 = h(x_{0}, b)` 

    2. `F_{n+1} = \frac{\partial F_{n}}{\partial x} + \frac{\partial F_{n}}{\partial y}F_{1}` 

 

    Examples 

    ======== 

 

    >>> from sympy import Function, Derivative, pprint, exp 

    >>> from sympy.solvers.ode import dsolve 

    >>> from sympy.abc import x 

    >>> f = Function('f') 

    >>> eq = exp(x)*(f(x).diff(x)) - f(x) 

    >>> pprint(dsolve(eq, hint='1st_power_series')) 

                           3       4       5 

                       C1*x    C1*x    C1*x     / 6\ 

    f(x) = C1 + C1*x - ----- + ----- + ----- + O\x / 

                         6       24      60 

 

 

    References 

    ========== 

 

    - Travis W. Walker, Analytic power series technique for solving first-order 

      differential equations, p.p 17, 18 

 

    """ 

    x = func.args[0] 

    y = match['y'] 

    f = func.func 

    h = -match[match['d']]/match[match['e']] 

    point = match.get('f0') 

    value = match.get('f0val') 

    terms = match.get('terms') 

 

    # First term 

    F = h 

    if not h: 

        return Eq(f(x), value) 

 

    # Initialisation 

    series = value 

    if terms > 1: 

        hc = h.subs({x: point, y: value}) 

        if hc.has(oo) or hc.has(NaN) or hc.has(zoo): 

            # Derivative does not exist, not analytic 

            return Eq(f(x), oo) 

        elif hc: 

            series += hc*(x - point) 

 

    for factcount in range(2, terms): 

        Fnew = F.diff(x) + F.diff(y)*h 

        Fnewc = Fnew.subs({x: point, y: value}) 

        # Same logic as above 

        if Fnewc.has(oo) or Fnewc.has(NaN) or Fnewc.has(-oo) or Fnewc.has(zoo): 

            return Eq(f(x), oo) 

        series += Fnewc*((x - point)**factcount)/factorial(factcount) 

        F = Fnew 

    series += Order(x**terms) 

    return Eq(f(x), series) 

 

 

def ode_nth_linear_constant_coeff_homogeneous(eq, func, order, match, 

        returns='sol'): 

    r""" 

    Solves an `n`\th order linear homogeneous differential equation with 

    constant coefficients. 

 

    This is an equation of the form 

 

    .. math:: a_n f^{(n)}(x) + a_{n-1} f^{(n-1)}(x) + \cdots + a_1 f'(x) 

                + a_0 f(x) = 0\text{.} 

 

    These equations can be solved in a general manner, by taking the roots of 

    the characteristic equation `a_n m^n + a_{n-1} m^{n-1} + \cdots + a_1 m + 

    a_0 = 0`.  The solution will then be the sum of `C_n x^i e^{r x}` terms, 

    for each where `C_n` is an arbitrary constant, `r` is a root of the 

    characteristic equation and `i` is one of each from 0 to the multiplicity 

    of the root - 1 (for example, a root 3 of multiplicity 2 would create the 

    terms `C_1 e^{3 x} + C_2 x e^{3 x}`).  The exponential is usually expanded 

    for complex roots using Euler's equation `e^{I x} = \cos(x) + I \sin(x)`. 

    Complex roots always come in conjugate pairs in polynomials with real 

    coefficients, so the two roots will be represented (after simplifying the 

    constants) as `e^{a x} \left(C_1 \cos(b x) + C_2 \sin(b x)\right)`. 

 

    If SymPy cannot find exact roots to the characteristic equation, a 

    :py:class:`~sympy.polys.rootoftools.RootOf` instance will be return 

    instead. 

 

    >>> from sympy import Function, dsolve, Eq 

    >>> from sympy.abc import x 

    >>> f = Function('f') 

    >>> dsolve(f(x).diff(x, 5) + 10*f(x).diff(x) - 2*f(x), f(x), 

    ... hint='nth_linear_constant_coeff_homogeneous') 

    ... # doctest: +NORMALIZE_WHITESPACE 

    Eq(f(x), C1*exp(x*RootOf(_x**5 + 10*_x - 2, 0)) + 

    C2*exp(x*RootOf(_x**5 + 10*_x - 2, 1)) + 

    C3*exp(x*RootOf(_x**5 + 10*_x - 2, 2)) + 

    C4*exp(x*RootOf(_x**5 + 10*_x - 2, 3)) + 

    C5*exp(x*RootOf(_x**5 + 10*_x - 2, 4))) 

 

    Note that because this method does not involve integration, there is no 

    ``nth_linear_constant_coeff_homogeneous_Integral`` hint. 

 

    The following is for internal use: 

 

    - ``returns = 'sol'`` returns the solution to the ODE. 

    - ``returns = 'list'`` returns a list of linearly independent solutions, 

      for use with non homogeneous solution methods like variation of 

      parameters and undetermined coefficients.  Note that, though the 

      solutions should be linearly independent, this function does not 

      explicitly check that.  You can do ``assert simplify(wronskian(sollist)) 

      != 0`` to check for linear independence.  Also, ``assert len(sollist) == 

      order`` will need to pass. 

    - ``returns = 'both'``, return a dictionary ``{'sol': <solution to ODE>, 

      'list': <list of linearly independent solutions>}``. 

 

    Examples 

    ======== 

 

    >>> from sympy import Function, dsolve, pprint 

    >>> from sympy.abc import x 

    >>> f = Function('f') 

    >>> pprint(dsolve(f(x).diff(x, 4) + 2*f(x).diff(x, 3) - 

    ... 2*f(x).diff(x, 2) - 6*f(x).diff(x) + 5*f(x), f(x), 

    ... hint='nth_linear_constant_coeff_homogeneous')) 

                        x                            -2*x 

    f(x) = (C1 + C2*x)*e  + (C3*sin(x) + C4*cos(x))*e 

 

    References 

    ========== 

 

    - http://en.wikipedia.org/wiki/Linear_differential_equation section: 

      Nonhomogeneous_equation_with_constant_coefficients 

    - M. Tenenbaum & H. Pollard, "Ordinary Differential Equations", 

      Dover 1963, pp. 211 

 

    # indirect doctest 

 

    """ 

    x = func.args[0] 

    f = func.func 

    r = match 

 

    # First, set up characteristic equation. 

    chareq, symbol = S.Zero, Dummy('x') 

 

    for i in r.keys(): 

        if type(i) == str or i < 0: 

            pass 

        else: 

            chareq += r[i]*symbol**i 

 

    chareq = Poly(chareq, symbol) 

    chareqroots = [RootOf(chareq, k) for k in range(chareq.degree())] 

    chareq_is_complex = not all([i.is_real for i in chareq.all_coeffs()]) 

 

    # A generator of constants 

    constants = list(get_numbered_constants(eq, num=chareq.degree()*2)) 

 

    # Create a dict root: multiplicity or charroots 

    charroots = defaultdict(int) 

    for root in chareqroots: 

        charroots[root] += 1 

    gsol = S(0) 

    # We need to keep track of terms so we can run collect() at the end. 

    # This is necessary for constantsimp to work properly. 

    global collectterms 

    collectterms = [] 

    gensols = [] 

    conjugate_roots = [] # used to prevent double-use of conjugate roots 

    for root, multiplicity in charroots.items(): 

        for i in range(multiplicity): 

            if isinstance(root, RootOf): 

                gensols.append(exp(root*x)) 

                if multiplicity != 1: 

                    raise ValueError("Value should be 1") 

                # This ordering is important 

                collectterms = [(0, root, 0)] + collectterms 

            else: 

                if chareq_is_complex: 

                    gensols.append(x**i*exp(root*x)) 

                    collectterms = [(i, root, 0)] + collectterms 

                    continue 

                reroot = re(root) 

                imroot = im(root) 

                if imroot.has(atan2) and reroot.has(atan2): 

                    # Remove this condition when re and im stop returning 

                    # circular atan2 usages. 

                    gensols.append(x**i*exp(root*x)) 

                    collectterms = [(i, root, 0)] + collectterms 

                else: 

                    if root in conjugate_roots: 

                        collectterms = [(i, reroot, imroot)] + collectterms 

                        continue 

                    if imroot == 0: 

                        gensols.append(x**i*exp(reroot*x)) 

                        collectterms = [(i, reroot, 0)] + collectterms 

                        continue 

                    conjugate_roots.append(conjugate(root)) 

                    gensols.append(x**i*exp(reroot*x) * sin(abs(imroot) * x)) 

                    gensols.append(x**i*exp(reroot*x) * cos(    imroot  * x)) 

 

                    # This ordering is important 

                    collectterms = [(i, reroot, imroot)] + collectterms 

    if returns == 'list': 

        return gensols 

    elif returns in ('sol' 'both'): 

        gsol = Add(*[i*j for (i,j) in zip(constants, gensols)]) 

        if returns == 'sol': 

            return Eq(f(x), gsol) 

        else: 

            return {'sol': Eq(f(x), gsol), 'list': gensols} 

    else: 

        raise ValueError('Unknown value for key "returns".') 

 

 

def ode_nth_linear_constant_coeff_undetermined_coefficients(eq, func, order, match): 

    r""" 

    Solves an `n`\th order linear differential equation with constant 

    coefficients using the method of undetermined coefficients. 

 

    This method works on differential equations of the form 

 

    .. math:: a_n f^{(n)}(x) + a_{n-1} f^{(n-1)}(x) + \cdots + a_1 f'(x) 

                + a_0 f(x) = P(x)\text{,} 

 

    where `P(x)` is a function that has a finite number of linearly 

    independent derivatives. 

 

    Functions that fit this requirement are finite sums functions of the form 

    `a x^i e^{b x} \sin(c x + d)` or `a x^i e^{b x} \cos(c x + d)`, where `i` 

    is a non-negative integer and `a`, `b`, `c`, and `d` are constants.  For 

    example any polynomial in `x`, functions like `x^2 e^{2 x}`, `x \sin(x)`, 

    and `e^x \cos(x)` can all be used.  Products of `\sin`'s and `\cos`'s have 

    a finite number of derivatives, because they can be expanded into `\sin(a 

    x)` and `\cos(b x)` terms.  However, SymPy currently cannot do that 

    expansion, so you will need to manually rewrite the expression in terms of 

    the above to use this method.  So, for example, you will need to manually 

    convert `\sin^2(x)` into `(1 + \cos(2 x))/2` to properly apply the method 

    of undetermined coefficients on it. 

 

    This method works by creating a trial function from the expression and all 

    of its linear independent derivatives and substituting them into the 

    original ODE.  The coefficients for each term will be a system of linear 

    equations, which are be solved for and substituted, giving the solution. 

    If any of the trial functions are linearly dependent on the solution to 

    the homogeneous equation, they are multiplied by sufficient `x` to make 

    them linearly independent. 

 

    Examples 

    ======== 

 

    >>> from sympy import Function, dsolve, pprint, exp, cos 

    >>> from sympy.abc import x 

    >>> f = Function('f') 

    >>> pprint(dsolve(f(x).diff(x, 2) + 2*f(x).diff(x) + f(x) - 

    ... 4*exp(-x)*x**2 + cos(2*x), f(x), 

    ... hint='nth_linear_constant_coeff_undetermined_coefficients')) 

           /             4\ 

           |            x |  -x   4*sin(2*x)   3*cos(2*x) 

    f(x) = |C1 + C2*x + --|*e   - ---------- + ---------- 

           \            3 /           25           25 

 

    References 

    ========== 

 

    - http://en.wikipedia.org/wiki/Method_of_undetermined_coefficients 

    - M. Tenenbaum & H. Pollard, "Ordinary Differential Equations", 

      Dover 1963, pp. 221 

 

    # indirect doctest 

 

    """ 

    gensol = ode_nth_linear_constant_coeff_homogeneous(eq, func, order, match, 

        returns='both') 

    match.update(gensol) 

    return _solve_undetermined_coefficients(eq, func, order, match) 

 

 

def _solve_undetermined_coefficients(eq, func, order, match): 

    r""" 

    Helper function for the method of undetermined coefficients. 

 

    See the 

    :py:meth:`~sympy.solvers.ode.ode_nth_linear_constant_coeff_undetermined_coefficients` 

    docstring for more information on this method. 

 

    The parameter ``match`` should be a dictionary that has the following 

    keys: 

 

    ``list`` 

      A list of solutions to the homogeneous equation, such as the list 

      returned by 

      ``ode_nth_linear_constant_coeff_homogeneous(returns='list')``. 

 

    ``sol`` 

      The general solution, such as the solution returned by 

      ``ode_nth_linear_constant_coeff_homogeneous(returns='sol')``. 

 

    ``trialset`` 

      The set of trial functions as returned by 

      ``_undetermined_coefficients_match()['trialset']``. 

 

    """ 

    x = func.args[0] 

    f = func.func 

    r = match 

    coeffs = numbered_symbols('a', cls=Dummy) 

    coefflist = [] 

    gensols = r['list'] 

    gsol = r['sol'] 

    trialset = r['trialset'] 

    notneedset = set([]) 

    newtrialset = set([]) 

    global collectterms 

    if len(gensols) != order: 

        raise NotImplementedError("Cannot find " + str(order) + 

        " solutions to the homogeneous equation necessary to apply" + 

        " undetermined coefficients to " + str(eq) + 

        " (number of terms != order)") 

    usedsin = set([]) 

    mult = 0  # The multiplicity of the root 

    getmult = True 

    for i, reroot, imroot in collectterms: 

        if getmult: 

            mult = i + 1 

            getmult = False 

        if i == 0: 

            getmult = True 

        if imroot: 

            # Alternate between sin and cos 

            if (i, reroot) in usedsin: 

                check = x**i*exp(reroot*x)*cos(imroot*x) 

            else: 

                check = x**i*exp(reroot*x)*sin(abs(imroot)*x) 

                usedsin.add((i, reroot)) 

        else: 

            check = x**i*exp(reroot*x) 

 

        if check in trialset: 

            # If an element of the trial function is already part of the 

            # homogeneous solution, we need to multiply by sufficient x to 

            # make it linearly independent.  We also don't need to bother 

            # checking for the coefficients on those elements, since we 

            # already know it will be 0. 

            while True: 

                if check*x**mult in trialset: 

                    mult += 1 

                else: 

                    break 

            trialset.add(check*x**mult) 

            notneedset.add(check) 

 

    newtrialset = trialset - notneedset 

 

    trialfunc = 0 

    for i in newtrialset: 

        c = next(coeffs) 

        coefflist.append(c) 

        trialfunc += c*i 

 

    eqs = sub_func_doit(eq, f(x), trialfunc) 

 

    coeffsdict = dict(list(zip(trialset, [0]*(len(trialset) + 1)))) 

 

    eqs = expand_mul(eqs) 

 

    for i in Add.make_args(eqs): 

        s = separatevars(i, dict=True, symbols=[x]) 

        coeffsdict[s[x]] += s['coeff'] 

 

    coeffvals = solve(list(coeffsdict.values()), coefflist) 

 

    if not coeffvals: 

        raise NotImplementedError( 

            "Could not solve `%s` using the " 

            "method of undetermined coefficients " 

            "(unable to solve for coefficients)." % eq) 

 

    psol = trialfunc.subs(coeffvals) 

 

    return Eq(f(x), gsol.rhs + psol) 

 

 

def _undetermined_coefficients_match(expr, x): 

    r""" 

    Returns a trial function match if undetermined coefficients can be applied 

    to ``expr``, and ``None`` otherwise. 

 

    A trial expression can be found for an expression for use with the method 

    of undetermined coefficients if the expression is an 

    additive/multiplicative combination of constants, polynomials in `x` (the 

    independent variable of expr), `\sin(a x + b)`, `\cos(a x + b)`, and 

    `e^{a x}` terms (in other words, it has a finite number of linearly 

    independent derivatives). 

 

    Note that you may still need to multiply each term returned here by 

    sufficient `x` to make it linearly independent with the solutions to the 

    homogeneous equation. 

 

    This is intended for internal use by ``undetermined_coefficients`` hints. 

 

    SymPy currently has no way to convert `\sin^n(x) \cos^m(y)` into a sum of 

    only `\sin(a x)` and `\cos(b x)` terms, so these are not implemented.  So, 

    for example, you will need to manually convert `\sin^2(x)` into `[1 + 

    \cos(2 x)]/2` to properly apply the method of undetermined coefficients on 

    it. 

 

    Examples 

    ======== 

 

    >>> from sympy import log, exp 

    >>> from sympy.solvers.ode import _undetermined_coefficients_match 

    >>> from sympy.abc import x 

    >>> _undetermined_coefficients_match(9*x*exp(x) + exp(-x), x) 

    {'test': True, 'trialset': set([x*exp(x), exp(-x), exp(x)])} 

    >>> _undetermined_coefficients_match(log(x), x) 

    {'test': False} 

 

    """ 

    a = Wild('a', exclude=[x]) 

    b = Wild('b', exclude=[x]) 

    expr = powsimp(expr, combine='exp')  # exp(x)*exp(2*x + 1) => exp(3*x + 1) 

    retdict = {} 

 

    def _test_term(expr, x): 

        r""" 

        Test if ``expr`` fits the proper form for undetermined coefficients. 

        """ 

        if expr.is_Add: 

            return all(_test_term(i, x) for i in expr.args) 

        elif expr.is_Mul: 

            if expr.has(sin, cos): 

                foundtrig = False 

                # Make sure that there is only one trig function in the args. 

                # See the docstring. 

                for i in expr.args: 

                    if i.has(sin, cos): 

                        if foundtrig: 

                            return False 

                        else: 

                            foundtrig = True 

            return all(_test_term(i, x) for i in expr.args) 

        elif expr.is_Function: 

            if expr.func in (sin, cos, exp): 

                if expr.args[0].match(a*x + b): 

                    return True 

                else: 

                    return False 

            else: 

                return False 

        elif expr.is_Pow and expr.base.is_Symbol and expr.exp.is_Integer and \ 

                expr.exp >= 0: 

            return True 

        elif expr.is_Pow and expr.base.is_number: 

            if expr.exp.match(a*x + b): 

                return True 

            else: 

                return False 

        elif expr.is_Symbol or expr.is_number: 

            return True 

        else: 

            return False 

 

    def _get_trial_set(expr, x, exprs=set([])): 

        r""" 

        Returns a set of trial terms for undetermined coefficients. 

 

        The idea behind undetermined coefficients is that the terms expression 

        repeat themselves after a finite number of derivatives, except for the 

        coefficients (they are linearly dependent).  So if we collect these, 

        we should have the terms of our trial function. 

        """ 

        def _remove_coefficient(expr, x): 

            r""" 

            Returns the expression without a coefficient. 

 

            Similar to expr.as_independent(x)[1], except it only works 

            multiplicatively. 

            """ 

            term = S.One 

            if expr.is_Mul: 

                for i in expr.args: 

                    if i.has(x): 

                        term *= i 

            elif expr.has(x): 

                term = expr 

            return term 

 

        expr = expand_mul(expr) 

        if expr.is_Add: 

            for term in expr.args: 

                if _remove_coefficient(term, x) in exprs: 

                    pass 

                else: 

                    exprs.add(_remove_coefficient(term, x)) 

                    exprs = exprs.union(_get_trial_set(term, x, exprs)) 

        else: 

            term = _remove_coefficient(expr, x) 

            tmpset = exprs.union(set([term])) 

            oldset = set([]) 

            while tmpset != oldset: 

                # If you get stuck in this loop, then _test_term is probably 

                # broken 

                oldset = tmpset.copy() 

                expr = expr.diff(x) 

                term = _remove_coefficient(expr, x) 

                if term.is_Add: 

                    tmpset = tmpset.union(_get_trial_set(term, x, tmpset)) 

                else: 

                    tmpset.add(term) 

            exprs = tmpset 

        return exprs 

 

    retdict['test'] = _test_term(expr, x) 

    if retdict['test']: 

        # Try to generate a list of trial solutions that will have the 

        # undetermined coefficients. Note that if any of these are not linearly 

        # independent with any of the solutions to the homogeneous equation, 

        # then they will need to be multiplied by sufficient x to make them so. 

        # This function DOES NOT do that (it doesn't even look at the 

        # homogeneous equation). 

        retdict['trialset'] = _get_trial_set(expr, x) 

 

    return retdict 

 

 

def ode_nth_linear_constant_coeff_variation_of_parameters(eq, func, order, match): 

    r""" 

    Solves an `n`\th order linear differential equation with constant 

    coefficients using the method of variation of parameters. 

 

    This method works on any differential equations of the form 

 

    .. math:: f^{(n)}(x) + a_{n-1} f^{(n-1)}(x) + \cdots + a_1 f'(x) + a_0 

                f(x) = P(x)\text{.} 

 

    This method works by assuming that the particular solution takes the form 

 

    .. math:: \sum_{x=1}^{n} c_i(x) y_i(x)\text{,} 

 

    where `y_i` is the `i`\th solution to the homogeneous equation.  The 

    solution is then solved using Wronskian's and Cramer's Rule.  The 

    particular solution is given by 

 

    .. math:: \sum_{x=1}^n \left( \int \frac{W_i(x)}{W(x)} \,dx 

                \right) y_i(x) \text{,} 

 

    where `W(x)` is the Wronskian of the fundamental system (the system of `n` 

    linearly independent solutions to the homogeneous equation), and `W_i(x)` 

    is the Wronskian of the fundamental system with the `i`\th column replaced 

    with `[0, 0, \cdots, 0, P(x)]`. 

 

    This method is general enough to solve any `n`\th order inhomogeneous 

    linear differential equation with constant coefficients, but sometimes 

    SymPy cannot simplify the Wronskian well enough to integrate it.  If this 

    method hangs, try using the 

    ``nth_linear_constant_coeff_variation_of_parameters_Integral`` hint and 

    simplifying the integrals manually.  Also, prefer using 

    ``nth_linear_constant_coeff_undetermined_coefficients`` when it 

    applies, because it doesn't use integration, making it faster and more 

    reliable. 

 

    Warning, using simplify=False with 

    'nth_linear_constant_coeff_variation_of_parameters' in 

    :py:meth:`~sympy.solvers.ode.dsolve` may cause it to hang, because it will 

    not attempt to simplify the Wronskian before integrating.  It is 

    recommended that you only use simplify=False with 

    'nth_linear_constant_coeff_variation_of_parameters_Integral' for this 

    method, especially if the solution to the homogeneous equation has 

    trigonometric functions in it. 

 

    Examples 

    ======== 

 

    >>> from sympy import Function, dsolve, pprint, exp, log 

    >>> from sympy.abc import x 

    >>> f = Function('f') 

    >>> pprint(dsolve(f(x).diff(x, 3) - 3*f(x).diff(x, 2) + 

    ... 3*f(x).diff(x) - f(x) - exp(x)*log(x), f(x), 

    ... hint='nth_linear_constant_coeff_variation_of_parameters')) 

           /                     3                \ 

           |                2   x *(6*log(x) - 11)|  x 

    f(x) = |C1 + C2*x + C3*x  + ------------------|*e 

           \                            36        / 

 

    References 

    ========== 

 

    - http://en.wikipedia.org/wiki/Variation_of_parameters 

    - http://planetmath.org/VariationOfParameters 

    - M. Tenenbaum & H. Pollard, "Ordinary Differential Equations", 

      Dover 1963, pp. 233 

 

    # indirect doctest 

 

    """ 

 

    gensol = ode_nth_linear_constant_coeff_homogeneous(eq, func, order, match, 

        returns='both') 

    match.update(gensol) 

    return _solve_variation_of_parameters(eq, func, order, match) 

 

 

def _solve_variation_of_parameters(eq, func, order, match): 

    r""" 

    Helper function for the method of variation of parameters and nonhomogeneous euler eq. 

 

    See the 

    :py:meth:`~sympy.solvers.ode.ode_nth_linear_constant_coeff_variation_of_parameters` 

    docstring for more information on this method. 

 

    The parameter ``match`` should be a dictionary that has the following 

    keys: 

 

    ``list`` 

      A list of solutions to the homogeneous equation, such as the list 

      returned by 

      ``ode_nth_linear_constant_coeff_homogeneous(returns='list')``. 

 

    ``sol`` 

      The general solution, such as the solution returned by 

      ``ode_nth_linear_constant_coeff_homogeneous(returns='sol')``. 

 

    """ 

 

    x = func.args[0] 

    f = func.func 

    r = match 

    psol = 0 

    gensols = r['list'] 

    gsol = r['sol'] 

    wr = wronskian(gensols, x) 

 

    if r.get('simplify', True): 

        wr = simplify(wr)  # We need much better simplification for 

                           # some ODEs. See issue 4662, for example. 

 

        # To reduce commonly occuring sin(x)**2 + cos(x)**2 to 1 

        wr = trigsimp(wr, deep=True, recursive=True) 

    if not wr: 

        # The wronskian will be 0 iff the solutions are not linearly 

        # independent. 

        raise NotImplementedError("Cannot find " + str(order) + 

        " solutions to the homogeneous equation nessesary to apply " + 

        "variation of parameters to " + str(eq) + " (Wronskian == 0)") 

    if len(gensols) != order: 

        raise NotImplementedError("Cannot find " + str(order) + 

        " solutions to the homogeneous equation nessesary to apply " + 

        "variation of parameters to " + 

        str(eq) + " (number of terms != order)") 

    negoneterm = (-1)**(order) 

    for i in gensols: 

        psol += negoneterm*Integral(wronskian([sol for sol in gensols if sol != i], x)*r[-1]/wr, x)*i/r[order] 

        negoneterm *= -1 

 

    if r.get('simplify', True): 

        psol = simplify(psol) 

        psol = trigsimp(psol, deep=True) 

    return Eq(f(x), gsol.rhs + psol) 

 

 

def ode_separable(eq, func, order, match): 

    r""" 

    Solves separable 1st order differential equations. 

 

    This is any differential equation that can be written as `P(y) 

    \tfrac{dy}{dx} = Q(x)`.  The solution can then just be found by 

    rearranging terms and integrating: `\int P(y) \,dy = \int Q(x) \,dx`. 

    This hint uses :py:meth:`sympy.simplify.simplify.separatevars` as its back 

    end, so if a separable equation is not caught by this solver, it is most 

    likely the fault of that function. 

    :py:meth:`~sympy.simplify.simplify.separatevars` is 

    smart enough to do most expansion and factoring necessary to convert a 

    separable equation `F(x, y)` into the proper form `P(x)\cdot{}Q(y)`.  The 

    general solution is:: 

 

        >>> from sympy import Function, dsolve, Eq, pprint 

        >>> from sympy.abc import x 

        >>> a, b, c, d, f = map(Function, ['a', 'b', 'c', 'd', 'f']) 

        >>> genform = Eq(a(x)*b(f(x))*f(x).diff(x), c(x)*d(f(x))) 

        >>> pprint(genform) 

                     d 

        a(x)*b(f(x))*--(f(x)) = c(x)*d(f(x)) 

                     dx 

        >>> pprint(dsolve(genform, f(x), hint='separable_Integral')) 

             f(x) 

           /                  / 

          |                  | 

          |  b(y)            | c(x) 

          |  ---- dy = C1 +  | ---- dx 

          |  d(y)            | a(x) 

          |                  | 

         /                  / 

 

    Examples 

    ======== 

 

    >>> from sympy import Function, dsolve, Eq 

    >>> from sympy.abc import x 

    >>> f = Function('f') 

    >>> pprint(dsolve(Eq(f(x)*f(x).diff(x) + x, 3*x*f(x)**2), f(x), 

    ... hint='separable', simplify=False)) 

       /   2       \         2 

    log\3*f (x) - 1/        x 

    ---------------- = C1 + -- 

           6                2 

 

    References 

    ========== 

 

    - M. Tenenbaum & H. Pollard, "Ordinary Differential Equations", 

      Dover 1963, pp. 52 

 

    # indirect doctest 

 

    """ 

    x = func.args[0] 

    f = func.func 

    C1 = get_numbered_constants(eq, num=1) 

    r = match  # {'m1':m1, 'm2':m2, 'y':y} 

    u = r.get('hint', f(x))  # get u from separable_reduced else get f(x) 

    return Eq(Integral(r['m2']['coeff']*r['m2'][r['y']]/r['m1'][r['y']], 

        (r['y'], None, u)), Integral(-r['m1']['coeff']*r['m1'][x]/ 

        r['m2'][x], x) + C1) 

 

 

def checkinfsol(eq, infinitesimals, func=None, order=None): 

    r""" 

    This function is used to check if the given infinitesimals are the 

    actual infinitesimals of the given first order differential equation. 

    This method is specific to the Lie Group Solver of ODEs. 

 

    As of now, it simply checks, by substituting the infinitesimals in the 

    partial differential equation. 

 

 

    .. math:: \frac{\partial \eta}{\partial x} + \left(\frac{\partial \eta}{\partial y} 

                - \frac{\partial \xi}{\partial x}\right)*h 

                - \frac{\partial \xi}{\partial y}*h^{2} 

                - \xi\frac{\partial h}{\partial x} - \eta\frac{\partial h}{\partial y} = 0 

 

 

    where `\eta`, and `\xi` are the infinitesimals and `h(x,y) = \frac{dy}{dx}` 

 

    The infinitesimals should be given in the form of a list of dicts 

    ``[{xi(x, y): inf, eta(x, y): inf}]``, corresponding to the 

    output of the function infinitesimals. It returns a list 

    of values of the form ``[(True/False, sol)]`` where ``sol`` is the value 

    obtained after substituting the infinitesimals in the PDE. If it 

    is ``True``, then ``sol`` would be 0. 

 

    """ 

    if isinstance(eq, Equality): 

        eq = eq.lhs - eq.rhs 

    if not func: 

        eq, func = _preprocess(eq) 

    variables = func.args 

    if len(variables) != 1: 

        raise ValueError("ODE's have only one independent variable") 

    else: 

        x = variables[0] 

        if not order: 

            order = ode_order(eq, func) 

        if order != 1: 

            raise NotImplementedError("Lie groups solver has been implemented " 

            "only for first order differential equations") 

        else: 

            df = func.diff(x) 

            a = Wild('a', exclude = [df]) 

            b = Wild('b', exclude = [df]) 

            match = collect(expand(eq), df).match(a*df + b) 

 

            if match: 

                h = -simplify(match[b]/match[a]) 

            else: 

                try: 

                    sol = solve(eq, df) 

                except NotImplementedError: 

                    raise NotImplementedError("Infinitesimals for the " 

                        "first order ODE could not be found") 

                else: 

                    h = sol[0]  # Find infinitesimals for one solution 

 

            y = Dummy('y') 

            h = h.subs(func, y) 

            xi = Function('xi')(x, y) 

            eta = Function('eta')(x, y) 

            dxi = Function('xi')(x, func) 

            deta = Function('eta')(x, func) 

            pde = (eta.diff(x) + (eta.diff(y) - xi.diff(x))*h - 

                (xi.diff(y))*h**2 - xi*(h.diff(x)) - eta*(h.diff(y))) 

            soltup = [] 

            for sol in infinitesimals: 

                tsol = {xi: S(sol[dxi]).subs(func, y), 

                    eta: S(sol[deta]).subs(func, y)} 

                sol = simplify(pde.subs(tsol).doit()) 

                if sol: 

                    soltup.append((False, sol.subs(y, func))) 

                else: 

                    soltup.append((True, 0)) 

            return soltup 

 

def ode_lie_group(eq, func, order, match): 

    r""" 

    This hint implements the Lie group method of solving first order differential 

    equations. The aim is to convert the given differential equation from the 

    given coordinate given system into another coordinate system where it becomes 

    invariant under the one-parameter Lie group of translations. The converted ODE is 

    quadrature and can be solved easily. It makes use of the 

    :py:meth:`sympy.solvers.ode.infinitesimals` function which returns the 

    infinitesimals of the transformation. 

 

    The coordinates `r` and `s` can be found by solving the following Partial 

    Differential Equations. 

 

    .. math :: \xi\frac{\partial r}{\partial x} + \eta\frac{\partial r}{\partial y} 

                  = 0 

 

    .. math :: \xi\frac{\partial s}{\partial x} + \eta\frac{\partial s}{\partial y} 

                  = 1 

 

    The differential equation becomes separable in the new coordinate system 

 

    .. math :: \frac{ds}{dr} = \frac{\frac{\partial s}{\partial x} + 

                 h(x, y)\frac{\partial s}{\partial y}}{ 

                 \frac{\partial r}{\partial x} + h(x, y)\frac{\partial r}{\partial y}} 

 

    After finding the solution by integration, it is then converted back to the original 

    coordinate system by subsituting `r` and `s` in terms of `x` and `y` again. 

 

    Examples 

    ======== 

 

    >>> from sympy import Function, dsolve, Eq, exp, pprint 

    >>> from sympy.abc import x 

    >>> f = Function('f') 

    >>> pprint(dsolve(f(x).diff(x) + 2*x*f(x) - x*exp(-x**2), f(x), 

    ... hint='lie_group')) 

           /      2\    2 

           |     x |  -x 

    f(x) = |C1 + --|*e 

           \     2 / 

 

 

    References 

    ========== 

 

    - Solving differential equations by Symmetry Groups, 

      John Starrett, pp. 1 - pp. 14 

 

    """ 

 

    heuristics = lie_heuristics 

    inf = {} 

    f = func.func 

    x = func.args[0] 

    df = func.diff(x) 

    xi = Function("xi") 

    eta = Function("eta") 

    a = Wild('a', exclude = [df]) 

    b = Wild('b', exclude = [df]) 

    xis = match.pop('xi') 

    etas = match.pop('eta') 

 

    if match: 

        h = -simplify(match[match['d']]/match[match['e']]) 

        y = match['y'] 

    else: 

        try: 

            sol = solve(eq, df) 

        except NotImplementedError: 

            raise NotImplementedError("Unable to solve the differential equation " + 

                str(eq) + " by the lie group method") 

        else: 

            y = Dummy("y") 

            h = sol[0].subs(func, y) 

 

    if xis is not None and etas is not None: 

        inf = [{xi(x, f(x)): S(xis), eta(x, f(x)): S(etas)}] 

 

        if not checkinfsol(eq, inf, func=f(x), order=1)[0][0]: 

            raise ValueError("The given infinitesimals xi and eta" 

                " are not the infinitesimals to the given equation") 

        else: 

            heuristics = ["user_defined"] 

 

    match = {'h': h, 'y': y} 

 

    # This is done so that if: 

    # a] solve raises a NotImplementedError. 

    # b] any heuristic raises a ValueError 

    # another heuristic can be used. 

    tempsol = []  # Used by solve below 

    for heuristic in heuristics: 

        try: 

            if not inf: 

                inf = infinitesimals(eq, hint=heuristic, func=func, order=1, match=match) 

        except ValueError: 

            continue 

        else: 

            for infsim in inf: 

                xiinf = (infsim[xi(x, func)]).subs(func, y) 

                etainf = (infsim[eta(x, func)]).subs(func, y) 

                # This condition creates recursion while using pdsolve. 

                # Since the first step while solving a PDE of form 

                # a*(f(x, y).diff(x)) + b*(f(x, y).diff(y)) + c = 0 

                # is to solve the ODE dy/dx = b/a 

                if simplify(etainf/xiinf) == h: 

                    continue 

                rpde = f(x, y).diff(x)*xiinf + f(x, y).diff(y)*etainf 

                r = pdsolve(rpde, func=f(x, y)).rhs 

                s = pdsolve(rpde - 1, func=f(x, y)).rhs 

                newcoord = [_lie_group_remove(coord) for coord in [r, s]] 

                r = Dummy("r") 

                s = Dummy("s") 

                C1 = Symbol("C1") 

                rcoord = newcoord[0] 

                scoord = newcoord[-1] 

                try: 

                    sol = solve([r - rcoord, s - scoord], x, y, dict=True) 

                except NotImplementedError: 

                    continue 

                else: 

                    sol = sol[0] 

                    xsub = sol[x] 

                    ysub = sol[y] 

                    num = simplify(scoord.diff(x) + scoord.diff(y)*h) 

                    denom = simplify(rcoord.diff(x) + rcoord.diff(y)*h) 

                    if num and denom: 

                        diffeq = simplify((num/denom).subs([(x, xsub), (y, ysub)])) 

                        sep = separatevars(diffeq, symbols=[r, s], dict=True) 

                        if sep: 

                            # Trying to separate, r and s coordinates 

                            deq = integrate((1/sep[s]), s) + C1 - integrate(sep['coeff']*sep[r], r) 

                            # Substituting and reverting back to original coordinates 

                            deq = deq.subs([(r, rcoord), (s, scoord)]) 

                            try: 

                                sdeq = solve(deq, y) 

                            except NotImplementedError: 

                                tempsol.append(deq) 

                            else: 

                                if len(sdeq) == 1: 

                                    return Eq(f(x), sdeq.pop()) 

                                else: 

                                    return [Eq(f(x), sol) for sol in sdeq] 

 

 

                    elif denom: # (ds/dr) is zero which means s is constant 

                        return Eq(f(x), solve(scoord - C1, y)[0]) 

 

                    elif num: # (dr/ds) is zero which means r is constant 

                        return Eq(f(x), solve(rcoord - C1, y)[0]) 

 

    # If nothing works, return solution as it is, without solving for y 

    if tempsol: 

        if len(tempsol) == 1: 

            return Eq(tempsol.pop().subs(y, f(x)), 0) 

        else: 

            return [Eq(sol.subs(y, f(x)), 0) for sol in tempsol] 

 

    raise NotImplementedError("The given ODE " + str(eq) + " cannot be solved by" 

        + " the lie group method") 

 

 

def _lie_group_remove(coords): 

    r""" 

    This function is strictly meant for internal use by the Lie group ODE solving 

    method. It replaces arbitrary functions returned by pdsolve with either 0 or 1 or the 

    args of the arbitrary function. 

 

    The algorithm used is: 

    1] If coords is an instance of an Undefined Function, then the args are returned 

    2] If the arbitrary function is present in an Add object, it is replaced by zero. 

    3] If the arbitrary function is present in an Mul object, it is replaced by one. 

    4] If coords has no Undefined Function, it is returned as it is. 

 

    Examples 

    ======== 

 

    >>> from sympy.solvers.ode import _lie_group_remove 

    >>> from sympy import Function 

    >>> from sympy.abc import x, y 

    >>> F = Function("F") 

    >>> eq = x**2*y 

    >>> _lie_group_remove(eq) 

    x**2*y 

    >>> eq = F(x**2*y) 

    >>> _lie_group_remove(eq) 

    x**2*y 

    >>> eq = y**2*x + F(x**3) 

    >>> _lie_group_remove(eq) 

    x*y**2 

    >>> eq = (F(x**3) + y)*x**4 

    >>> _lie_group_remove(eq) 

    x**4*y 

 

    """ 

    if isinstance(coords, AppliedUndef): 

        return coords.args[0] 

    elif coords.is_Add: 

        subfunc = coords.atoms(AppliedUndef) 

        if subfunc: 

            for func in subfunc: 

                coords = coords.subs(func, 0) 

        return coords 

    elif coords.is_Pow: 

        base, expr = coords.as_base_exp() 

        base = _lie_group_remove(base) 

        expr = _lie_group_remove(expr) 

        return base**expr 

    elif coords.is_Mul: 

        mulargs = [] 

        coordargs = coords.args 

        for arg in coordargs: 

            if not isinstance(coords, AppliedUndef): 

                mulargs.append(_lie_group_remove(arg)) 

        return Mul(*mulargs) 

    return coords 

 

def infinitesimals(eq, func=None, order=None, hint='default', match=None): 

    r""" 

    The infinitesimal functions of an ordinary differential equation, `\xi(x,y)` 

    and `\eta(x,y)`, are the infinitesimals of the Lie group of point transformations 

    for which the differential equation is invariant. So, the ODE `y'=f(x,y)` 

    would admit a Lie group `x^*=X(x,y;\varepsilon)=x+\varepsilon\xi(x,y)`, 

    `y^*=Y(x,y;\varepsilon)=y+\varepsilon\eta(x,y)` such that `(y^*)'=f(x^*, y^*)`. 

    A change of coordinates, to `r(x,y)` and `s(x,y)`, can be performed so this Lie group 

    becomes the translation group, `r^*=r` and `s^*=s+\varepsilon`. 

    They are tangents to the coordinate curves of the new system. 

 

    Consider the transformation `(x, y) \to (X, Y)` such that the 

    differential equation remains invariant. `\xi` and `\eta` are the tangents to 

    the transformed coordinates `X` and `Y`, at `\varepsilon=0`. 

 

    .. math:: \left(\frac{\partial X(x,y;\varepsilon)}{\partial\varepsilon 

                }\right)|_{\varepsilon=0} = \xi, 

              \left(\frac{\partial Y(x,y;\varepsilon)}{\partial\varepsilon 

                }\right)|_{\varepsilon=0} = \eta, 

 

    The infinitesimals can be found by solving the following PDE: 

 

        >>> from sympy import Function, diff, Eq, pprint 

        >>> from sympy.abc import x, y 

        >>> xi, eta, h = map(Function, ['xi', 'eta', 'h']) 

        >>> h = h(x, y)  # dy/dx = h 

        >>> eta = eta(x, y) 

        >>> xi = xi(x, y) 

        >>> genform = Eq(eta.diff(x) + (eta.diff(y) - xi.diff(x))*h 

        ... - (xi.diff(y))*h**2 - xi*(h.diff(x)) - eta*(h.diff(y)), 0) 

        >>> pprint(genform) 

        /d               d           \                     d              2       d 

        |--(eta(x, y)) - --(xi(x, y))|*h(x, y) - eta(x, y)*--(h(x, y)) - h (x, y)*--(x 

        \dy              dx          /                     dy                     dy 

        <BLANKLINE> 

                            d             d 

        i(x, y)) - xi(x, y)*--(h(x, y)) + --(eta(x, y)) = 0 

                            dx            dx 

 

    Solving the above mentioned PDE is not trivial, and can be solved only by 

    making intelligent assumptions for `\xi` and `\eta` (heuristics). Once an 

    infinitesimal is found, the attempt to find more heuristics stops. This is done to 

    optimise the speed of solving the differential equation. If a list of all the 

    infinitesimals is needed, ``hint`` should be flagged as ``all``, which gives 

    the complete list of infinitesimals. If the infinitesimals for a particular 

    heuristic needs to be found, it can be passed as a flag to ``hint``. 

 

    Examples 

    ======== 

 

    >>> from sympy import Function, diff 

    >>> from sympy.solvers.ode import infinitesimals 

    >>> from sympy.abc import x 

    >>> f = Function('f') 

    >>> eq = f(x).diff(x) - x**2*f(x) 

    >>> infinitesimals(eq) 

    [{eta(x, f(x)): exp(x**3/3), xi(x, f(x)): 0}] 

 

    References 

    ========== 

 

    - Solving differential equations by Symmetry Groups, 

      John Starrett, pp. 1 - pp. 14 

 

    """ 

 

    if isinstance(eq, Equality): 

        eq = eq.lhs - eq.rhs 

    if not func: 

        eq, func = _preprocess(eq) 

    variables = func.args 

    if len(variables) != 1: 

        raise ValueError("ODE's have only one independent variable") 

    else: 

        x = variables[0] 

        if not order: 

            order = ode_order(eq, func) 

        if order != 1: 

            raise NotImplementedError("Infinitesimals for only " 

                "first order ODE's have been implemented") 

        else: 

            df = func.diff(x) 

            # Matching differential equation of the form a*df + b 

            a = Wild('a', exclude = [df]) 

            b = Wild('b', exclude = [df]) 

            if match:  # Used by lie_group hint 

                h = match['h'] 

                y = match['y'] 

            else: 

                match = collect(expand(eq), df).match(a*df + b) 

                if match: 

                    h = -simplify(match[b]/match[a]) 

                else: 

                    try: 

                        sol = solve(eq, df) 

                    except NotImplementedError: 

                        raise NotImplementedError("Infinitesimals for the " 

                            "first order ODE could not be found") 

                    else: 

                        h = sol[0]  # Find infinitesimals for one solution 

                y = Dummy("y") 

                h = h.subs(func, y) 

 

            u = Dummy("u") 

            hx = h.diff(x) 

            hy = h.diff(y) 

            hinv = ((1/h).subs([(x, u), (y, x)])).subs(u, y)  # Inverse ODE 

            match = {'h': h, 'func': func, 'hx': hx, 'hy': hy, 'y': y, 'hinv': hinv} 

            if hint == 'all': 

                xieta = [] 

                for heuristic in lie_heuristics: 

                    function = globals()['lie_heuristic_' + heuristic] 

                    inflist = function(match, comp=True) 

                    if inflist: 

                        xieta.extend([inf for inf in inflist if inf not in xieta]) 

                if xieta: 

                    return xieta 

                else: 

                    raise NotImplementedError("Infinitesimals could not be found for " 

                        "the given ODE") 

 

            elif hint == 'default': 

                for heuristic in lie_heuristics: 

                    function = globals()['lie_heuristic_' + heuristic] 

                    xieta = function(match, comp=False) 

                    if xieta: 

                        return xieta 

 

                raise NotImplementedError("Infinitesimals could not be found for" 

                    " the given ODE") 

 

            elif hint not in lie_heuristics: 

                 raise ValueError("Heuristic not recognized: " + hint) 

 

            else: 

                 function = globals()['lie_heuristic_' + hint] 

                 xieta = function(match, comp=True) 

                 if xieta: 

                     return xieta 

                 else: 

                     raise ValueError("Infinitesimals could not be found using the" 

                         " given heuristic") 

 

 

def lie_heuristic_abaco1_simple(match, comp=False): 

    r""" 

    The first heuristic uses the following four sets of 

    assumptions on `\xi` and `\eta` 

 

    .. math:: \xi = 0, \eta = f(x) 

 

    .. math:: \xi = 0, \eta = f(y) 

 

    .. math:: \xi = f(x), \eta = 0 

 

    .. math:: \xi = f(y), \eta = 0 

 

    The success of this heuristic is determined by algebraic factorisation. 

    For the first assumption `\xi = 0` and `\eta` to be a function of `x`, the PDE 

 

    .. math:: \frac{\partial \eta}{\partial x} + (\frac{\partial \eta}{\partial y} 

                - \frac{\partial \xi}{\partial x})*h 

                - \frac{\partial \xi}{\partial y}*h^{2} 

                - \xi*\frac{\partial h}{\partial x} - \eta*\frac{\partial h}{\partial y} = 0 

 

    reduces to `f'(x) - f\frac{\partial h}{\partial y} = 0` 

    If `\frac{\partial h}{\partial y}` is a function of `x`, then this can usually 

    be integrated easily. A similar idea is applied to the other 3 assumptions as well. 

 

 

    References 

    ========== 

 

    - E.S Cheb-Terrab, L.G.S Duarte and L.A,C.P da Mota, Computer Algebra 

      Solving of First Order ODEs Using Symmetry Methods, pp. 8 

 

 

    """ 

 

    xieta = [] 

    y = match['y'] 

    h = match['h'] 

    func = match['func'] 

    x = func.args[0] 

    hx = match['hx'] 

    hy = match['hy'] 

    xi = Function('xi')(x, func) 

    eta = Function('eta')(x, func) 

 

    hysym = hy.free_symbols 

    if y not in hysym: 

        try: 

            fx = exp(integrate(hy, x)) 

        except NotImplementedError: 

            pass 

        else: 

            inf = {xi: S(0), eta: fx} 

            if not comp: 

                return [inf] 

            if comp and inf not in xieta: 

                xieta.append(inf) 

 

    factor = hy/h 

    facsym = factor.free_symbols 

    if x not in facsym: 

        try: 

            fy = exp(integrate(factor, y)) 

        except NotImplementedError: 

            pass 

        else: 

            inf = {xi: S(0), eta: fy.subs(y, func)} 

            if not comp: 

                return [inf] 

            if comp and inf not in xieta: 

                xieta.append(inf) 

 

    factor = -hx/h 

    facsym = factor.free_symbols 

    if y not in facsym: 

        try: 

            fx = exp(integrate(factor, x)) 

        except NotImplementedError: 

            pass 

        else: 

            inf = {xi: fx, eta: S(0)} 

            if not comp: 

                return [inf] 

            if comp and inf not in xieta: 

                xieta.append(inf) 

 

    factor = -hx/(h**2) 

    facsym = factor.free_symbols 

    if x not in facsym: 

        try: 

            fy = exp(integrate(factor, y)) 

        except NotImplementedError: 

            pass 

        else: 

            inf = {xi: fy.subs(y, func), eta: S(0)} 

            if not comp: 

                return [inf] 

            if comp and inf not in xieta: 

                xieta.append(inf) 

 

    if xieta: 

        return xieta 

 

def lie_heuristic_abaco1_product(match, comp=False): 

    r""" 

    The second heuristic uses the following two assumptions on `\xi` and `\eta` 

 

    .. math:: \eta = 0, \xi = f(x)*g(y) 

 

    .. math:: \eta = f(x)*g(y), \xi = 0 

 

    The first assumption of this heuristic holds good if 

    `\frac{1}{h^{2}}\frac{\partial^2}{\partial x \partial y}\log(h)` is 

    separable in `x` and `y`, then the separated factors containing `x` 

    is `f(x)`, and `g(y)` is obtained by 

 

    .. math:: e^{\int f\frac{\partial}{\partial x}\left(\frac{1}{f*h}\right)\,dy} 

 

    provided `f\frac{\partial}{\partial x}\left(\frac{1}{f*h}\right)` is a function 

    of `y` only. 

 

    The second assumption holds good if `\frac{dy}{dx} = h(x, y)` is rewritten as 

    `\frac{dy}{dx} = \frac{1}{h(y, x)}` and the same properties of the first assumption 

    satisifes. After obtaining `f(x)` and `g(y)`, the coordinates are again 

    interchanged, to get `\eta` as `f(x)*g(y)` 

 

 

    References 

    ========== 

    - E.S. Cheb-Terrab, A.D. Roche, Symmetries and First Order 

      ODE Patterns, pp. 7 - pp. 8 

 

    """ 

 

    xieta = [] 

    y = match['y'] 

    h = match['h'] 

    hinv = match['hinv'] 

    func = match['func'] 

    x = func.args[0] 

    xi = Function('xi')(x, func) 

    eta = Function('eta')(x, func) 

 

 

    inf = separatevars(((log(h).diff(y)).diff(x))/h**2, dict=True, symbols=[x, y]) 

    if inf and inf['coeff']: 

        fx = inf[x] 

        gy = simplify(fx*((1/(fx*h)).diff(x))) 

        gysyms = gy.free_symbols 

        if x not in gysyms: 

            gy = exp(integrate(gy, y)) 

            inf = {eta: S(0), xi: (fx*gy).subs(y, func)} 

            if not comp: 

                return [inf] 

            if comp and inf not in xieta: 

                xieta.append(inf) 

 

    u1 = Dummy("u1") 

    inf = separatevars(((log(hinv).diff(y)).diff(x))/hinv**2, dict=True, symbols=[x, y]) 

    if inf and inf['coeff']: 

        fx = inf[x] 

        gy = simplify(fx*((1/(fx*hinv)).diff(x))) 

        gysyms = gy.free_symbols 

        if x not in gysyms: 

            gy = exp(integrate(gy, y)) 

            etaval = fx*gy 

            etaval = (etaval.subs([(x, u1), (y, x)])).subs(u1, y) 

            inf = {eta: etaval.subs(y, func), xi: S(0)} 

            if not comp: 

                return [inf] 

            if comp and inf not in xieta: 

                xieta.append(inf) 

 

    if xieta: 

        return xieta 

 

def lie_heuristic_bivariate(match, comp=False): 

    r""" 

    The third heuristic assumes the infinitesimals `\xi` and `\eta` 

    to be bi-variate polynomials in `x` and `y`. The assumption made here 

    for the logic below is that `h` is a rational function in `x` and `y` 

    though that may not be necessary for the infinitesimals to be 

    bivariate polynomials. The coefficients of the infinitesimals 

    are found out by substituting them in the PDE and grouping similar terms 

    that are polynomials and since they form a linear system, solve and check 

    for non trivial solutions. The degree of the assumed bivariates 

    are increased till a certain maximum value. 

 

    References 

    ========== 

    - Lie Groups and Differential Equations 

      pp. 327 - pp. 329 

 

    """ 

 

    h = match['h'] 

    hx = match['hx'] 

    hy = match['hy'] 

    func = match['func'] 

    x = func.args[0] 

    y = match['y'] 

    xi = Function('xi')(x, func) 

    eta = Function('eta')(x, func) 

 

    if h.is_rational_function(): 

        # The maximum degree that the infinitesimals can take is 

        # calculated by this technique. 

        etax, etay, etad, xix, xiy, xid = symbols("etax etay etad xix xiy xid") 

        ipde = etax + (etay - xix)*h - xiy*h**2 - xid*hx - etad*hy 

        num, denom = cancel(ipde).as_numer_denom() 

        deg = Poly(num, x, y).total_degree() 

        deta = Function('deta')(x, y) 

        dxi = Function('dxi')(x, y) 

        ipde = (deta.diff(x) + (deta.diff(y) - dxi.diff(x))*h - (dxi.diff(y))*h**2 

            - dxi*hx - deta*hy) 

        xieq = Symbol("xi0") 

        etaeq = Symbol("eta0") 

 

        for i in range(deg + 1): 

            if i: 

                xieq += Add(*[ 

                    Symbol("xi_" + str(power) + "_" + str(i - power))*x**power*y**(i - power) 

                    for power in range(i + 1)]) 

                etaeq += Add(*[ 

                    Symbol("eta_" + str(power) + "_" + str(i - power))*x**power*y**(i - power) 

                    for power in range(i + 1)]) 

            pden, denom = (ipde.subs({dxi: xieq, deta: etaeq}).doit()).as_numer_denom() 

            pden = expand(pden) 

 

            # If the individual terms are monomials, the coefficients 

            # are grouped 

            if pden.is_polynomial(x, y) and pden.is_Add: 

                polyy = Poly(pden, x, y).as_dict() 

            if polyy: 

                symset = xieq.free_symbols.union(etaeq.free_symbols) - set([x, y]) 

                soldict = solve(polyy.values(), *symset) 

                if isinstance(soldict, list): 

                    soldict = soldict[0] 

                if any(x for x in soldict.values()): 

                    xired = xieq.subs(soldict) 

                    etared = etaeq.subs(soldict) 

                    # Scaling is done by substituting one for the parameters 

                    # This can be any number except zero. 

                    dict_ = dict((sym, 1) for sym in symset) 

                    inf = {eta: etared.subs(dict_).subs(y, func), 

                        xi: xired.subs(dict_).subs(y, func)} 

                    return [inf] 

 

def lie_heuristic_chi(match, comp=False): 

    r""" 

    The aim of the fourth heuristic is to find the function `\chi(x, y)` 

    that satisifies the PDE `\frac{d\chi}{dx} + h\frac{d\chi}{dx} 

    - \frac{\partial h}{\partial y}\chi = 0`. 

 

    This assumes `\chi` to be a bivariate polynomial in `x` and `y`. By intution, 

    `h` should be a rational function in `x` and `y`. The method used here is 

    to substitute a general binomial for `\chi` up to a certain maximum degree 

    is reached. The coefficients of the polynomials, are calculated by by collecting 

    terms of the same order in `x` and `y`. 

 

    After finding `\chi`, the next step is to use `\eta = \xi*h + \chi`, to 

    determine `\xi` and `\eta`. This can be done by dividing `\chi` by `h` 

    which would give `-\xi` as the quotient and `\eta` as the remainder. 

 

 

    References 

    ========== 

    - E.S Cheb-Terrab, L.G.S Duarte and L.A,C.P da Mota, Computer Algebra 

      Solving of First Order ODEs Using Symmetry Methods, pp. 8 

 

    """ 

 

    h = match['h'] 

    hx = match['hx'] 

    hy = match['hy'] 

    func = match['func'] 

    x = func.args[0] 

    y = match['y'] 

    xi = Function('xi')(x, func) 

    eta = Function('eta')(x, func) 

 

    if h.is_rational_function(): 

        schi, schix, schiy = symbols("schi, schix, schiy") 

        cpde = schix + h*schiy - hy*schi 

        num, denom = cancel(cpde).as_numer_denom() 

        deg = Poly(num, x, y).total_degree() 

 

        chi = Function('chi')(x, y) 

        chix = chi.diff(x) 

        chiy = chi.diff(y) 

        cpde = chix + h*chiy - hy*chi 

        chieq = Symbol("chi") 

        for i in range(1, deg + 1): 

            chieq += Add(*[ 

                Symbol("chi_" + str(power) + "_" + str(i - power))*x**power*y**(i - power) 

                for power in range(i + 1)]) 

            cnum, cden = cancel(cpde.subs({chi : chieq}).doit()).as_numer_denom() 

            cnum = expand(cnum) 

            if cnum.is_polynomial(x, y) and cnum.is_Add: 

                cpoly = Poly(cnum, x, y).as_dict() 

                if cpoly: 

                    solsyms = chieq.free_symbols - set([x, y]) 

                    soldict = solve(cpoly.values(), *solsyms) 

                    if isinstance(soldict, list): 

                        soldict = soldict[0] 

                    if any(x for x in soldict.values()): 

                        chieq = chieq.subs(soldict) 

                        dict_ = dict((sym, 1) for sym in solsyms) 

                        chieq = chieq.subs(dict_) 

                        # After finding chi, the main aim is to find out 

                        # eta, xi by the equation eta = xi*h + chi 

                        # One method to set xi, would be rearranging it to 

                        # (eta/h) - xi = (chi/h). This would mean dividing 

                        # chi by h would give -xi as the quotient and eta 

                        # as the remainder. Thanks to Sean Vig for suggesting 

                        # this method. 

                        xic, etac = div(chieq, h) 

                        inf = {eta: etac.subs(y, func), xi: -xic.subs(y, func)} 

                        return [inf] 

 

def lie_heuristic_function_sum(match, comp=False): 

    r""" 

    This heuristic uses the following two assumptions on `\xi` and `\eta` 

 

    .. math:: \eta = 0, \xi = f(x) + g(y) 

 

    .. math:: \eta = f(x) + g(y), \xi = 0 

 

    The first assumption of this heuristic holds good if 

 

    .. math:: \frac{\partial}{\partial y}[(h\frac{\partial^{2}}{ 

                \partial x^{2}}(h^{-1}))^{-1}] 

 

    is separable in `x` and `y`, 

 

    1. The separated factors containing `y` is `\frac{\partial g}{\partial y}`. 

       From this `g(y)` can be determined. 

    2. The separated factors containing `x` is `f''(x)`. 

    3. `h\frac{\partial^{2}}{\partial x^{2}}(h^{-1})` equals 

       `\frac{f''(x)}{f(x) + g(y)}`. From this `f(x)` can be determined. 

 

    The second assumption holds good if `\frac{dy}{dx} = h(x, y)` is rewritten as 

    `\frac{dy}{dx} = \frac{1}{h(y, x)}` and the same properties of the first 

    assumption satisifes. After obtaining `f(x)` and `g(y)`, the coordinates 

    are again interchanged, to get `\eta` as `f(x) + g(y)`. 

 

    For both assumptions, the constant factors are separated among `g(y)` 

    and `f''(x)`, such that `f''(x)` obtained from 3] is the same as that 

    obtained from 2]. If not possible, then this heuristic fails. 

 

 

    References 

    ========== 

    - E.S. Cheb-Terrab, A.D. Roche, Symmetries and First Order 

      ODE Patterns, pp. 7 - pp. 8 

 

    """ 

 

    xieta = [] 

    h = match['h'] 

    hx = match['hx'] 

    hy = match['hy'] 

    func = match['func'] 

    hinv = match['hinv'] 

    x = func.args[0] 

    y = match['y'] 

    xi = Function('xi')(x, func) 

    eta = Function('eta')(x, func) 

 

    for odefac in [h, hinv]: 

        factor = odefac*((1/odefac).diff(x, 2)) 

        sep = separatevars((1/factor).diff(y), dict=True, symbols=[x, y]) 

        if sep and sep['coeff'] and sep[x].has(x) and sep[y].has(y): 

            k = Dummy("k") 

            try: 

                gy = k*integrate(sep[y], y) 

            except NotImplementedError: 

                pass 

            else: 

                fdd = 1/(k*sep[x]*sep['coeff']) 

                fx = simplify(fdd/factor - gy) 

                check = simplify(fx.diff(x, 2) - fdd) 

                if fx: 

                    if not check: 

                        fx = fx.subs(k, 1) 

                        gy = (gy/k) 

                    else: 

                        sol = solve(check, k) 

                        if sol: 

                            sol = sol[0] 

                            fx = fx.subs(k, sol) 

                            gy = (gy/k)*sol 

                        else: 

                            continue 

                    if odefac == hinv:  # Inverse ODE 

                        fx = fx.subs(x, y) 

                        gy = gy.subs(y, x) 

                    etaval = factor_terms(fx + gy) 

                    if etaval.is_Mul: 

                        etaval = Mul(*[arg for arg in etaval.args if arg.has(x, y)]) 

                    if odefac == hinv:  # Inverse ODE 

                        inf = {eta: etaval.subs(y, func), xi : S(0)} 

                    else: 

                        inf = {xi: etaval.subs(y, func), eta : S(0)} 

                    if not comp: 

                        return [inf] 

                    else: 

                        xieta.append(inf) 

 

        if xieta: 

            return xieta 

 

def lie_heuristic_abaco2_similar(match, comp=False): 

    r""" 

    This heuristic uses the following two assumptions on `\xi` and `\eta` 

 

    .. math:: \eta = g(x), \xi = f(x) 

 

    .. math:: \eta = f(y), \xi = g(y) 

 

    For the first assumption, 

 

    1. First `\frac{\frac{\partial h}{\partial y}}{\frac{\partial^{2} h}{ 

       \partial yy}}` is calculated. Let us say this value is A 

 

    2. If this is constant, then `h` is matched to the form `A(x) + B(x)e^{ 

       \frac{y}{C}}` then, `\frac{e^{\int \frac{A(x)}{C} \,dx}}{B(x)}` gives `f(x)` 

       and `A(x)*f(x)` gives `g(x)` 

 

    3. Otherwise `\frac{\frac{\partial A}{\partial X}}{\frac{\partial A}{ 

       \partial Y}} = \gamma` is calculated. If 

 

       a] `\gamma` is a function of `x` alone 

 

       b] `\frac{\gamma\frac{\partial h}{\partial y} - \gamma'(x) - \frac{ 

       \partial h}{\partial x}}{h + \gamma} = G` is a function of `x` alone. 

       then, `e^{\int G \,dx}` gives `f(x)` and `-\gamma*f(x)` gives `g(x)` 

 

    The second assumption holds good if `\frac{dy}{dx} = h(x, y)` is rewritten as 

    `\frac{dy}{dx} = \frac{1}{h(y, x)}` and the same properties of the first assumption 

    satisifes. After obtaining `f(x)` and `g(x)`, the coordinates are again 

    interchanged, to get `\xi` as `f(x^*)` and `\eta` as `g(y^*)` 

 

    References 

    ========== 

    - E.S. Cheb-Terrab, A.D. Roche, Symmetries and First Order 

      ODE Patterns, pp. 10 - pp. 12 

 

    """ 

 

    xieta = [] 

    h = match['h'] 

    hx = match['hx'] 

    hy = match['hy'] 

    func = match['func'] 

    hinv = match['hinv'] 

    x = func.args[0] 

    y = match['y'] 

    xi = Function('xi')(x, func) 

    eta = Function('eta')(x, func) 

 

    factor = cancel(h.diff(y)/h.diff(y, 2)) 

    factorx = factor.diff(x) 

    factory = factor.diff(y) 

    if not factor.has(x) and not factor.has(y): 

        A = Wild('A', exclude=[y]) 

        B = Wild('B', exclude=[y]) 

        C = Wild('C', exclude=[x, y]) 

        match = h.match(A + B*exp(y/C)) 

        try: 

            tau = exp(-integrate(match[A]/match[C]), x)/match[B] 

        except NotImplementedError: 

            pass 

        else: 

            gx = match[A]*tau 

            return [{xi: tau, eta: gx}] 

 

    else: 

        gamma = cancel(factorx/factory) 

        if not gamma.has(y): 

            tauint = cancel((gamma*hy - gamma.diff(x) - hx)/(h + gamma)) 

            if not tauint.has(y): 

                try: 

                    tau = exp(integrate(tauint, x)) 

                except NotImplementedError: 

                    pass 

                else: 

                    gx = -tau*gamma 

                    return [{xi: tau, eta: gx}] 

 

    factor = cancel(hinv.diff(y)/hinv.diff(y, 2)) 

    factorx = factor.diff(x) 

    factory = factor.diff(y) 

    if not factor.has(x) and not factor.has(y): 

        A = Wild('A', exclude=[y]) 

        B = Wild('B', exclude=[y]) 

        C = Wild('C', exclude=[x, y]) 

        match = h.match(A + B*exp(y/C)) 

        try: 

            tau = exp(-integrate(match[A]/match[C]), x)/match[B] 

        except NotImplementedError: 

            pass 

        else: 

            gx = match[A]*tau 

            return [{eta: tau.subs(x, func), xi: gx.subs(x, func)}] 

 

    else: 

        gamma = cancel(factorx/factory) 

        if not gamma.has(y): 

            tauint = cancel((gamma*hinv.diff(y) - gamma.diff(x) - hinv.diff(x))/( 

                hinv + gamma)) 

            if not tauint.has(y): 

                try: 

                    tau = exp(integrate(tauint, x)) 

                except NotImplementedError: 

                    pass 

                else: 

                    gx = -tau*gamma 

                    return [{eta: tau.subs(x, func), xi: gx.subs(x, func)}] 

 

 

def lie_heuristic_abaco2_unique_unknown(match, comp=False): 

    r""" 

    This heuristic assumes the presence of unknown functions or known functions 

    with non-integer powers. 

 

    1. A list of all functions and non-integer powers containing x and y 

    2. Loop over each element `f` in the list, find `\frac{\frac{\partial f}{\partial x}}{ 

       \frac{\partial f}{\partial x}} = R` 

 

       If it is separable in `x` and `y`, let `X` be the factors containing `x`. Then 

 

       a] Check if `\xi = X` and `\eta = -\frac{X}{R}` satisfy the PDE. If yes, then return 

          `\xi` and `\eta` 

       b] Check if `\xi = \frac{-R}{X}` and `\eta = -\frac{1}{X}` satisfy the PDE. 

           If yes, then return `\xi` and `\eta` 

 

       If not, then check if 

 

       a] :math:`\xi = -R,\eta = 1` 

 

       b] :math:`\xi = 1, \eta = -\frac{1}{R}` 

 

       are solutions. 

 

    References 

    ========== 

    - E.S. Cheb-Terrab, A.D. Roche, Symmetries and First Order 

      ODE Patterns, pp. 10 - pp. 12 

 

    """ 

 

    xieta = [] 

    h = match['h'] 

    hx = match['hx'] 

    hy = match['hy'] 

    func = match['func'] 

    hinv = match['hinv'] 

    x = func.args[0] 

    y = match['y'] 

    xi = Function('xi')(x, func) 

    eta = Function('eta')(x, func) 

 

    funclist = [] 

    for atom in h.atoms(Pow): 

        base, exp = atom.as_base_exp() 

        if base.has(x) and base.has(y): 

            if not exp.is_Integer: 

                funclist.append(atom) 

 

    for function in h.atoms(AppliedUndef): 

        syms = function.free_symbols 

        if x in syms and y in syms: 

            funclist.append(function) 

 

    for f in funclist: 

        frac = cancel(f.diff(y)/f.diff(x)) 

        sep = separatevars(frac, dict=True, symbols=[x, y]) 

        if sep and sep['coeff']: 

            xitry1 = sep[x] 

            etatry1 = -1/(sep[y]*sep['coeff']) 

            pde1 = etatry1.diff(y)*h - xitry1.diff(x)*h - xitry1*hx - etatry1*hy 

            if not simplify(pde1): 

                return [{xi: xitry1, eta: etatry1.subs(y, func)}] 

            xitry2 = 1/etatry1 

            etatry2 = 1/xitry1 

            pde2 = etatry2.diff(x) - (xitry2.diff(y))*h**2 - xitry2*hx - etatry2*hy 

            if not simplify(expand(pde2)): 

                return [{xi: xitry2.subs(y, func), eta: etatry2}] 

 

        else: 

            etatry = -1/frac 

            pde = etatry.diff(x) + etatry.diff(y)*h - hx - etatry*hy 

            if not simplify(pde): 

                return [{xi: S(1), eta: etatry.subs(y, func)}] 

            xitry = -frac 

            pde = -xitry.diff(x)*h -xitry.diff(y)*h**2 - xitry*hx -hy 

            if not simplify(expand(pde)): 

                return [{xi: xitry.subs(y, func), eta: S(1)}] 

 

 

def lie_heuristic_abaco2_unique_general(match, comp=False): 

    r""" 

    This heuristic finds if infinitesimals of the form `\eta = f(x)`, `\xi = g(y)` 

    without making any assumptions on `h`. 

 

    The complete sequence of steps is given in the paper mentioned below. 

 

    References 

    ========== 

    - E.S. Cheb-Terrab, A.D. Roche, Symmetries and First Order 

      ODE Patterns, pp. 10 - pp. 12 

 

    """ 

    xieta = [] 

    h = match['h'] 

    hx = match['hx'] 

    hy = match['hy'] 

    func = match['func'] 

    hinv = match['hinv'] 

    x = func.args[0] 

    y = match['y'] 

    xi = Function('xi')(x, func) 

    eta = Function('eta')(x, func) 

 

    C = S(0) 

    A = hx.diff(y) 

    B = hy.diff(y) + hy**2 

    C = hx.diff(x) - hx**2 

 

    if not (A and B and C): 

        return 

 

    Ax = A.diff(x) 

    Ay = A.diff(y) 

    Axy = Ax.diff(y) 

    Axx = Ax.diff(x) 

    Ayy = Ay.diff(y) 

    D = simplify(2*Axy + hx*Ay - Ax*hy + (hx*hy + 2*A)*A)*A - 3*Ax*Ay 

    if not D: 

        E1 = simplify(3*Ax**2 + ((hx**2 + 2*C)*A - 2*Axx)*A) 

        if E1: 

            E2 = simplify((2*Ayy + (2*B - hy**2)*A)*A - 3*Ay**2) 

            if not E2: 

                E3 = simplify( 

                    E1*((28*Ax + 4*hx*A)*A**3 - E1*(hy*A + Ay)) - E1.diff(x)*8*A**4) 

                if not E3: 

                    etaval = cancel((4*A**3*(Ax - hx*A) + E1*(hy*A - Ay))/(S(2)*A*E1)) 

                    if x not in etaval: 

                        try: 

                            etaval = exp(integrate(etaval, y)) 

                        except NotImplementedError: 

                            pass 

                        else: 

                            xival = -4*A**3*etaval/E1 

                            if y not in xival: 

                                return [{xi: xival, eta: etaval.subs(y, func)}] 

 

    else: 

        E1 = simplify((2*Ayy + (2*B - hy**2)*A)*A - 3*Ay**2) 

        if E1: 

            E2 = simplify( 

                4*A**3*D - D**2 + E1*((2*Axx - (hx**2 + 2*C)*A)*A - 3*Ax**2)) 

            if not E2: 

                E3 = simplify( 

                   -(A*D)*E1.diff(y) + ((E1.diff(x) - hy*D)*A + 3*Ay*D + 

                    (A*hx - 3*Ax)*E1)*E1) 

                if not E3: 

                    etaval = cancel(((A*hx - Ax)*E1 - (Ay + A*hy)*D)/(S(2)*A*D)) 

                    if x not in etaval: 

                        try: 

                            etaval = exp(integrate(etaval, y)) 

                        except NotImplementedError: 

                            pass 

                        else: 

                            xival = -E1*etaval/D 

                            if y not in xival: 

                                return [{xi: xival, eta: etaval.subs(y, func)}] 

 

 

def lie_heuristic_linear(match, comp=False): 

    r""" 

    This heuristic assumes 

 

    1. `\xi = ax + by + c` and 

    2. `\eta = fx + gy + h` 

 

    After substituting the following assumptions in the determining PDE, it 

    reduces to 

 

    .. math:: f + (g - a)h - bh^{2} - (ax + by + c)\frac{\partial h}{\partial x} 

                 - (fx + gy + c)\frac{\partial h}{\partial y} 

 

    Solving the reduced PDE obtained, using the method of characteristics, becomes 

    impractical. The method followed is grouping similar terms and solving the system 

    of linear equations obtained. The difference between the bivariate heuristic is that 

    `h` need not be a rational function in this case. 

 

    References 

    ========== 

    - E.S. Cheb-Terrab, A.D. Roche, Symmetries and First Order 

      ODE Patterns, pp. 10 - pp. 12 

 

    """ 

    xieta = [] 

    h = match['h'] 

    hx = match['hx'] 

    hy = match['hy'] 

    func = match['func'] 

    hinv = match['hinv'] 

    x = func.args[0] 

    y = match['y'] 

    xi = Function('xi')(x, func) 

    eta = Function('eta')(x, func) 

 

    coeffdict = {} 

    symbols = numbered_symbols("c", cls=Dummy) 

    symlist = [next(symbols) for i in islice(symbols, 6)] 

    C0, C1, C2, C3, C4, C5 = symlist 

    pde = C3 + (C4 - C0)*h -(C0*x + C1*y + C2)*hx - (C3*x + C4*y + C5)*hy - C1*h**2 

    pde, denom = pde.as_numer_denom() 

    pde = powsimp(expand(pde)) 

    if pde.is_Add: 

        terms = pde.args 

        for term in terms: 

            if term.is_Mul: 

                rem = Mul(*[m for m in term.args if not m.has(x, y)]) 

                xypart = term/rem 

                if xypart not in coeffdict: 

                    coeffdict[xypart] = rem 

                else: 

                    coeffdict[xypart] += rem 

            else: 

                if term not in coeffdict: 

                    coeffdict[term] = S(1) 

                else: 

                    coeffdict[term] += S(1) 

 

    sollist = coeffdict.values() 

    soldict = solve(sollist, symlist) 

    if soldict: 

        if isinstance(soldict, list): 

            soldict = soldict[0] 

        subval = soldict.values() 

        if any(t for t in subval): 

            onedict = dict(zip(symlist, [1]*6)) 

            xival = C0*x + C1*func + C2 

            etaval = C3*x + C4*func + C5 

            xival = xival.subs(soldict) 

            etaval = etaval.subs(soldict) 

            xival = xival.subs(onedict) 

            etaval = etaval.subs(onedict) 

            return [{xi: xival, eta: etaval}] 

 

 

def sysode_linear_2eq_order1(match_): 

    C1, C2, C3, C4 = symbols('C1:5') 

    x = match_['func'][0].func 

    y = match_['func'][1].func 

    func = match_['func'] 

    fc = match_['func_coeff'] 

    eq = match_['eq'] 

    r = dict() 

    t = list(list(eq[0].atoms(Derivative))[0].atoms(Symbol))[0] 

    for i in range(2): 

        eqs = 0 

        for terms in Add.make_args(eq[i]): 

            eqs += terms/fc[i,func[i],1] 

        eq[i] = eqs 

 

    # for equations Eq(a1*diff(x(t),t), a*x(t) + b*y(t) + k1) 

    # and Eq(a2*diff(x(t),t), c*x(t) + d*y(t) + k2) 

    r['a'] = -fc[0,x(t),0]/fc[0,x(t),1] 

    r['c'] = -fc[1,x(t),0]/fc[1,y(t),1] 

    r['b'] = -fc[0,y(t),0]/fc[0,x(t),1] 

    r['d'] = -fc[1,y(t),0]/fc[1,y(t),1] 

    forcing = [S(0),S(0)] 

    for i in range(2): 

        for j in Add.make_args(eq[i]): 

            if not j.has(x(t), y(t)): 

                forcing[i] += j 

    if not (forcing[0].has(t) or forcing[1].has(t)): 

        r['k1'] = forcing[0] 

        r['k2'] = forcing[1] 

    else: 

        raise NotImplementedError("Only homogeneous problems are supported" + 

                                  " (and constant inhomogeneity)") 

 

    if match_['type_of_equation'] == 'type1': 

        sol = _linear_2eq_order1_type1(x, y, t, r) 

    if match_['type_of_equation'] == 'type2': 

        gsol = _linear_2eq_order1_type1(x, y, t, r) 

        psol = _linear_2eq_order1_type2(x, y, t, r) 

        sol = [Eq(x(t), gsol[0].rhs+psol[0]), Eq(y(t), gsol[1].rhs+psol[1])] 

    if match_['type_of_equation'] == 'type3': 

        sol = _linear_2eq_order1_type3(x, y, t, r) 

    if match_['type_of_equation'] == 'type4': 

        sol = _linear_2eq_order1_type4(x, y, t, r) 

    if match_['type_of_equation'] == 'type5': 

        sol = _linear_2eq_order1_type5(x, y, t, r) 

    if match_['type_of_equation'] == 'type6': 

        sol = _linear_2eq_order1_type6(x, y, t, r) 

    if match_['type_of_equation'] == 'type7': 

        sol = _linear_2eq_order1_type7(x, y, t, r) 

    return sol 

 

def _linear_2eq_order1_type1(x, y, t, r): 

    r""" 

    It is classified under system of two linear homogeneous first-order constant-coefficient 

    ordinary differential equations. 

 

    The equations which come under this type are 

 

    .. math:: x' = ax + by, 

 

    .. math:: y' = cx + dy 

 

    The characteristics equation is written as 

 

    .. math:: \lambda^{2} + (a+d) \lambda + ad - bc = 0 

 

    and its discriminant is `D = (a-d)^{2} + 4bc`. There are several cases 

 

    1. Case when `ad - bc \neq 0`. The origin of coordinates, `x = y = 0`, 

    is the only stationary point; it is 

    - a node if `D = 0` 

    - a node if `D > 0` and `ad - bc > 0` 

    - a saddle if `D > 0` and `ad - bc < 0` 

    - a focus if `D < 0` and `a + d \neq 0` 

    - a centre if `D < 0` and `a + d \neq 0`. 

 

    1.1. If `D > 0`. The characteristic equation has two distinct real roots 

    `\lambda_1` and `\lambda_ 2` . The general solution of the system in question is expressed as 

 

    .. math:: x = C_1 b e^{\lambda_1 t} + C_2 b e^{\lambda_2 t} 

 

    .. math:: y = C_1 (\lambda_1 - a) e^{\lambda_1 t} + C_2 (\lambda_2 - a) e^{\lambda_2 t} 

 

    where `C_1` and `C_2` being arbitary constants 

 

    1.2. If `D < 0`. The characteristics equation has two conjugate 

    roots, `\lambda_1 = \sigma + i \beta` and `\lambda_2 = \sigma - i \beta`. 

    The general solution of the system is given by 

 

    .. math:: x = b e^{\sigma t} (C_1 \sin(\beta t) + C_2 \cos(\beta t)) 

 

    .. math:: y = e^{\sigma t} ([(\sigma - a) C_1 - \beta C_2] \sin(\beta t) + [\beta C_1 + (\sigma - a) C_2 \cos(\beta t)]) 

 

    1.3. If `D = 0` and `a \neq d`. The characteristic equation has 

    two equal roots, `\lambda_1 = \lambda_2`. The general solution of the system is written as 

 

    .. math:: x = 2b (C_1 + \frac{C_2}{a-d} + C_2 t) e^{\frac{a+d}{2} t} 

 

    .. math:: y = [(d - a) C_1 + C_2 + (d - a) C_2 t] e^{\frac{a+d}{2} t} 

 

    1.4. If `D = 0` and `a = d \neq 0` and `b = 0` 

 

    .. math:: x = C_1 e^{a t} , y = (c C_1 t + C_2) e^{a t} 

 

    1.5. If `D = 0` and `a = d \neq 0` and `c = 0` 

 

    .. math:: x = (b C_1 t + C_2) e^{a t} , y = C_1 e^{a t} 

 

    2. Case when `ad - bc = 0` and `a^{2} + b^{2} > 0`. The whole straight 

    line `ax + by = 0` consists of singular points. The orginal system of differential 

    equaitons can be rewritten as 

 

    .. math:: x' = ax + by , y' = k (ax + by) 

 

    2.1 If `a + bk \neq 0`, solution will be 

 

    .. math:: x = b C_1 + C_2 e^{(a + bk) t} , y = -a C_1 + k C_2 e^{(a + bk) t} 

 

    2.2 If `a + bk = 0`, solution will be 

 

    .. math:: x = C_1 (bk t - 1) + b C_2 t , y = k^{2} b C_1 t + (b k^{2} t + 1) C_2 

 

    """ 

    # FIXME: at least some of these can fail to give two linearly 

    # independent solutions e.g., because they make assumptions about 

    # zero/nonzero of certain coefficients.  I've "fixed" one and 

    # raised NotImplementedError in another.  I think this should probably 

    # just be re-written in terms of eigenvectors... 

 

    l = Dummy('l') 

    C1, C2, C3, C4 = symbols('C1:5') 

    l1 = RootOf(l**2 - (r['a']+r['d'])*l + r['a']*r['d'] - r['b']*r['c'], l, 0) 

    l2 = RootOf(l**2 - (r['a']+r['d'])*l + r['a']*r['d'] - r['b']*r['c'], l, 1) 

    D = (r['a'] - r['d'])**2 + 4*r['b']*r['c'] 

    if (r['a']*r['d'] - r['b']*r['c']) != 0: 

        if D > 0: 

            if r['b'].is_zero: 

                # tempting to use this in all cases, but does not guarantee linearly independent eigenvectors 

                gsol1 = C1*(l1 - r['d'] + r['b'])*exp(l1*t) + C2*(l2 - r['d'] + r['b'])*exp(l2*t) 

                gsol2 = C1*(l1 - r['a'] + r['c'])*exp(l1*t) + C2*(l2 - r['a'] + r['c'])*exp(l2*t) 

            else: 

                gsol1 = C1*r['b']*exp(l1*t) + C2*r['b']*exp(l2*t) 

                gsol2 = C1*(l1 - r['a'])*exp(l1*t) + C2*(l2 - r['a'])*exp(l2*t) 

        if D < 0: 

            sigma = re(l1) 

            if im(l1).is_positive: 

                beta = im(l1) 

            else: 

                beta = im(l2) 

            if r['b'].is_zero: 

                raise NotImplementedError('b == 0 case not implemented') 

            gsol1 = r['b']*exp(sigma*t)*(C1*sin(beta*t)+C2*cos(beta*t)) 

            gsol2 = exp(sigma*t)*(((C1*(sigma-r['a'])-C2*beta)*sin(beta*t)+(C1*beta+(sigma-r['a'])*C2)*cos(beta*t))) 

        if D == 0: 

            if r['a']!=r['d']: 

                gsol1 = 2*r['b']*(C1 + C2/(r['a']-r['d'])+C2*t)*exp((r['a']+r['d'])*t/2) 

                gsol2 = ((r['d']-r['a'])*C1+C2+(r['d']-r['a'])*C2*t)*exp((r['a']+r['d'])*t/2) 

            if r['a']==r['d'] and r['a']!=0 and r['b']==0: 

                gsol1 = C1*exp(r['a']*t) 

                gsol2 = (r['c']*C1*t+C2)*exp(r['a']*t) 

            if r['a']==r['d'] and r['a']!=0 and r['c']==0: 

                gsol1 = (r['b']*C1*t+C2)*exp(r['a']*t) 

                gsol2 = C1*exp(r['a']*t) 

    elif (r['a']*r['d'] - r['b']*r['c']) == 0 and (r['a']**2+r['b']**2) > 0: 

        k = r['c']/r['a'] 

        if r['a']+r['b']*k != 0: 

            gsol1 = r['b']*C1 + C2*exp((r['a']+r['b']*k)*t) 

            gsol2 = -r['a']*C1 + k*C2*exp((r['a']+r['b']*k)*t) 

        else: 

            gsol1 = C1*(r['b']*k*t-1)+r['b']*C2*t 

            gsol2 = k**2*r['b']*C1*t+(r['b']*k**2*t+1)*C2 

    return [Eq(x(t), gsol1), Eq(y(t), gsol2)] 

 

def _linear_2eq_order1_type2(x, y, t, r): 

    r""" 

    The equations of this type are 

 

    .. math:: x' = ax + by + k1 , y' = cx + dy + k2 

 

    The general solution of this system is given by sum of its particular solution and the 

    general solution of the corresponding homogeneous system is obtained from type1. 

 

    1. When `ad - bc \neq 0`. The particular solution will be 

    `x = x_0` and `y = y_0` where `x_0` and `y_0` are determined by solving linear system of equations 

 

    .. math:: a x_0 + b y_0 + k1 = 0 , c x_0 + d y_0 + k2 = 0 

 

    2. When `ad - bc = 0` and `a^{2} + b^{2} > 0`. In this case, the system of equation becomes 

 

    .. math:: x' = ax + by + k_1 , y' = k (ax + by) + k_2 

 

    2.1 If `\sigma = a + bk \neq 0`, particular solution is given by 

 

    .. math:: x = b \sigma^{-1} (c_1 k - c_2) t - \sigma^{-2} (a c_1 + b c_2) 

 

    .. math:: y = kx + (c_2 - c_1 k) t 

 

    2.2 If `\sigma = a + bk = 0`, particular solution is given by 

 

    .. math:: x = \frac{1}{2} b (c_2 - c_1 k) t^{2} + c_1 t 

 

    .. math:: y = kx + (c_2 - c_1 k) t 

 

    """ 

    r['k1'] = -r['k1']; r['k2'] = -r['k2'] 

    if (r['a']*r['d'] - r['b']*r['c']) != 0: 

        x0, y0 = symbols('x0, y0', cls=Dummy) 

        sol = solve((r['a']*x0+r['b']*y0+r['k1'], r['c']*x0+r['d']*y0+r['k2']), x0, y0) 

        psol = [sol[x0], sol[y0]] 

    elif (r['a']*r['d'] - r['b']*r['c']) == 0 and (r['a']**2+r['b']**2) > 0: 

        k = r['c']/r['a'] 

        sigma = r['a'] + r['b']*k 

        if sigma != 0: 

            sol1 = r['b']*sigma**-1*(r['k1']*k-r['k2'])*t - sigma**-2*(r['a']*r['k1']+r['b']*r['k2']) 

            sol2 = k*sol1 + (r['k2']-r['k1']*k)*t 

        else: 

            # FIXME: a previous typo fix shows this is not covered by tests 

            sol1 = r['b']*(r['k2']-r['k1']*k)*t**2 + r['k1']*t 

            sol2 = k*sol1 + (r['k2']-r['k1']*k)*t 

        psol = [sol1, sol2] 

    return psol 

 

def _linear_2eq_order1_type3(x, y, t, r): 

    r""" 

    The equations of this type of ode are 

 

    .. math:: x' = f(t) x + g(t) y 

 

    .. math:: y' = g(t) x + f(t) y 

 

    The solution of such equations is given by 

 

    .. math:: x = e^{F} (C_1 e^{G} + C_2 e^{-G}) , y = e^{F} (C_1 e^{G} - C_2 e^{-G}) 

 

    where `C_1` and `C_2` are arbitary constants, and 

 

    .. math:: F = \int f(t) \,dt , G = \int g(t) \,dt 

 

    """ 

    C1, C2, C3, C4 = symbols('C1:5') 

    F = Integral(r['a'], t) 

    G = Integral(r['b'], t) 

    sol1 = exp(F)*(C1*exp(G) + C2*exp(-G)) 

    sol2 = exp(F)*(C1*exp(G) - C2*exp(-G)) 

    return [Eq(x(t), sol1), Eq(y(t), sol2)] 

 

def _linear_2eq_order1_type4(x, y, t, r): 

    r""" 

    The equations of this type of ode are . 

 

    .. math:: x' = f(t) x + g(t) y 

 

    .. math:: y' = -g(t) x + f(t) y 

 

    The solution is given by 

 

    .. math:: x = F (C_1 \cos(G) + C_2 \sin(G)), y = F (-C_1 \sin(G) + C_2 \cos(G)) 

 

    where `C_1` and `C_2` are arbitary constants, and 

 

    .. math:: F = \int f(t) \,dt , G = \int g(t) \,dt 

 

    """ 

    C1, C2, C3, C4 = symbols('C1:5') 

    if r['b'] == -r['c']: 

        F = exp(Integral(r['a'], t)) 

        G = Integral(r['b'], t) 

        sol1 = F*(C1*cos(G) + C2*sin(G)) 

        sol2 = F*(-C1*sin(G) + C2*cos(G)) 

    elif r['d'] == -r['a']: 

        F = exp(Integral(r['c'], t)) 

        G = Integral(r['d'], t) 

        sol1 = F*(-C1*sin(G) + C2*cos(G)) 

        sol2 = F*(C1*cos(G) + C2*sin(G)) 

    return [Eq(x(t), sol1), Eq(y(t), sol2)] 

 

def _linear_2eq_order1_type5(x, y, t, r): 

    r""" 

    The equations of this type of ode are . 

 

    .. math:: x' = f(t) x + g(t) y 

 

    .. math:: y' = a g(t) x + [f(t) + b g(t)] y 

 

    The transformation of 

 

    .. math:: x = e^{\int f(t) \,dt} u , y = e^{\int f(t) \,dt} v , T = \int g(t) \,dt 

 

    leads to a system of constant coefficient linear differential equations 

 

    .. math:: u'(T) = v , v'(T) = au + bv 

 

    """ 

    C1, C2, C3, C4 = symbols('C1:5') 

    u, v = symbols('u, v', function=True) 

    T = Symbol('T') 

    if not cancel(r['c']/r['b']).has(t): 

        p = cancel(r['c']/r['b']) 

        q = cancel((r['d']-r['a'])/r['b']) 

        eq = (Eq(diff(u(T),T), v(T)), Eq(diff(v(T),T), p*u(T)+q*v(T))) 

        sol = dsolve(eq) 

        sol1 = exp(Integral(r['a'], t))*sol[0].rhs.subs(T, Integral(r['b'],t)) 

        sol2 = exp(Integral(r['a'], t))*sol[1].rhs.subs(T, Integral(r['b'],t)) 

    if not cancel(r['a']/r['d']).has(t): 

        p = cancel(r['a']/r['d']) 

        q = cancel((r['b']-r['c'])/r['d']) 

        sol = dsolve(Eq(diff(u(T),T), v(T)), Eq(diff(v(T),T), p*u(T)+q*v(T))) 

        sol1 = exp(Integral(r['c'], t))*sol[1].rhs.subs(T, Integral(r['d'],t)) 

        sol2 = exp(Integral(r['c'], t))*sol[0].rhs.subs(T, Integral(r['d'],t)) 

    return [Eq(x(t), sol1), Eq(y(t), sol2)] 

 

def _linear_2eq_order1_type6(x, y, t, r): 

    r""" 

    The equations of this type of ode are . 

 

    .. math:: x' = f(t) x + g(t) y 

 

    .. math:: y' = a [f(t) + a h(t)] x + a [g(t) - h(t)] y 

 

    This is solved by first multiplying the first equation by `-a` and adding 

    it to the second equation to obtain 

 

    .. math:: y' - a x' = -a h(t) (y - a x) 

 

    Setting `U = y - ax` and integrating the equation we arrive at 

 

    .. math:: y - ax = C_1 e^{-a \int h(t) \,dt} 

 

    and on substituing the value of y in first equation give rise to first order ODEs. After solving for 

    `x`, we can obtain `y` by substituting the value of `x` in second equation. 

 

    """ 

    C1, C2, C3, C4 = symbols('C1:5') 

    p = 0 

    q = 0 

    p1 = cancel(r['c']/cancel(r['c']/r['d']).as_numer_denom()[0]) 

    p2 = cancel(r['a']/cancel(r['a']/r['b']).as_numer_denom()[0]) 

    for n, i in enumerate([p1, p2]): 

        for j in Mul.make_args(collect_const(i)): 

            if not j.has(t): 

                q = j 

            if q!=0 and n==0: 

                if ((r['c']/j - r['a'])/(r['b'] - r['d']/j)) == j: 

                    p = 1 

                    s = j 

                    break 

            if q!=0 and n==1: 

                if ((r['a']/j - r['c'])/(r['d'] - r['b']/j)) == j: 

                    p = 2 

                    s = j 

                    break 

    if p == 1: 

        equ = diff(x(t),t) - r['a']*x(t) - r['b']*(s*x(t) + C1*exp(-s*Integral(r['b'] - r['d']/s, t))) 

        hint1 = classify_ode(equ)[1] 

        sol1 = dsolve(equ, hint=hint1+'_Integral').rhs 

        sol2 = s*sol1 + C1*exp(-s*Integral(r['b'] - r['d']/s, t)) 

    elif p ==2: 

        equ = diff(y(t),t) - r['c']*y(t) - r['d']*s*y(t) + C1*exp(-s*Integral(r['d'] - r['b']/s, t)) 

        hint1 = classify_ode(equ)[1] 

        sol2 = dsolve(equ, hint=hint1+'_Integral').rhs 

        sol1 = s*sol2 + C1*exp(-s*Integral(r['d'] - r['b']/s, t)) 

    return [Eq(x(t), sol1), Eq(y(t), sol2)] 

 

def _linear_2eq_order1_type7(x, y, t, r): 

    r""" 

    The equations of this type of ode are . 

 

    .. math:: x' = f(t) x + g(t) y 

 

    .. math:: y' = h(t) x + p(t) y 

 

    Differentiating the first equation and substituting the value of `y` 

    from second equation will give a second-order linear equation 

 

    .. math:: g x'' - (fg + gp + g') x' + (fgp - g^{2} h + f g' - f' g) x = 0 

 

    This above equation can be easily integrated if following conditions are satisfied. 

 

    1. `fgp - g^{2} h + f g' - f' g = 0` 

 

    2. `fgp - g^{2} h + f g' - f' g = ag, fg + gp + g' = bg` 

 

    If first condition is satisfied then it is solved by current dsolve solver and in second case it becomes 

    a constant cofficient differential equation which is also solved by current solver. 

 

    Otherwise if the above condition fails then, 

    a particular solution is assumed as `x = x_0(t)` and `y = y_0(t)` 

    Then the general solution is expressed as 

 

    .. math:: x = C_1 x_0(t) + C_2 x_0(t) \int \frac{g(t) F(t) P(t)}{x_0^{2}(t)} \,dt 

 

    .. math:: y = C_1 y_0(t) + C_2 [\frac{F(t) P(t)}{x_0(t)} + y_0(t) \int \frac{g(t) F(t) P(t)}{x_0^{2}(t)} \,dt] 

 

    where C1 and C2 are arbitary constants and 

 

    .. math:: F(t) = e^{\int f(t) \,dt} , P(t) = e^{\int p(t) \,dt} 

 

    """ 

    C1, C2, C3, C4 = symbols('C1:5') 

    e1 = r['a']*r['b']*r['c'] - r['b']**2*r['c'] + r['a']*diff(r['b'],t) - diff(r['a'],t)*r['b'] 

    e2 = r['a']*r['c']*r['d'] - r['b']*r['c']**2 + diff(r['c'],t)*r['d'] - r['c']*diff(r['d'],t) 

    m1 = r['a']*r['b'] + r['b']*r['d'] + diff(r['b'],t) 

    m2 = r['a']*r['c'] + r['c']*r['d'] + diff(r['c'],t) 

    if e1 == 0: 

        sol1 = dsolve(r['b']*diff(x(t),t,t) - m1*diff(x(t),t)).rhs 

        sol2 = dsolve(diff(y(t),t) - r['c']*sol1 - r['d']*y(t)).rhs 

    elif e2 == 0: 

        sol2 = dsolve(r['c']*diff(y(t),t,t) - m2*diff(y(t),t)).rhs 

        sol1 = dsolve(diff(x(t),t) - r['a']*x(t) - r['b']*sol2).rhs 

    elif not (e1/r['b']).has(t) and not (m1/r['b']).has(t): 

        sol1 = dsolve(diff(x(t),t,t) - (m1/r['b'])*diff(x(t),t) - (e1/r['b'])*x(t)).rhs 

        sol2 = dsolve(diff(y(t),t) - r['c']*sol1 - r['d']*y(t)).rhs 

    elif not (e2/r['c']).has(t) and not (m2/r['c']).has(t): 

        sol2 = dsolve(diff(y(t),t,t) - (m2/r['c'])*diff(y(t),t) - (e2/r['c'])*y(t)).rhs 

        sol1 = dsolve(diff(x(t),t) - r['a']*x(t) - r['b']*sol2).rhs 

    else: 

        x0, y0 = symbols('x0, y0')              #x0 and y0 being particular solutions 

        F = exp(Integral(r['a'],t)) 

        P = exp(Integral(r['d'],t)) 

        sol1 = C1*x0 + C2*x0*Integral(r['b']*F*P/x0**2, t) 

        sol2 = C1*y0 + C2(F*P/x0 + y0*Integral(r['b']*F*P/x0**2, t)) 

    return [Eq(x(t), sol1), Eq(y(t), sol2)] 

 

 

def sysode_linear_2eq_order2(match_): 

    C1, C2, C3, C4 = symbols('C1:5') 

    x = match_['func'][0].func 

    y = match_['func'][1].func 

    func = match_['func'] 

    fc = match_['func_coeff'] 

    eq = match_['eq'] 

    r = dict() 

    t = list(list(eq[0].atoms(Derivative))[0].atoms(Symbol))[0] 

    for i in range(2): 

        eqs = 0 

        for terms in Add.make_args(eq[i]): 

            eqs += terms/fc[i,func[i],2] 

        eq[i] = eqs 

    # for equations Eq(diff(x(t),t,t), a1*diff(x(t),t)+b1*diff(y(t),t)+c1*x(t)+d1*y(t)+e1) 

    # and Eq(a2*diff(y(t),t,t), a2*diff(x(t),t)+b2*diff(y(t),t)+c2*x(t)+d2*y(t)+e2) 

    r['a1'] = -fc[0,x(t),1]/fc[0,x(t),2] ; r['a2'] = -fc[1,x(t),1]/fc[1,y(t),2] 

    r['b1'] = -fc[0,y(t),1]/fc[0,x(t),2] ; r['b2'] = -fc[1,y(t),1]/fc[1,y(t),2] 

    r['c1'] = -fc[0,x(t),0]/fc[0,x(t),2] ; r['c2'] = -fc[1,x(t),0]/fc[1,y(t),2] 

    r['d1'] = -fc[0,y(t),0]/fc[0,x(t),2] ; r['d2'] = -fc[1,y(t),0]/fc[1,y(t),2] 

    const = [S(0), S(0)] 

    for i in range(2): 

        for j in Add.make_args(eq[i]): 

            if not (j.has(x(t)) or j.has(y(t))): 

                const[i] += j 

    r['e1'] = -const[0] 

    r['e2'] = -const[1] 

    if match_['type_of_equation'] == 'type1': 

        sol = _linear_2eq_order2_type1(x, y, t, r) 

    elif match_['type_of_equation'] == 'type2': 

        gsol = _linear_2eq_order2_type1(x, y, t, r) 

        psol = _linear_2eq_order2_type2(x, y, t, r) 

        sol = [Eq(x(t), gsol[0].rhs+psol[0]), Eq(y(t), gsol[1].rhs+psol[1])] 

    elif match_['type_of_equation'] == 'type3': 

        sol = _linear_2eq_order2_type3(x, y, t, r) 

    elif match_['type_of_equation'] == 'type4': 

        sol = _linear_2eq_order2_type4(x, y, t, r) 

    elif match_['type_of_equation'] == 'type5': 

        sol = _linear_2eq_order2_type5(x, y, t, r) 

    elif match_['type_of_equation'] == 'type6': 

        sol = _linear_2eq_order2_type6(x, y, t, r) 

    elif match_['type_of_equation'] == 'type7': 

        sol = _linear_2eq_order2_type7(x, y, t, r) 

    elif match_['type_of_equation'] == 'type8': 

        sol = _linear_2eq_order2_type8(x, y, t, r) 

    elif match_['type_of_equation'] == 'type9': 

        sol = _linear_2eq_order2_type9(x, y, t, r) 

    elif match_['type_of_equation'] == 'type10': 

        sol = _linear_2eq_order2_type10(x, y, t, r) 

    elif match_['type_of_equation'] == 'type11': 

        sol = _linear_2eq_order2_type11(x, y, t, r) 

    return sol 

 

def _linear_2eq_order2_type1(x, y, t, r): 

    r""" 

    System of two constant-coefficient second-order linear homogeneous differential equations 

 

    .. math:: x'' = ax + by 

 

    .. math:: y'' = cx + dy 

 

    The charecteristic equation for above equations 

 

    .. math:: \lambda^4 - (a + d) \lambda^2 + ad - bc = 0 

 

    whose discriminant is `D = (a - d)^2 + 4bc \neq 0` 

 

    1. When `ad - bc \neq 0` 

 

    1.1. If `D \neq 0`. The characteristic equation has four distict roots, `\lambda_1, \lambda_2, \lambda_3, \lambda_4`. 

    The general solution of the system is 

 

    .. math:: x = C_1 b e^{\lambda_1 t} + C_2 b e^{\lambda_2 t} + C_3 b e^{\lambda_3 t} + C_4 b e^{\lambda_4 t} 

 

    .. math:: y = C_1 (\lambda_1^{2} - a) e^{\lambda_1 t} + C_2 (\lambda_2^{2} - a) e^{\lambda_2 t} + C_3 (\lambda_3^{2} - a) e^{\lambda_3 t} + C_4 (\lambda_4^{2} - a) e^{\lambda_4 t} 

 

    where `C_1,..., C_4` are arbitary constants. 

 

    1.2. If `D = 0` and `a \neq d`: 

 

    .. math:: x = 2 C_1 (bt + \frac{2bk}{a - d}) e^{\frac{kt}{2}} + 2 C_2 (bt + \frac{2bk}{a - d}) e^{\frac{-kt}{2}} + 2b C_3 t e^{\frac{kt}{2}} + 2b C_4 t e^{\frac{-kt}{2}} 

 

    .. math:: y = C_1 (d - a) t e^{\frac{kt}{2}} + C_2 (d - a) t e^{\frac{-kt}{2}} + C_3 [(d - a) t + 2k] e^{\frac{kt}{2}} + C_4 [(d - a) t - 2k] e^{\frac{-kt}{2}} 

 

    where `C_1,..., C_4` are arbitary constants and `k = \sqrt{2 (a + d)}` 

 

    1.3. If `D = 0` and `a = d \neq 0` and `b = 0`: 

 

    .. math:: x = 2 \sqrt{a} C_1 e^{\sqrt{a} t} + 2 \sqrt{a} C_2 e^{-\sqrt{a} t} 

 

    .. math:: y = c C_1 t e^{\sqrt{a} t} - c C_2 t e^{-\sqrt{a} t} + C_3 e^{\sqrt{a} t} + C_4 e^{-\sqrt{a} t} 

 

    1.4. If `D = 0` and `a = d \neq 0` and `c = 0`: 

 

    .. math:: x = b C_1 t e^{\sqrt{a} t} - b C_2 t e^{-\sqrt{a} t} + C_3 e^{\sqrt{a} t} + C_4 e^{-\sqrt{a} t} 

 

    .. math:: y = 2 \sqrt{a} C_1 e^{\sqrt{a} t} + 2 \sqrt{a} C_2 e^{-\sqrt{a} t} 

 

    2. When `ad - bc = 0` and `a^2 + b^2 > 0`. Then the original system becomes 

 

    .. math:: x'' = ax + by 

 

    .. math:: y'' = k (ax + by) 

 

    2.1. If `a + bk \neq 0`: 

 

    .. math:: x = C_1 e^{t \sqrt{a + bk}} + C_2 e^{-t \sqrt{a + bk}} + C_3 bt + C_4 b 

 

    .. math:: y = C_1 k e^{t \sqrt{a + bk}} + C_2 k e^{-t \sqrt{a + bk}} - C_3 at - C_4 a 

 

    2.2. If `a + bk = 0`: 

 

    .. math:: x = C_1 b t^3 + C_2 b t^2 + C_3 t + C_4 

 

    .. math:: y = kx + 6 C_1 t + 2 C_2 

 

    """ 

    r['a'] = r['c1'] 

    r['b'] = r['d1'] 

    r['c'] = r['c2'] 

    r['d'] = r['d2'] 

    l = Symbol('l') 

    C1, C2, C3, C4 = symbols('C1:5') 

    chara_eq = l**4 - (r['a']+r['d'])*l**2 + r['a']*r['d'] - r['b']*r['c'] 

    l1 = RootOf(chara_eq, 0) 

    l2 = RootOf(chara_eq, 1) 

    l3 = RootOf(chara_eq, 2) 

    l4 = RootOf(chara_eq, 3) 

    D = (r['a'] - r['d'])**2 + 4*r['b']*r['c'] 

    if (r['a']*r['d'] - r['b']*r['c']) != 0: 

        if D != 0: 

            gsol1 = C1*r['b']*exp(l1*t) + C2*r['b']*exp(l2*t) + C3*r['b']*exp(l3*t) \ 

            + C4*r['b']*exp(l4*t) 

            gsol2 = C1*(l1**2-r['a'])*exp(l1*t) + C2*(l2**2-r['a'])*exp(l2*t) + \ 

            C3*(l3**2-r['a'])*exp(l3*t) + C4*(l4**2-r['a'])*exp(l4*t) 

        else: 

            if r['a'] != r['d']: 

                k = sqrt(2*(r['a']+r['d'])) 

                mid = r['b']*t+2*r['b']*k/(r['a']-r['d']) 

                gsol1 = 2*C1*mid*exp(k*t/2) + 2*C2*mid*exp(-k*t/2) + \ 

                2*r['b']*C3*t*exp(k*t/2) + 2*r['b']*C4*t*exp(-k*t/2) 

                gsol2 = C1*(r['d']-r['a'])*t*exp(k*t/2) + C2*(r['d']-r['a'])*t*exp(-k*t/2) + \ 

                C3*((r['d']-r['a'])*t+2*k)*exp(k*t/2) + C4*((r['d']-r['a'])*t-2*k)*exp(-k*t/2) 

            elif r['a'] == r['d'] != 0 and r['b'] == 0: 

                sa = sqrt(r['a']) 

                gsol1 = 2*sa*C1*exp(sa*t) + 2*sa*C2*exp(-sa*t) 

                gsol2 = r['c']*C1*t*exp(sa*t)-r['c']*C2*t*exp(-sa*t)+C3*exp(sa*t)+C4*exp(-sa*t) 

            elif r['a'] == r['d'] != 0 and r['c'] == 0: 

                sa = sqrt(r['a']) 

                gsol1 = r['b']*C1*t*exp(sa*t)-r['b']*C2*t*exp(-sa*t)+C3*exp(sa*t)+C4*exp(-sa*t) 

                gsol2 = 2*sa*C1*exp(sa*t) + 2*sa*C2*exp(-sa*t) 

    elif (r['a']*r['d'] - r['b']*r['c']) == 0 and (r['a']**2 + r['b']**2) > 0: 

        k = r['c']/r['a'] 

        if r['a'] + r['b']*k != 0: 

            mid = sqrt(r['a'] + r['b']*k) 

            gsol1 = C1*exp(mid*t) + C2*exp(-mid*t) + C3*r['b']*t + C4*r['b'] 

            gsol2 = C1*k*exp(mid*t) + C2*k*exp(-mid*t) - C3*r['a']*t - C4*r['a'] 

        else: 

            gsol1 = C1*r['b']*t**3 + C2*r['b']*t**2 + C3*t + C4 

            gsol2 = k*gsol1 + 6*C1*t + 2*C2 

    return [Eq(x(t), gsol1), Eq(y(t), gsol2)] 

 

def _linear_2eq_order2_type2(x, y, t, r): 

    r""" 

    The equations in this type are 

 

    .. math:: x'' = a_1 x + b_1 y + c_1 

 

    .. math:: y'' = a_2 x + b_2 y + c_2 

 

    The general solution of this system is given by the sum of its particular solution 

    and the general solution of the homogeneous system. The general solution is given 

    by the linear system of 2 equation of order 2 and type 1 

 

    1. If `a_1 b_2 - a_2 b_1 \neq 0`. A particular solution will be `x = x_0` and `y = y_0` 

    where the constants `x_0` and `y_0` are determined by solving the linear algebraic system 

 

    .. math:: a_1 x_0 + b_1 y_0 + c_1 = 0, a_2 x_0 + b_2 y_0 + c_2 = 0 

 

    2. If `a_1 b_2 - a_2 b_1 = 0` and `a_1^2 + b_1^2 > 0`. In this case, the system in question becomes 

 

    .. math:: x'' = ax + by + c_1, y'' = k (ax + by) + c_2 

 

    2.1. If `\sigma = a + bk \neq 0`, the particular solution will be 

 

    .. math:: x = \frac{1}{2} b \sigma^{-1} (c_1 k - c_2) t^2 - \sigma^{-2} (a c_1 + b c_2) 

 

    .. math:: y = kx + \frac{1}{2} (c_2 - c_1 k) t^2 

 

    2.2. If `\sigma = a + bk = 0`, the particular solution will be 

 

    .. math:: x = \frac{1}{24} b (c_2 - c_1 k) t^4 + \frac{1}{2} c_1 t^2 

 

    .. math:: y = kx + \frac{1}{2} (c_2 - c_1 k) t^2 

 

    """ 

    x0, y0 = symbols('x0, y0') 

    if r['c1']*r['d2'] - r['c2']*r['d1'] != 0: 

        sol = solve((r['c1']*x0+r['d1']*y0+r['e1'], r['c2']*x0+r['d2']*y0+r['e2']), x0, y0) 

        psol = [sol[x0], sol[y0]] 

    elif r['c1']*r['d2'] - r['c2']*r['d1'] == 0 and (r['c1']**2 + r['d1']**2) > 0: 

        k = r['c2']/r['c1'] 

        sig = r['c1'] + r['d1']*k 

        if sig != 0: 

            psol1 = r['d1']*sig**-1*(r['e1']*k-r['e2'])*t**2/2 - \ 

            sig**-2*(r['c1']*r['e1']+r['d1']*r['e2']) 

            psol2 = k*psol1  + (r['e2'] - r['e1']*k)*t**2/2 

            psol = [psol1, psol2] 

        else: 

            psol1 = r['d1']*(r['e2']-r['e1']*k)*t**4/24 + r['e1']*t**2/2 

            psol2 = k*psol1 + (r['e2']-r['e1']*k)*t**2/2 

            psol = [psol1, psol2] 

    return psol 

 

def _linear_2eq_order2_type3(x, y, t, r): 

    r""" 

    These type of equation is used for describing the horizontal motion of a pendulum 

    taking into account the Earth rotation. 

    The solution is given with `a^2 + 4b > 0`: 

 

    .. math:: x = C_1 \cos(\alpha t) + C_2 \sin(\alpha t) + C_3 \cos(\beta t) + C_4 \sin(\beta t) 

 

    .. math:: y = -C_1 \sin(\alpha t) + C_2 \cos(\alpha t) - C_3 \sin(\beta t) + C_4 \cos(\beta t) 

 

    where `C_1,...,C_4` and 

 

    .. math:: \alpha = \frac{1}{2} a + \frac{1}{2} \sqrt{a^2 + 4b}, \beta = \frac{1}{2} a - \frac{1}{2} \sqrt{a^2 + 4b} 

 

    """ 

    C1, C2, C3, C4 = symbols('C1:5') 

    if r['b1']**2 - 4*r['c1'] > 0: 

        r['a'] = r['b1'] ; r['b'] = -r['c1'] 

        alpha = r['a']/2 + sqrt(r['a']**2 + 4*r['b'])/2 

        beta = r['a']/2 - sqrt(r['a']**2 + 4*r['b'])/2 

        sol1 = C1*cos(alpha*t) + C2*sin(alpha*t) + C3*cos(beta*t) + C4*sin(beta*t) 

        sol2 = -C1*sin(alpha*t) + C2*cos(alpha*t) - C3*sin(beta*t) + C4*cos(beta*t) 

    return [Eq(x(t), sol1), Eq(y(t), sol2)] 

 

def _linear_2eq_order2_type4(x, y, t, r): 

    r""" 

    These equations are found in the theory of oscillations 

 

    .. math:: x'' + a_1 x' + b_1 y' + c_1 x + d_1 y = k_1 e^{i \omega t} 

 

    .. math:: y'' + a_2 x' + b_2 y' + c_2 x + d_2 y = k_2 e^{i \omega t} 

 

    The general solution of this linear nonhomogeneous system of constant-coefficient 

    differential equations is given by the sum of its particular solution and the 

    general solution of the corresponding homogeneous system (with `k_1 = k_2 = 0`) 

 

    1. A particular solution is obtained by the method of undetermined coefficients: 

 

    .. math:: x = A_* e^{i \omega t}, y = B_* e^{i \omega t} 

 

    On substituting these expressions into the original system of differential equations, 

    one arrive at a linear nonhomogeneous system of algebraic equations for the 

    coefficients `A` and `B`. 

 

    2. The general solution of the homogeneous system of differential equations is determined 

    by a linear combination of linearly independent particular solutions determined by 

    the method of undetermined coefficients in the form of exponentials: 

 

    .. math:: x = A e^{\lambda t}, y = B e^{\lambda t} 

 

    On substituting these expressions into the original system and colleting the 

    coefficients of the unknown `A` and `B`, one obtains 

 

    .. math:: (\lambda^{2} + a_1 \lambda + c_1) A + (b_1 \lambda + d_1) B = 0 

 

    .. math:: (a_2 \lambda + c_2) A + (\lambda^{2} + b_2 \lambda + d_2) B = 0 

 

    The determinant of this system must vanish for nontrivial solutions A, B to exist. 

    This requirement results in the following characteristic equation for `\lambda` 

 

    .. math:: (\lambda^2 + a_1 \lambda + c_1) (\lambda^2 + b_2 \lambda + d_2) - (b_1 \lambda + d_1) (a_2 \lambda + c_2) = 0 

 

    If all roots `k_1,...,k_4` of this equation are distict, the general solution of the original 

    system of the differential equations has the form 

 

    .. math:: x = C_1 (b_1 \lambda_1 + d_1) e^{\lambda_1 t} - C_2 (b_1 \lambda_2 + d_1) e^{\lambda_2 t} - C_3 (b_1 \lambda_3 + d_1) e^{\lambda_3 t} - C_4 (b_1 \lambda_4 + d_1) e^{\lambda_4 t} 

 

    .. math:: y = C_1 (\lambda_1^{2} + a_1 \lambda_1 + c_1) e^{\lambda_1 t} + C_2 (\lambda_2^{2} + a_1 \lambda_2 + c_1) e^{\lambda_2 t} + C_3 (\lambda_3^{2} + a_1 \lambda_3 + c_1) e^{\lambda_3 t} + C_4 (\lambda_4^{2} + a_1 \lambda_4 + c_1) e^{\lambda_4 t} 

 

    """ 

    C1, C2, C3, C4 = symbols('C1:5') 

    k = Symbol('k') 

    Ra, Ca, Rb, Cb = symbols('Ra, Ca, Rb, Cb') 

    a1 = r['a1'] ; a2 = r['a2'] 

    b1 = r['b1'] ; b2 = r['b2'] 

    c1 = r['c1'] ; c2 = r['c2'] 

    d1 = r['d1'] ; d2 = r['d2'] 

    k1 = r['e1'].expand().as_independent(t)[0] 

    k2 = r['e2'].expand().as_independent(t)[0] 

    ew1 = r['e1'].expand().as_independent(t)[1] 

    ew2 = powdenest(ew1).as_base_exp()[1] 

    ew3 = collect(ew2, t).coeff(t) 

    w = cancel(ew3/I) 

    # The particular solution is assumed to be (Ra+I*Ca)*exp(I*w*t) and 

    # (Rb+I*Cb)*exp(I*w*t) for x(t) and y(t) respectively 

    peq1 = (-w**2+c1)*Ra - a1*w*Ca + d1*Rb - b1*w*Cb - k1 

    peq2 = a1*w*Ra + (-w**2+c1)*Ca + b1*w*Rb + d1*Cb 

    peq3 = c2*Ra - a2*w*Ca + (-w**2+d2)*Rb - b2*w*Cb - k2 

    peq4 = a2*w*Ra + c2*Ca + b2*w*Rb + (-w**2+d2)*Cb 

    # FIXME: solve for what in what?  Ra, Rb, etc I guess 

    # but then psol not used for anything? 

    psol = solve([peq1, peq2, peq3, peq4]) 

 

    chareq = (k**2+a1*k+c1)*(k**2+b2*k+d2) - (b1*k+d1)*(a2*k+c2) 

    [k1, k2, k3, k4] = roots_quartic(Poly(chareq)) 

    sol1 = -C1*(b1*k1+d1)*exp(k1*t) - C2*(b1*k2+d1)*exp(k2*t) - \ 

    C3*(b1*k3+d1)*exp(k3*t) - C4*(b1*k4+d1)*exp(k4*t) + (Ra+I*Ca)*exp(I*w*t) 

 

    a1_ = (a1-1) 

    sol2 = C1*(k1**2+a1_*k1+c1)*exp(k1*t) + C2*(k2**2+a1_*k2+c1)*exp(k2*t) + \ 

    C3*(k3**2+a1_*k3+c1)*exp(k3*t) + C4*(k4**2+a1_*k4+c1)*exp(k4*t) + (Rb+I*Cb)*exp(I*w*t) 

    return [Eq(x(t), sol1), Eq(y(t), sol2)] 

 

def _linear_2eq_order2_type5(x, y, t, r): 

    r""" 

    The equation which come under this catagory are 

 

    .. math:: x'' = a (t y' - y) 

 

    .. math:: y'' = b (t x' - x) 

 

    The transformation 

 

    .. math:: u = t x' - x, b = t y' - y 

 

    leads to the first-order system 

 

    .. math:: u' = atv, v' = btu 

 

    The general solution of this system is given by 

 

    If `ab > 0`: 

 

    .. math:: u = C_1 a e^{\frac{1}{2} \sqrt{ab} t^2} + C_2 a e^{-\frac{1}{2} \sqrt{ab} t^2} 

 

    .. math:: v = C_1 \sqrt{ab} e^{\frac{1}{2} \sqrt{ab} t^2} - C_2 \sqrt{ab} e^{-\frac{1}{2} \sqrt{ab} t^2} 

 

    If `ab < 0`: 

 

    .. math:: u = C_1 a \cos(\frac{1}{2} \sqrt{\left|ab\right|} t^2) + C_2 a \sin(-\frac{1}{2} \sqrt{\left|ab\right|} t^2) 

 

    .. math:: v = C_1 \sqrt{\left|ab\right|} \sin(\frac{1}{2} \sqrt{\left|ab\right|} t^2) + C_2 \sqrt{\left|ab\right|} \cos(-\frac{1}{2} \sqrt{\left|ab\right|} t^2) 

 

    where `C_1` and `C_2` are arbitary constants. On substituting the value of `u` and `v` 

    in above equations and integrating the resulting expressions, the general solution will become 

 

    .. math:: x = C_3 t + t \int \frac{u}{t^2} \,dt, y = C_4 t + t \int \frac{u}{t^2} \,dt 

 

    where `C_3` and `C_4` are arbitrary constants. 

 

    """ 

    C1, C2, C3, C4 = symbols('C1:5') 

    r['a'] = -r['d1'] ; r['b'] = -r['c2'] 

    mul = sqrt(abs(r['a']*r['b'])) 

    if r['a']*r['b'] > 0: 

        u = C1*r['a']*exp(mul*t**2/2) + C2*r['a']*exp(-mul*t**2/2) 

        v = C1*mul*exp(mul*t**2/2) - C2*mul*exp(-mul*t**2/2) 

    else: 

        u = C1*r['a']*cos(mul*t**2/2) + C2*r['a']*sin(mul*t**2/2) 

        v = -C1*mul*sin(mul*t**2/2) + C2*mul*cos(mul*t**2/2) 

    sol1 = C3*t + t*Integral(u/t**2, t) 

    sol2 = C4*t + t*Integral(v/t**2, t) 

    return [Eq(x(t), sol1), Eq(y(t), sol2)] 

 

def _linear_2eq_order2_type6(x, y, t, r): 

    r""" 

    The equations are 

 

    .. math:: x'' = f(t) (a_1 x + b_1 y) 

 

    .. math:: y'' = f(t) (a_2 x + b_2 y) 

 

    If `k_1` and `k_2` are roots of the quadratic equation 

 

    .. math:: k^2 - (a_1 + b_2) k + a_1 b_2 - a_2 b_1 = 0 

 

    Then by multiplying appropriate constants and adding together original equations 

    we obtain two independent equations: 

 

    .. math:: z_1'' = k_1 f(t) z_1, z_1 = a_2 x + (k_1 - a_1) y 

 

    .. math:: z_2'' = k_2 f(t) z_2, z_2 = a_2 x + (k_2 - a_1) y 

 

    Solving the equations will give the values of `x` and `y` after obtaining the value 

    of `z_1` and `z_2` by solving the differential equation and substuting the result. 

 

    """ 

    C1, C2, C3, C4 = symbols('C1:5') 

    k = Symbol('k') 

    z = Function('z') 

    num, denum = cancel((r['c1']*x(t) + r['d1']*y(t))/(r['c2']*x(t) + r['d2']*y(t))).as_numer_denom() 

    f = r['c1']/num.coeff(x(t)) 

    a1 = num.coeff(x(t)) 

    b1 = num.coeff(y(t)) 

    a2 = denum.coeff(x(t)) 

    b2 = denum.coeff(y(t)) 

    chareq = k**2 - (a1 + b2)*k + a1*b2 - a2*b1 

    [k1, k2] = [RootOf(chareq, k) for k in range(Poly(chareq).degree())] 

    z1 = dsolve(diff(z(t),t,t) - k1*f*z(t)).rhs 

    z2 = dsolve(diff(z(t),t,t) - k2*f*z(t)).rhs 

    sol1 = (k1*z2 - k2*z1 + a1*(z1 - z2))/(a2*(k1-k2)) 

    sol2 = (z1 - z2)/(k1 - k2) 

    return [Eq(x(t), sol1), Eq(y(t), sol2)] 

 

def _linear_2eq_order2_type7(x, y, t, r): 

    r""" 

    The equations are given as 

 

    .. math:: x'' = f(t) (a_1 x' + b_1 y') 

 

    .. math:: y'' = f(t) (a_2 x' + b_2 y') 

 

    If `k_1` and 'k_2` are roots of the quadratic equation 

 

    .. math:: k^2 - (a_1 + b_2) k + a_1 b_2 - a_2 b_1 = 0 

 

    Then the system can be reduced by adding together the two equations multiplied 

    by appropriate constants give following two independent equations: 

 

    .. math:: z_1'' = k_1 f(t) z_1', z_1 = a_2 x + (k_1 - a_1) y 

 

    .. math:: z_2'' = k_2 f(t) z_2', z_2 = a_2 x + (k_2 - a_1) y 

 

    Integrating these and returning to the original variables, one arrives at a linear 

    algebraic system for the unknowns `x` and `y`: 

 

    .. math:: a_2 x + (k_1 - a_1) y = C_1 \int e^{k_1 F(t)} \,dt + C_2 

 

    .. math:: a_2 x + (k_2 - a_1) y = C_3 \int e^{k_2 F(t)} \,dt + C_4 

 

    where `C_1,...,C_4` are arbitrary constants and `F(t) = \int f(t) \,dt` 

 

    """ 

    C1, C2, C3, C4 = symbols('C1:5') 

    k = Symbol('k') 

    num, denum = cancel((r['a1']*x(t) + r['b1']*y(t))/(r['a2']*x(t) + r['b2']*y(t))).as_numer_denom() 

    f = r['a1']/num.coeff(x(t)) 

    a1 = num.coeff(x(t)) 

    b1 = num.coeff(y(t)) 

    a2 = denum.coeff(x(t)) 

    b2 = denum.coeff(y(t)) 

    chareq = k**2 - (a1 + b2)*k + a1*b2 - a2*b1 

    [k1, k2] = [RootOf(chareq, k) for k in range(Poly(chareq).degree())] 

    F = Integral(f, t) 

    z1 = C1*Integral(exp(k1*F), t) + C2 

    z2 = C3*Integral(exp(k2*F), t) + C4 

    sol1 = (k1*z2 - k2*z1 + a1*(z1 - z2))/(a2*(k1-k2)) 

    sol2 = (z1 - z2)/(k1 - k2) 

    return [Eq(x(t), sol1), Eq(y(t), sol2)] 

 

def _linear_2eq_order2_type8(x, y, t, r): 

    r""" 

    The equation of this catagory are 

 

    .. math:: x'' = a f(t) (t y' - y) 

 

    .. math:: y'' = b f(t) (t x' - x) 

 

    The transformation 

 

    .. math:: u = t x' - x, v = t y' - y 

 

    leads to the system of first-order equations 

 

    .. math:: u' = a t f(t) v, v' = b t f(t) u 

 

    The general solution of this system has the form 

 

    If `ab > 0`: 

 

    .. math:: u = C_1 a e^{\sqrt{ab} \int t f(t) \,dt} + C_2 a e^{-\sqrt{ab} \int t f(t) \,dt} 

 

    .. math:: v = C_1 \sqrt{ab} e^{\sqrt{ab} \int t f(t) \,dt} - C_2 \sqrt{ab} e^{-\sqrt{ab} \int t f(t) \,dt} 

 

    If `ab < 0`: 

 

    .. math:: u = C_1 a \cos(\sqrt{\left|ab\right|} \int t f(t) \,dt) + C_2 a \sin(-\sqrt{\left|ab\right|} \int t f(t) \,dt) 

 

    .. math:: v = C_1 \sqrt{\left|ab\right|} \sin(\sqrt{\left|ab\right|} \int t f(t) \,dt) + C_2 \sqrt{\left|ab\right|} \cos(-\sqrt{\left|ab\right|} \int t f(t) \,dt) 

 

    where `C_1` and `C_2` are arbitary constants. On substituting the value of `u` and `v` 

    in above equations and integrating the resulting expressions, the general solution will become 

 

    .. math:: x = C_3 t + t \int \frac{u}{t^2} \,dt, y = C_4 t + t \int \frac{u}{t^2} \,dt 

 

    where `C_3` and `C_4` are arbitrary constants. 

 

    """ 

    C1, C2, C3, C4 = symbols('C1:5') 

    num, denum = cancel(r['d1']/r['c2']).as_numer_denom() 

    f = -r['d1']/num 

    a = num 

    b = denum 

    mul = sqrt(abs(a*b)) 

    Igral = Integral(t*f, t) 

    if a*b > 0: 

        u = C1*a*exp(mul*Igral) + C2*a*exp(-mul*Igral) 

        v = C1*mul*exp(mul*Igral) - C2*mul*exp(-mul*Igral) 

    else: 

        u = C1*a*cos(mul*Igral) + C2*a*sin(mul*Igral) 

        v = -C1*mul*sin(mul*Igral) + C2*mul*cos(mul*Igral) 

    sol1 = C3*t + t*Integral(u/t**2, t) 

    sol2 = C4*t + t*Integral(v/t**2, t) 

    return [Eq(x(t), sol1), Eq(y(t), sol2)] 

 

def _linear_2eq_order2_type9(x, y, t, r): 

    r""" 

    .. math:: t^2 x'' + a_1 t x' + b_1 t y' + c_1 x + d_1 y = 0 

 

    .. math:: t^2 y'' + a_2 t x' + b_2 t y' + c_2 x + d_2 y = 0 

 

    These system of equations are euler type. 

 

    The substitution of `t = \sigma e^{\tau} (\sigma \neq 0)` leads to the system of constant 

    coefficient linear differential equations 

 

    .. math:: x'' + (a_1 - 1) x' + b_1 y' + c_1 x + d_1 y = 0 

 

    .. math:: y'' + a_2 x' + (b_2 - 1) y' + c_2 x + d_2 y = 0 

 

    The general solution of the homogeneous system of differential equations is determined 

    by a linear combination of linearly independent particular solutions determined by 

    the method of undetermined coefficients in the form of exponentials 

 

    .. math:: x = A e^{\lambda t}, y = B e^{\lambda t} 

 

    On substituting these expressions into the original system and colleting the 

    coefficients of the unknown `A` and `B`, one obtains 

 

    .. math:: (\lambda^{2} + (a_1 - 1) \lambda + c_1) A + (b_1 \lambda + d_1) B = 0 

 

    .. math:: (a_2 \lambda + c_2) A + (\lambda^{2} + (b_2 - 1) \lambda + d_2) B = 0 

 

    The determinant of this system must vanish for nontrivial solutions A, B to exist. 

    This requirement results in the following characteristic equation for `\lambda` 

 

    .. math:: (\lambda^2 + (a_1 - 1) \lambda + c_1) (\lambda^2 + (b_2 - 1) \lambda + d_2) - (b_1 \lambda + d_1) (a_2 \lambda + c_2) = 0 

 

    If all roots `k_1,...,k_4` of this equation are distict, the general solution of the original 

    system of the differential equations has the form 

 

    .. math:: x = C_1 (b_1 \lambda_1 + d_1) e^{\lambda_1 t} - C_2 (b_1 \lambda_2 + d_1) e^{\lambda_2 t} - C_3 (b_1 \lambda_3 + d_1) e^{\lambda_3 t} - C_4 (b_1 \lambda_4 + d_1) e^{\lambda_4 t} 

 

    .. math:: y = C_1 (\lambda_1^{2} + (a_1 - 1) \lambda_1 + c_1) e^{\lambda_1 t} + C_2 (\lambda_2^{2} + (a_1 - 1) \lambda_2 + c_1) e^{\lambda_2 t} + C_3 (\lambda_3^{2} + (a_1 - 1) \lambda_3 + c_1) e^{\lambda_3 t} + C_4 (\lambda_4^{2} + (a_1 - 1) \lambda_4 + c_1) e^{\lambda_4 t} 

 

    """ 

    C1, C2, C3, C4 = symbols('C1:5') 

    k = Symbol('k') 

    a1 = -r['a1']*t; a2 = -r['a2']*t 

    b1 = -r['b1']*t; b2 = -r['b2']*t 

    c1 = -r['c1']*t**2; c2 = -r['c2']*t**2 

    d1 = -r['d1']*t**2; d2 = -r['d2']*t**2 

    eq = (k**2+(a1-1)*k+c1)*(k**2+(b2-1)*k+d2)-(b1*k+d1)*(a2*k+c2) 

    [k1, k2, k3, k4] = roots_quartic(Poly(eq)) 

    sol1 = -C1*(b1*k1+d1)*exp(k1*log(t)) - C2*(b1*k2+d1)*exp(k2*log(t)) - \ 

    C3*(b1*k3+d1)*exp(k3*log(t)) - C4*(b1*k4+d1)*exp(k4*log(t)) 

 

    a1_ = (a1-1) 

    sol2 = C1*(k1**2+a1_*k1+c1)*exp(k1*log(t)) + C2*(k2**2+a1_*k2+c1)*exp(k2*log(t)) \ 

    + C3*(k3**2+a1_*k3+c1)*exp(k3*log(t)) + C4*(k4**2+a1_*k4+c1)*exp(k4*log(t)) 

 

    return [Eq(x(t), sol1), Eq(y(t), sol2)] 

 

def _linear_2eq_order2_type10(x, y, t, r): 

    r""" 

    The equation of this catagory are 

 

    .. math:: (\alpha t^2 + \beta t + \gamma)^{2} x'' = ax + by 

 

    .. math:: (\alpha t^2 + \beta t + \gamma)^{2} y'' = cx + dy 

 

    The transformation 

 

    .. math:: \tau = \int \frac{1}{\alpha t^2 + \beta t + \gamma} \,dt , u = \frac{x}{\sqrt{\left|\alpha t^2 + \beta t + \gamma\right|}} , v = \frac{y}{\sqrt{\left|\alpha t^2 + \beta t + \gamma\right|}} 

 

    leads to a constant coefficient linear system of equations 

 

    .. math:: u'' = (a - \alpha \gamma + \frac{1}{4} \beta^{2}) u + b v 

 

    .. math:: v'' = c u + (d - \alpha \gamma + \frac{1}{4} \beta^{2}) v 

 

    These system of equations obtained can be solved by type1 of System of two 

    constant-coefficient second-order linear homogeneous differential equations. 

 

    """ 

    C1, C2, C3, C4 = symbols('C1:5') 

    u, v = symbols('u, v', function=True) 

    T = Symbol('T') 

    p = Wild('p', exclude=[t, t**2]) 

    q = Wild('q', exclude=[t, t**2]) 

    s = Wild('s', exclude=[t, t**2]) 

    n = Wild('n', exclude=[t, t**2]) 

    num, denum = r['c1'].as_numer_denom() 

    dic = denum.match((n*(p*t**2+q*t+s)**2).expand()) 

    eqz = dic[p]*t**2 + dic[q]*t + dic[s] 

    a = num/dic[n] 

    b = cancel(r['d1']*eqz**2) 

    c = cancel(r['c2']*eqz**2) 

    d = cancel(r['d2']*eqz**2) 

    [msol1, msol2] = dsolve([Eq(diff(u(t), t, t), (a - dic[p]*dic[s] + dic[q]**2/4)*u(t) \ 

    + b*v(t)), Eq(diff(v(t),t,t), c*u(t) + (d - dic[p]*dic[s] + dic[q]**2/4)*v(t))]) 

    sol1 = (msol1.rhs*sqrt(abs(eqz))).subs(t, Integral(1/eqz, t)) 

    sol2 = (msol2.rhs*sqrt(abs(eqz))).subs(t, Integral(1/eqz, t)) 

    return [Eq(x(t), sol1), Eq(y(t), sol2)] 

 

def _linear_2eq_order2_type11(x, y, t, r): 

    r""" 

    The equations which comes under this type are 

 

    .. math:: x'' = f(t) (t x' - x) + g(t) (t y' - y) 

 

    .. math:: y'' = h(t) (t x' - x) + p(t) (t y' - y) 

 

    The transformation 

 

    .. math:: u = t x' - x, v = t y' - y 

 

    leads to the linear system of first-order equations 

 

    .. math:: u' = t f(t) u + t g(t) v, v' = t h(t) u + t p(t) v 

 

    On substituting the value of `u` and `v` in transformed equation gives value of `x` and `y` as 

 

    .. math:: x = C_3 t + t \int \frac{u}{t^2} \,dt , y = C_4 t + t \int \frac{v}{t^2} \,dt. 

 

    where `C_3` and `C_4` are arbitrary constants. 

 

    """ 

    C1, C2, C3, C4 = symbols('C1:5') 

    u, v = symbols('u, v', function=True) 

    f = -r['c1'] ; g = -r['d1'] 

    h = -r['c2'] ; p = -r['d2'] 

    [msol1, msol2] = dsolve([Eq(diff(u(t),t), t*f*u(t) + t*g*v(t)), Eq(diff(v(t),t), t*h*u(t) + t*p*v(t))]) 

    sol1 = C3*t + t*Integral(msol1.rhs/t**2, t) 

    sol2 = C4*t + t*Integral(msol2.rhs/t**2, t) 

    return [Eq(x(t), sol1), Eq(y(t), sol2)] 

 

def sysode_linear_3eq_order1(match_): 

    C1, C2, C3, C4 = symbols('C1:5') 

    x = match_['func'][0].func 

    y = match_['func'][1].func 

    z = match_['func'][2].func 

    func = match_['func'] 

    fc = match_['func_coeff'] 

    eq = match_['eq'] 

    r = dict() 

    t = list(list(eq[0].atoms(Derivative))[0].atoms(Symbol))[0] 

    for i in range(3): 

        eqs = 0 

        for terms in Add.make_args(eq[i]): 

            eqs += terms/fc[i,func[i],1] 

        eq[i] = eqs 

    # for equations: 

    #   Eq(g1*diff(x(t),t), a1*x(t)+b1*y(t)+c1*z(t)+d1), 

    #   Eq(g2*diff(y(t),t), a2*x(t)+b2*y(t)+c2*z(t)+d2), and 

    #   Eq(g3*diff(z(t),t), a3*x(t)+b3*y(t)+c3*z(t)+d3) 

    r['a1'] = fc[0,x(t),0]/fc[0,x(t),1]; r['a2'] = fc[1,x(t),0]/fc[1,y(t),1]; 

    r['a3'] = fc[2,x(t),0]/fc[2,z(t),1] 

    r['b1'] = fc[0,y(t),0]/fc[0,x(t),1]; r['b2'] = fc[1,y(t),0]/fc[1,y(t),1]; 

    r['b3'] = fc[2,y(t),0]/fc[2,z(t),1] 

    r['c1'] = fc[0,z(t),0]/fc[0,x(t),1]; r['c2'] = fc[1,z(t),0]/fc[1,y(t),1]; 

    r['c3'] = fc[2,z(t),0]/fc[2,z(t),1] 

    forcing = [S(0), S(0), S(0)] 

    for i in range(3): 

        for j in Add.make_args(eq[i]): 

            if not j.has(x(t), y(t), z(t)): 

                forcing[i] += j 

    if not (forcing[0].has(t) or forcing[1].has(t) or forcing[2].has(t)): 

        # We can handle homogeneous case and simple constant forcings 

        r['d1'] = -forcing[0] 

        r['d2'] = -forcing[1] 

        r['d3'] = -forcing[2] 

    else: 

        # Issue #9244: nonhomogeneous linear systems are not supported 

        raise NotImplementedError("Only homogeneous problems are supported" + 

                                  " (and constant inhomogeneity)") 

 

    if match_['type_of_equation'] == 'type1': 

        sol = _linear_3eq_order1_type1(x, y, z, t, r) 

    if match_['type_of_equation'] == 'type2': 

        sol = _linear_3eq_order1_type2(x, y, z, t, r) 

    if match_['type_of_equation'] == 'type3': 

        sol = _linear_3eq_order1_type3(x, y, z, t, r) 

    if match_['type_of_equation'] == 'type4': 

        sol = _linear_3eq_order1_type4(x, y, z, t, r) 

    if match_['type_of_equation'] == 'type6': 

        sol = _linear_neq_order1_type1(match_) 

    return sol 

 

def _linear_3eq_order1_type1(x, y, z, t, r): 

    r""" 

    .. math:: x' = ax 

 

    .. math:: y' = bx + cy 

 

    .. math:: z' = dx + ky + pz 

 

    Solution of such equations are forward substitution. Solving first equations 

    gives the value of `x`, substituting it in second and third equation and 

    solving second equation gives `y` and similarly substituting `y` in third 

    equation give `z`. 

 

    .. math:: x = C_1 e^{at} 

 

    .. math:: y = \frac{b C_1}{a - c} e^{at} + C_2 e^{ct} 

 

    .. math:: z = \frac{C_1}{a - p} (d + \frac{bk}{a - c}) e^{at} + \frac{k C_2}{c - p} e^{ct} + C_3 e^{pt} 

 

    where `C_1, C_2` and `C_3` are arbitrary constants. 

 

    """ 

    C1, C2, C3, C4 = symbols('C1:5') 

    a = -r['a1']; b = -r['a2']; c = -r['b2'] 

    d = -r['a3']; k = -r['b3']; p = -r['c3'] 

    sol1 = C1*exp(a*t) 

    sol2 = b*C1*exp(a*t)/(a-c) + C2*exp(c*t) 

    sol3 = C1*(d+b*k/(a-c))*exp(a*t)/(a-p) + k*C2*exp(c*t)/(c-p) + C3*exp(p*t) 

    return [Eq(x(t), sol1), Eq(y(t), sol2), Eq(z(t), sol3)] 

 

def _linear_3eq_order1_type2(x, y, z, t, r): 

    r""" 

    The equations of this type are 

 

    .. math:: x' = cy - bz 

 

    .. math:: y' = az - cx 

 

    .. math:: z' = bx - ay 

 

    1. First integral: 

 

    .. math:: ax + by + cz = A             \qquad - (1) 

 

    .. math:: x^2 + y^2 + z^2 = B^2        \qquad - (2) 

 

    where `A` and `B` are arbitrary constants. It follows from these integrals 

    that the integral lines are circles formed by the intersection of the planes 

    `(1)` and sphere `(2)` 

 

    2. Solution: 

 

    .. math:: x = a C_0 + k C_1 \cos(kt) + (c C_2 - b C_3) \sin(kt) 

 

    .. math:: y = b C_0 + k C_2 \cos(kt) + (a C_2 - c C_3) \sin(kt) 

 

    .. math:: z = c C_0 + k C_3 \cos(kt) + (b C_2 - a C_3) \sin(kt) 

 

    where `k = \sqrt{a^2 + b^2 + c^2}` and the four constants of integration, 

    `C_1,...,C_4` are constrained by a single relation, 

 

    .. math:: a C_1 + b C_2 + c C_3 = 0 

 

    """ 

    C0, C1, C2, C3 = symbols('C0:4') 

    a = -r['c2']; b = -r['a3']; c = -r['b1'] 

    k = sqrt(a**2 + b**2 + c**2) 

    C3 = (-a*C1 - b*C2)/c 

    sol1 = a*C0 + k*C1*cos(k*t) + (c*C2-b*C3)*sin(k*t) 

    sol2 = b*C0 + k*C2*cos(k*t) + (a*C3-c*C1)*sin(k*t) 

    sol3 = c*C0 + k*C3*cos(k*t) + (b*C1-a*C2)*sin(k*t) 

    return [Eq(x(t), sol1), Eq(y(t), sol2), Eq(z(t), sol3)] 

 

def _linear_3eq_order1_type3(x, y, z, t, r): 

    r""" 

    Equations of this system of ODEs 

 

    .. math:: a x' = bc (y - z) 

 

    .. math:: b y' = ac (z - x) 

 

    .. math:: c z' = ab (x - y) 

 

    1. First integral: 

 

    .. math:: a^2 x + b^2 y + c^2 z = A 

 

    where A is an arbitary constant. It follows that the integral lines are plane curves. 

 

    2. Solution: 

 

    .. math:: x = C_0 + k C_1 \cos(kt) + a^{-1} bc (C_2 - C_3) \sin(kt) 

 

    .. math:: y = C_0 + k C_2 \cos(kt) + a b^{-1} c (C_3 - C_1) \sin(kt) 

 

    .. math:: z = C_0 + k C_3 \cos(kt) + ab c^{-1} (C_1 - C_2) \sin(kt) 

 

    where `k = \sqrt{a^2 + b^2 + c^2}` and the four constants of integration, 

    `C_1,...,C_4` are constrained by a single relation 

 

    .. math:: a^2 C_1 + b^2 C_2 + c^2 C_3 = 0 

 

    """ 

    C0, C1, C2, C3 = symbols('C0:4') 

    c = sqrt(r['b1']*r['c2']) 

    b = sqrt(r['b1']*r['a3']) 

    a = sqrt(r['c2']*r['a3']) 

    C3 = (-a**2*C1-b**2*C2)/c**2 

    k = sqrt(a**2 + b**2 + c**2) 

    sol1 = C0 + k*C1*cos(k*t) + a**-1*b*c*(C2-C3)*sin(k*t) 

    sol2 = C0 + k*C2*cos(k*t) + a*b**-1*c*(C3-C1)*sin(k*t) 

    sol3 = C0 + k*C3*cos(k*t) + a*b*c**-1*(C1-C2)*sin(k*t) 

    return [Eq(x(t), sol1), Eq(y(t), sol2), Eq(z(t), sol3)] 

 

def _linear_3eq_order1_type4(x, y, z, t, r): 

    r""" 

    Equations: 

 

    .. math:: x' = (a_1 f(t) + g(t)) x + a_2 f(t) y + a_3 f(t) z 

 

    .. math:: y' = b_1 f(t) x + (b_2 f(t) + g(t)) y + b_3 f(t) z 

 

    .. math:: z' = c_1 f(t) x + c_2 f(t) y + (c_3 f(t) + g(t)) z 

 

    The transformation 

 

    .. math:: x = e^{\int g(t) \,dt} u, y = e^{\int g(t) \,dt} v, z = e^{\int g(t) \,dt} w, \tau = \int f(t) \,dt 

 

    leads to the system of constant coefficient linear differential equations 

 

    .. math:: u' = a_1 u + a_2 v + a_3 w 

 

    .. math:: v' = b_1 u + b_2 v + b_3 w 

 

    .. math:: w' = c_1 u + c_2 v + c_3 w 

 

    These system of equations are solved by homogeneous linear system of constant 

    coefficients of `n` equations of first order. Then substituting the value of 

    `u, v` and `w` in transformed equation gives value of `x, y` and `z`. 

 

    """ 

    u, v, w = symbols('u, v, w', function=True) 

    a2, a3 = cancel(r['b1']/r['c1']).as_numer_denom() 

    f = cancel(r['b1']/a2) 

    b1 = cancel(r['a2']/f); b3 = cancel(r['c2']/f) 

    c1 = cancel(r['a3']/f); c2 = cancel(r['b3']/f) 

    a1, g = div(r['a1'],f) 

    b2 = div(r['b2'],f)[0] 

    c3 = div(r['c3'],f)[0] 

    trans_eq = (diff(u(t),t)-a1*u(t)-a2*v(t)-a3*w(t), diff(v(t),t)-b1*u(t)-\ 

    b2*v(t)-b3*w(t), diff(w(t),t)-c1*u(t)-c2*v(t)-c3*w(t)) 

    sol = dsolve(trans_eq) 

    sol1 = exp(Integral(g,t))*((sol[0].rhs).subs(t, Integral(f,t))) 

    sol2 = exp(Integral(g,t))*((sol[1].rhs).subs(t, Integral(f,t))) 

    sol3 = exp(Integral(g,t))*((sol[2].rhs).subs(t, Integral(f,t))) 

    return [Eq(x(t), sol1), Eq(y(t), sol2), Eq(z(t), sol3)] 

 

def sysode_linear_neq_order1(match_): 

    sol = _linear_neq_order1_type1(match_) 

 

def _linear_neq_order1_type1(match_): 

    r""" 

    System of n first-order constant-coefficient linear nonhomogeneous differential equation 

 

    .. math:: y'_k = a_{k1} y_1 + a_{k2} y_2 +...+ a_{kn} y_n; k = 1,2,...,n 

 

    or that can be written as `\vec{y'} = A . \vec{y}` 

    where `\vec{y}` is matrix of `y_k` for `k = 1,2,...n` and `A` is a `n \times n` matrix. 

 

    Since these equations are equivalent to a first order homogeneous linear 

    differential equation. So the general solution will contain `n` linearly 

    independent parts and solution will consist some type of exponential 

    functions. Assuming `y = \vec{v} e^{rt}` is a solution of the system where 

    `\vec{v}` is a vector of coefficients of `y_1,...,y_n`. Substituting `y` and 

    `y' = r v e^{r t}` into the equation `\vec{y'} = A . \vec{y}`, we get 

 

    .. math:: r \vec{v} e^{rt} = A \vec{v} e^{rt} 

 

    .. math:: r \vec{v} = A \vec{v} 

 

    where `r` comes out to be eigenvalue of `A` and vector `\vec{v}` is the eigenvector 

    of `A` corresponding to `r`. There are three possiblities of eigenvalues of `A` 

 

    - `n` distinct real eigenvalues 

    - complex conjugate eigenvalues 

    - eigenvalues with multiplicity `k` 

 

    1. When all eigenvalues `r_1,..,r_n` are distinct with `n` different eigenvectors 

    `v_1,...v_n` then the solution is given by 

 

    .. math:: \vec{y} = C_1 e^{r_1 t} \vec{v_1} + C_2 e^{r_2 t} \vec{v_2} +...+ C_n e^{r_n t} \vec{v_n} 

 

    where `C_1,C_2,...,C_n` are arbitrary constants. 

 

    2. When some eigenvalues are complex then in order to make the solution real, 

    we take a llinear combination: if `r = a + bi` has an eigenvector 

    `\vec{v} = \vec{w_1} + i \vec{w_2}` then to obtain real-valued solutions to 

    the system, replace the complex-valued solutions `e^{rx} \vec{v}` 

    with real-valued solution `e^{ax} (\vec{w_1} \cos(bx) - \vec{w_2} \sin(bx))` 

    and for `r = a - bi` replace the solution `e^{-r x} \vec{v}` with 

    `e^{ax} (\vec{w_1} \sin(bx) + \vec{w_2} \cos(bx))` 

 

    3. If some eigenvalues are repeated. Then we get fewer than `n` linearly 

    independent eigenvectors, we miss some of the solutions and need to 

    construct the missing ones. We do this via generalized eigenvectors, vectors 

    which are not eigenvectors but are close enough that we can use to write 

    down the remaining solutions. For a eigenvalue `r` with eigenvector `\vec{w}` 

    we obtain `\vec{w_2},...,\vec{w_k}` using 

 

    .. math:: (A - r I) . \vec{w_2} = \vec{w} 

 

    .. math:: (A - r I) . \vec{w_3} = \vec{w_2} 

 

    .. math:: \vdots 

 

    .. math:: (A - r I) . \vec{w_k} = \vec{w_{k-1}} 

 

    Then the solutions to the system for the eigenspace are `e^{rt} [\vec{w}], 

    e^{rt} [t \vec{w} + \vec{w_2}], e^{rt} [\frac{t^2}{2} \vec{w} + t \vec{w_2} + \vec{w_3}], 

    ...,e^{rt} [\frac{t^{k-1}}{(k-1)!} \vec{w} + \frac{t^{k-2}}{(k-2)!} \vec{w_2} +...+ t \vec{w_{k-1}} 

    + \vec{w_k}]` 

 

    So, If `\vec{y_1},...,\vec{y_n}` are `n` solution of obtained from three 

    categories of `A`, then general solution to the system `\vec{y'} = A . \vec{y}` 

 

    .. math:: \vec{y} = C_1 \vec{y_1} + C_2 \vec{y_2} + \cdots + C_n \vec{y_n} 

 

    """ 

    eq = match_['eq'] 

    func = match_['func'] 

    fc = match_['func_coeff'] 

    n = len(eq) 

    t = list(list(eq[0].atoms(Derivative))[0].atoms(Symbol))[0] 

    constants = numbered_symbols(prefix='C', cls=Symbol, start=1) 

    M = Matrix(n,n,lambda i,j:-fc[i,func[j],0]) 

    evector = M.eigenvects(simplify=True) 

    def is_complex(mat, root): 

        return Matrix(n, 1, lambda i,j: re(mat[i])*cos(im(root)*t) - im(mat[i])*sin(im(root)*t)) 

    def is_complex_conjugate(mat, root): 

        return Matrix(n, 1, lambda i,j: re(mat[i])*sin(abs(im(root))*t) + im(mat[i])*cos(im(root)*t)*abs(im(root))/im(root)) 

    conjugate_root = [] 

    e_vector = zeros(n,1) 

    for evects in evector: 

        if evects[0] not in conjugate_root: 

            # If number of column of an eigenvector is not equal to the multiplicity 

            # of its eigenvalue then the legt eigenvectors are calculated 

            if len(evects[2])!=evects[1]: 

                var_mat = Matrix(n, 1, lambda i,j: Symbol('x'+str(i))) 

                Mnew = (M - evects[0]*eye(evects[2][-1].rows))*var_mat 

                w = [0 for i in range(evects[1])] 

                w[0] = evects[2][-1] 

                for r in range(1, evects[1]): 

                    w_ = Mnew - w[r-1] 

                    sol_dict = solve(list(w_), var_mat[1:]) 

                    sol_dict[var_mat[0]] = var_mat[0] 

                    for key, value in sol_dict.items(): 

                        sol_dict[key] = value.subs(var_mat[0],1) 

                    w[r] = Matrix(n, 1, lambda i,j: sol_dict[var_mat[i]]) 

                    evects[2].append(w[r]) 

            for i in range(evects[1]): 

                C = next(constants) 

                for j in range(i+1): 

                    if evects[0].has(I): 

                        evects[2][j] = simplify(evects[2][j]) 

                        e_vector += C*is_complex(evects[2][j], evects[0])*t**(i-j)*exp(re(evects[0])*t)/factorial(i-j) 

                        C = next(constants) 

                        e_vector += C*is_complex_conjugate(evects[2][j], evects[0])*t**(i-j)*exp(re(evects[0])*t)/factorial(i-j) 

                    else: 

                        e_vector += C*evects[2][j]*t**(i-j)*exp(evects[0]*t)/factorial(i-j) 

            if evects[0].has(I): 

                conjugate_root.append(conjugate(evects[0])) 

    sol = [] 

    for i in range(len(eq)): 

        sol.append(Eq(func[i],e_vector[i])) 

    return sol 

 

def sysode_nonlinear_2eq_order1(match_): 

    func = match_['func'] 

    eq = match_['eq'] 

    fc = match_['func_coeff'] 

    t = list(list(eq[0].atoms(Derivative))[0].atoms(Symbol))[0] 

    if match_['type_of_equation'] == 'type5': 

        sol = _nonlinear_2eq_order1_type5(func, t, eq) 

        return sol 

    x = func[0].func 

    y = func[1].func 

    for i in range(2): 

        eqs = 0 

        for terms in Add.make_args(eq[i]): 

            eqs += terms/fc[i,func[i],1] 

        eq[i] = eqs 

    if match_['type_of_equation'] == 'type1': 

        sol = _nonlinear_2eq_order1_type1(x, y, t, eq) 

    elif match_['type_of_equation'] == 'type2': 

        sol = _nonlinear_2eq_order1_type2(x, y, t, eq) 

    elif match_['type_of_equation'] == 'type3': 

        sol = _nonlinear_2eq_order1_type3(x, y, t, eq) 

    elif match_['type_of_equation'] == 'type4': 

        sol = _nonlinear_2eq_order1_type4(x, y, t, eq) 

    return sol 

 

def _nonlinear_2eq_order1_type1(x, y, t, eq): 

    r""" 

    Equations: 

 

    .. math:: x' = x^n F(x,y) 

 

    .. math:: y' = g(y) F(x,y) 

 

    Solution: 

 

    .. math:: x = \varphi(y), \int \frac{1}{g(y) F(\varphi(y),y)} \,dy = t + C_2 

 

    where 

 

    if `n \neq 1` 

 

    .. math:: \varphi = [C_1 + (1-n) \int \frac{1}{g(y)} \,dy]^{\frac{1}{1-n}} 

 

    if `n = 1` 

 

    .. math:: \varphi = C_1 e^{\int \frac{1}{g(y)} \,dy} 

 

    where `C_1` and `C_2` are arbitrary constants. 

 

    """ 

    C1, C2 = symbols('C1:3') 

    n = Wild('n', exclude=[x(t),y(t)]) 

    f = Wild('f') 

    u, v, phi = symbols('u, v, phi', function=True) 

    r = eq[0].match(diff(x(t),t) - x(t)**n*f) 

    g = ((diff(y(t),t) - eq[1])/r[f]).subs(y(t),v) 

    F = r[f].subs(x(t),u).subs(y(t),v) 

    n = r[n] 

    if n!=1: 

        phi = (C1 + (1-n)*Integral(1/g, v))**(1/(1-n)) 

    else: 

        phi = C1*exp(Integral(1/g, v)) 

    phi = phi.doit() 

    sol2 = solve(Integral(1/(g*F.subs(u,phi)), v).doit() - t - C2, v) 

    sol = [] 

    for sols in sol2: 

        sol.append(Eq(x(t),phi.subs(v, sols))) 

        sol.append(Eq(y(t), sols)) 

    return sol 

 

def _nonlinear_2eq_order1_type2(x, y, t, eq): 

    r""" 

    Equations: 

 

    .. math:: x' = e^{\lambda x} F(x,y) 

 

    .. math:: y' = g(y) F(x,y) 

 

    Solution: 

 

    .. math:: x = \varphi(y), \int \frac{1}{g(y) F(\varphi(y),y)} \,dy = t + C_2 

 

    where 

 

    if `\lambda \neq 0` 

 

    .. math:: \varphi = -\frac{1}{\lambda} log(C_1 - \lambda \int \frac{1}{g(y)} \,dy) 

 

    if `\lambda = 0` 

 

    .. math:: \varphi = C_1 + \int \frac{1}{g(y)} \,dy 

 

    where `C_1` and `C_2` are arbitrary constants. 

 

    """ 

    C1, C2 = symbols('C1:3') 

    n = Wild('n', exclude=[x(t),y(t)]) 

    f = Wild('f') 

    u, v, phi = symbols('u, v, phi', function=True) 

    r = eq[0].match(diff(x(t),t) - exp(n*x(t))*f) 

    g = ((diff(y(t),t) - eq[1])/r[f]).subs(y(t),v) 

    F = r[f].subs(x(t),u).subs(y(t),v) 

    n = r[n] 

    if n: 

        phi = -1/n*log(C1 - n*Integral(1/g, v)) 

    else: 

        phi = C1 + Integral(1/g, v) 

    phi = phi.doit() 

    sol2 = solve(Integral(1/(g*F.subs(u,phi)), v).doit() - t - C2, v) 

    sol = [] 

    for sols in sol2: 

        sol.append(Eq(x(t),phi.subs(v, sols))) 

        sol.append(Eq(y(t), sols)) 

    return sol 

 

def _nonlinear_2eq_order1_type3(x, y, t, eq): 

    r""" 

    Autonomous system of general form 

 

    .. math:: x' = F(x,y) 

 

    .. math:: y' = G(x,y) 

 

    Assuming `y = y(x, C_1)` where `C_1` is an arbitrary constant is the general 

    solution of the first-order equation 

 

    .. math:: F(x,y) y'_x = G(x,y) 

 

    Then the general solution of the original system of equations has the form 

 

    .. math:: \int \frac{1}{F(x,y(x,C_1))} \,dx = t + C_1 

 

    """ 

    C1, C2, C3, C4 = symbols('C1:5') 

    u, v = symbols('u, v', function=True) 

    f = Wild('f') 

    g = Wild('g') 

    r1 = eq[0].match(diff(x(t),t) - f) 

    r2 = eq[1].match(diff(y(t),t) - g) 

    F = r1[f].subs(x(t),u).subs(y(t),v) 

    G = r2[g].subs(x(t),u).subs(y(t),v) 

    sol2r = dsolve(Eq(diff(v(u),u), G.subs(v,v(u))/F.subs(v,v(u)))) 

    for sol2s in sol2r: 

        sol1 = solve(Integral(1/F.subs(v, sol2s.rhs), u).doit() - t - C2, u) 

    sol = [] 

    for sols in sol1: 

        sol.append(Eq(x(t), sols)) 

        sol.append(Eq(y(t), (sol2s.rhs).subs(u, sols))) 

    return sol 

 

def _nonlinear_2eq_order1_type4(x, y, t, eq): 

    r""" 

    Equation: 

 

    .. math:: x' = f_1(x) g_1(y) \phi(x,y,t) 

 

    .. math:: y' = f_2(x) g_2(y) \phi(x,y,t) 

 

    First integral: 

 

    .. math:: \int \frac{f_2(x)}{f_1(x)} \,dx - \int \frac{g_1(y)}{g_2(y)} \,dy = C 

 

    where `C` is an arbitrary constant. 

 

    On solving the first integral for `x` (resp., `y` ) and on substituting the 

    resulting expression into either equation of the original solution, one 

    arrives at a firs-order equation for determining `y` (resp., `x` ). 

 

    """ 

    C1, C2 = symbols('C1:3') 

    u, v = symbols('u, v') 

    f = Wild('f') 

    g = Wild('g') 

    f1 = Wild('f1', exclude=[v,t]) 

    f2 = Wild('f2', exclude=[v,t]) 

    g1 = Wild('g1', exclude=[u,t]) 

    g2 = Wild('g2', exclude=[u,t]) 

    r1 = eq[0].match(diff(x(t),t) - f) 

    r2 = eq[1].match(diff(y(t),t) - g) 

    num, denum = ((r1[f].subs(x(t),u).subs(y(t),v))/(r2[g].subs(x(t),u).subs(y(t),v))).as_numer_denom() 

    R1 = num.match(f1*g1) 

    R2 = denum.match(f2*g2) 

    phi = (r1[f].subs(x(t),u).subs(y(t),v))/num 

    F1 = R1[f1]; F2 = R2[f2] 

    G1 = R1[g1]; G2 = R2[g2] 

    sol1r = solve(Integral(F2/F1, u).doit() - Integral(G1/G2,v).doit() - C1, u) 

    sol2r = solve(Integral(F2/F1, u).doit() - Integral(G1/G2,v).doit() - C1, v) 

    sol = [] 

    for sols in sol1r: 

        sol.append(Eq(y(t), dsolve(diff(v(t),t) - F2.subs(u,sols).subs(v,v(t))*G2.subs(v,v(t))*phi.subs(u,sols).subs(v,v(t))).rhs)) 

    for sols in sol2r: 

        sol.append(Eq(x(t), dsolve(diff(u(t),t) - F1.subs(u,u(t))*G1.subs(v,sols).subs(u,u(t))*phi.subs(v,sols).subs(u,u(t))).rhs)) 

    return set(sol) 

 

def _nonlinear_2eq_order1_type5(func, t, eq): 

    r""" 

    Clairaut system of ODEs 

 

    .. math:: x = t x' + F(x',y') 

 

    .. math:: y = t y' + G(x',y') 

 

    The following are solutions of the system 

 

    `(i)` straight lines: 

 

    .. math:: x = C_1 t + F(C_1, C_2), y = C_2 t + G(C_1, C_2) 

 

    where `C_1` and `C_2` are arbitrary constants; 

 

    `(ii)` envelopes of the above lines; 

 

    `(iii)` continuously differentiable lines made up from segments of the lines 

    `(i)` and `(ii)`. 

 

    """ 

    C1, C2 = symbols('C1:3') 

    f = Wild('f') 

    g = Wild('g') 

    def check_type(x, y): 

        r1 = eq[0].match(t*diff(x(t),t) - x(t) + f) 

        r2 = eq[1].match(t*diff(y(t),t) - y(t) + g) 

        if not (r1 and r2): 

            r1 = eq[0].match(diff(x(t),t) - x(t)/t + f/t) 

            r2 = eq[1].match(diff(y(t),t) - y(t)/t + g/t) 

        if not (r1 and r2): 

            r1 = (-eq[0]).match(t*diff(x(t),t) - x(t) + f) 

            r2 = (-eq[1]).match(t*diff(y(t),t) - y(t) + g) 

        if not (r1 and r2): 

            r1 = (-eq[0]).match(diff(x(t),t) - x(t)/t + f/t) 

            r2 = (-eq[1]).match(diff(y(t),t) - y(t)/t + g/t) 

        return [r1, r2] 

    for func_ in func: 

        if isinstance(func_, list): 

            x = func[0][0].func 

            y = func[0][1].func 

            [r1, r2] = check_type(x, y) 

            if not (r1 and r2): 

                [r1, r2] = check_type(y, x) 

                x, y = y, x 

    x1 = diff(x(t),t); y1 = diff(y(t),t) 

    return set([Eq(x(t), C1*t + r1[f].subs(x1,C1).subs(y1,C2)), Eq(y(t), C2*t + r2[g].subs(x1,C1).subs(y1,C2))]) 

 

def sysode_nonlinear_3eq_order1(match_): 

    x = match_['func'][0].func 

    y = match_['func'][1].func 

    z = match_['func'][2].func 

    eq = match_['eq'] 

    fc = match_['func_coeff'] 

    func = match_['func'] 

    t = list(list(eq[0].atoms(Derivative))[0].atoms(Symbol))[0] 

    if match_['type_of_equation'] == 'type1': 

        sol = _nonlinear_3eq_order1_type1(x, y, z, t, eq) 

    if match_['type_of_equation'] == 'type2': 

        sol = _nonlinear_3eq_order1_type2(x, y, z, t, eq) 

    if match_['type_of_equation'] == 'type3': 

        sol = _nonlinear_3eq_order1_type3(x, y, z, t, eq) 

    if match_['type_of_equation'] == 'type4': 

        sol = _nonlinear_3eq_order1_type4(x, y, z, t, eq) 

    if match_['type_of_equation'] == 'type5': 

        sol = _nonlinear_3eq_order1_type5(x, y, z, t, eq) 

    return sol 

 

def _nonlinear_3eq_order1_type1(x, y, z, t, eq): 

    r""" 

    Equations: 

 

    .. math:: a x' = (b - c) y z, \enspace b y' = (c - a) z x, \enspace c z' = (a - b) x y 

 

    First Integrals: 

 

    .. math:: a x^{2} + b y^{2} + c z^{2} = C_1 

 

    .. math:: a^{2} x^{2} + b^{2} y^{2} + c^{2} z^{2} = C_2 

 

    where `C_1` and `C_2` are arbitrary constants. On solving the integrals for `y` and 

    `z` and on substituting the resulting expressions into the first equation of the 

    system, we arrives at a separable first-order equation on `x`. Similarly doing that 

    for other two equations, we will arrive at first order equation on `y` and `z` too. 

 

    References 

    ========== 

    -http://eqworld.ipmnet.ru/en/solutions/sysode/sode0401.pdf 

 

    """ 

    C1, C2 = symbols('C1:3') 

    u, v, w = symbols('u, v, w') 

    p = Wild('p', exclude=[x(t), y(t), z(t), t]) 

    q = Wild('q', exclude=[x(t), y(t), z(t), t]) 

    s = Wild('s', exclude=[x(t), y(t), z(t), t]) 

    r = (diff(x(t),t) - eq[0]).match(p*y(t)*z(t)) 

    r.update((diff(y(t),t) - eq[1]).match(q*z(t)*x(t))) 

    r.update((diff(z(t),t) - eq[2]).match(s*x(t)*y(t))) 

    n1, d1 = r[p].as_numer_denom() 

    n2, d2 = r[q].as_numer_denom() 

    n3, d3 = r[s].as_numer_denom() 

    val = solve([n1*u-d1*v+d1*w, d2*u+n2*v-d2*w, d3*u-d3*v-n3*w],[u,v]) 

    vals = [val[v], val[u]] 

    c = lcm(vals[0].as_numer_denom()[1], vals[1].as_numer_denom()[1]) 

    b = vals[0].subs(w,c) 

    a = vals[1].subs(w,c) 

    y_x = sqrt(((c*C1-C2) - a*(c-a)*x(t)**2)/(b*(c-b))) 

    z_x = sqrt(((b*C1-C2) - a*(b-a)*x(t)**2)/(c*(b-c))) 

    z_y = sqrt(((a*C1-C2) - b*(a-b)*y(t)**2)/(c*(a-c))) 

    x_y = sqrt(((c*C1-C2) - b*(c-b)*y(t)**2)/(a*(c-a))) 

    x_z = sqrt(((b*C1-C2) - c*(b-c)*z(t)**2)/(a*(b-a))) 

    y_z = sqrt(((a*C1-C2) - c*(a-c)*z(t)**2)/(b*(a-b))) 

    try: 

        sol1 = dsolve(a*diff(x(t),t) - (b-c)*y_x*z_x).rhs 

    except: 

        sol1 = dsolve(a*diff(x(t),t) - (b-c)*y_x*z_x, hint='separable_Integral') 

    try: 

        sol2 = dsolve(b*diff(y(t),t) - (c-a)*z_y*x_y).rhs 

    except: 

        sol2 = dsolve(b*diff(y(t),t) - (c-a)*z_y*x_y, hint='separable_Integral') 

    try: 

        sol3 = dsolve(c*diff(z(t),t) - (a-b)*x_z*y_z).rhs 

    except: 

        sol3 = dsolve(c*diff(z(t),t) - (a-b)*x_z*y_z, hint='separable_Integral') 

    return [Eq(x(t), sol1), Eq(y(t), sol2), Eq(z(t), sol3)] 

 

def _nonlinear_3eq_order1_type2(x, y, z, t, eq): 

    r""" 

    Equations: 

 

    .. math:: a x' = (b - c) y z f(x, y, z, t) 

 

    .. math:: b y' = (c - a) z x f(x, y, z, t) 

 

    .. math:: c z' = (a - b) x y f(x, y, z, t) 

 

    First Integrals: 

 

    .. math:: a x^{2} + b y^{2} + c z^{2} = C_1 

 

    .. math:: a^{2} x^{2} + b^{2} y^{2} + c^{2} z^{2} = C_2 

 

    where `C_1` and `C_2` are arbitrary constants. On solving the integrals for `y` and 

    `z` and on substituting the resulting expressions into the first equation of the 

    system, we arrives at a first-order differential equations on `x`. Similarly doing 

    that for other two equations we will arrive at first order equation on `y` and `z`. 

 

    References 

    ========== 

    -http://eqworld.ipmnet.ru/en/solutions/sysode/sode0402.pdf 

 

    """ 

    C1, C2 = symbols('C1:3') 

    u, v, w = symbols('u, v, w') 

    p = Wild('p', exclude=[x(t), y(t), z(t), t]) 

    q = Wild('q', exclude=[x(t), y(t), z(t), t]) 

    s = Wild('s', exclude=[x(t), y(t), z(t), t]) 

    f = Wild('f') 

    r1 = (diff(x(t),t) - eq[0]).match(y(t)*z(t)*f) 

    r = collect_const(r1[f]).match(p*f) 

    r.update(((diff(y(t),t) - eq[1])/r[f]).match(q*z(t)*x(t))) 

    r.update(((diff(z(t),t) - eq[2])/r[f]).match(s*x(t)*y(t))) 

    n1, d1 = r[p].as_numer_denom() 

    n2, d2 = r[q].as_numer_denom() 

    n3, d3 = r[s].as_numer_denom() 

    val = solve([n1*u-d1*v+d1*w, d2*u+n2*v-d2*w, -d3*u+d3*v+n3*w],[u,v]) 

    vals = [val[v], val[u]] 

    c = lcm(vals[0].as_numer_denom()[1], vals[1].as_numer_denom()[1]) 

    a = vals[0].subs(w,c) 

    b = vals[1].subs(w,c) 

    y_x = sqrt(((c*C1-C2) - a*(c-a)*x(t)**2)/(b*(c-b))) 

    z_x = sqrt(((b*C1-C2) - a*(b-a)*x(t)**2)/(c*(b-c))) 

    z_y = sqrt(((a*C1-C2) - b*(a-b)*y(t)**2)/(c*(a-c))) 

    x_y = sqrt(((c*C1-C2) - b*(c-b)*y(t)**2)/(a*(c-a))) 

    x_z = sqrt(((b*C1-C2) - c*(b-c)*z(t)**2)/(a*(b-a))) 

    y_z = sqrt(((a*C1-C2) - c*(a-c)*z(t)**2)/(b*(a-b))) 

    try: 

        sol1 = dsolve(a*diff(x(t),t) - (b-c)*y_x*z_x*r[f]).rhs 

    except: 

        sol1 = dsolve(a*diff(x(t),t) - (b-c)*y_x*z_x*r[f], hint='separable_Integral') 

    try: 

        sol2 = dsolve(b*diff(y(t),t) - (c-a)*z_y*x_y*r[f]).rhs 

    except: 

        sol2 = dsolve(b*diff(y(t),t) - (c-a)*z_y*x_y*r[f], hint='separable_Integral') 

    try: 

        sol3 = dsolve(c*diff(z(t),t) - (a-b)*x_z*y_z*r[f]).rhs 

    except: 

        sol3 = dsolve(c*diff(z(t),t) - (a-b)*x_z*y_z*r[f], hint='separable_Integral') 

    return [Eq(x(t), sol1), Eq(y(t), sol2), Eq(z(t), sol3)] 

 

def _nonlinear_3eq_order1_type3(x, y, z, t, eq): 

    r""" 

    Equations: 

 

    .. math:: x' = c F_2 - b F_3, \enspace y' = a F_3 - c F_1, \enspace z' = b F_1 - a F_2 

 

    where `F_n = F_n(x, y, z, t)`. 

 

    1. First Integral: 

 

    .. math:: a x + b y + c z = C_1, 

 

    where C is an arbitrary constant. 

 

    2. If we assume function `F_n` to be independent of `t`,i.e, `F_n` = `F_n (x, y, z)` 

    Then, on eliminating `t` and `z` from the first two equation of the system, one 

    arrives at the first-order equation 

 

    .. math:: \frac{dy}{dx} = \frac{a F_3 (x, y, z) - c F_1 (x, y, z)}{c F_2 (x, y, z) - 

                b F_3 (x, y, z)} 

 

    where `z = \frac{1}{c} (C_1 - a x - b y)` 

 

    References 

    ========== 

    -http://eqworld.ipmnet.ru/en/solutions/sysode/sode0404.pdf 

 

    """ 

    C1 = symbols('C1') 

    u, v, w = symbols('u, v, w') 

    p = Wild('p', exclude=[x(t), y(t), z(t), t]) 

    q = Wild('q', exclude=[x(t), y(t), z(t), t]) 

    s = Wild('s', exclude=[x(t), y(t), z(t), t]) 

    F1, F2, F3 = symbols('F1, F2, F3', cls=Wild) 

    r1 = (diff(x(t),t) - eq[0]).match(F2-F3) 

    r = collect_const(r1[F2]).match(s*F2) 

    r.update(collect_const(r1[F3]).match(q*F3)) 

    if eq[1].has(r[F2]) and not eq[1].has(r[F3]): 

        r[F2], r[F3] = r[F3], r[F2] 

        r[s], r[q] = -r[q], -r[s] 

    r.update((diff(y(t),t) - eq[1]).match(p*r[F3] - r[s]*F1)) 

    a = r[p]; b = r[q]; c = r[s] 

    F1 = r[F1].subs(x(t),u).subs(y(t),v).subs(z(t),w) 

    F2 = r[F2].subs(x(t),u).subs(y(t),v).subs(z(t),w) 

    F3 = r[F3].subs(x(t),u).subs(y(t),v).subs(z(t),w) 

    z_xy = (C1-a*u-b*v)/c 

    y_zx = (C1-a*u-c*w)/b 

    x_yz = (C1-b*v-c*w)/a 

    y_x = dsolve(diff(v(u),u) - ((a*F3-c*F1)/(c*F2-b*F3)).subs(w,z_xy).subs(v,v(u))).rhs 

    z_x = dsolve(diff(w(u),u) - ((b*F1-a*F2)/(c*F2-b*F3)).subs(v,y_zx).subs(w,w(u))).rhs 

    z_y = dsolve(diff(w(v),v) - ((b*F1-a*F2)/(a*F3-c*F1)).subs(u,x_yz).subs(w,w(v))).rhs 

    x_y = dsolve(diff(u(v),v) - ((c*F2-b*F3)/(a*F3-c*F1)).subs(w,z_xy).subs(u,u(v))).rhs 

    y_z = dsolve(diff(v(w),w) - ((a*F3-c*F1)/(b*F1-a*F2)).subs(u,x_yz).subs(v,v(w))).rhs 

    x_z = dsolve(diff(u(w),w) - ((c*F2-b*F3)/(b*F1-a*F2)).subs(v,y_zx).subs(u,u(w))).rhs 

    sol1 = dsolve(diff(u(t),t) - (c*F2 - b*F3).subs(v,y_x).subs(w,z_x).subs(u,u(t))).rhs 

    sol2 = dsolve(diff(v(t),t) - (a*F3 - c*F1).subs(u,x_y).subs(w,z_y).subs(v,v(t))).rhs 

    sol3 = dsolve(diff(w(t),t) - (b*F1 - a*F2).subs(u,x_z).subs(v,y_z).subs(w,w(t))).rhs 

    return [Eq(x(t), sol1), Eq(y(t), sol2), Eq(z(t), sol3)] 

 

def _nonlinear_3eq_order1_type4(x, y, z, t, eq): 

    r""" 

    Equations: 

 

    .. math:: x' = c z F_2 - b y F_3, \enspace y' = a x F_3 - c z F_1, \enspace z' = b y F_1 - a x F_2 

 

    where `F_n = F_n (x, y, z, t)` 

 

    1. First integral: 

 

    .. math:: a x^{2} + b y^{2} + c z^{2} = C_1 

 

    where `C` is an arbitrary constant. 

 

    2. Assuming the function `F_n` is independent of `t`: `F_n = F_n (x, y, z)`. Then on 

    eliminating `t` and `z` from the first two equations of the system, one arrives at 

    the first-order equation 

 

    .. math:: \frac{dy}{dx} = \frac{a x F_3 (x, y, z) - c z F_1 (x, y, z)} 

                {c z F_2 (x, y, z) - b y F_3 (x, y, z)} 

 

    where `z = \pm \sqrt{\frac{1}{c} (C_1 - a x^{2} - b y^{2})}` 

 

    References 

    ========== 

    -http://eqworld.ipmnet.ru/en/solutions/sysode/sode0405.pdf 

 

    """ 

    C1 = symbols('C1') 

    u, v, w = symbols('u, v, w') 

    p = Wild('p', exclude=[x(t), y(t), z(t), t]) 

    q = Wild('q', exclude=[x(t), y(t), z(t), t]) 

    s = Wild('s', exclude=[x(t), y(t), z(t), t]) 

    F1, F2, F3 = symbols('F1, F2, F3', cls=Wild) 

    r1 = eq[0].match(diff(x(t),t) - z(t)*F2 + y(t)*F3) 

    r = collect_const(r1[F2]).match(s*F2) 

    r.update(collect_const(r1[F3]).match(q*F3)) 

    if eq[1].has(r[F2]) and not eq[1].has(r[F3]): 

        r[F2], r[F3] = r[F3], r[F2] 

        r[s], r[q] = -r[q], -r[s] 

    r.update((diff(y(t),t) - eq[1]).match(p*x(t)*r[F3] - r[s]*z(t)*F1)) 

    a = r[p]; b = r[q]; c = r[s] 

    F1 = r[F1].subs(x(t),u).subs(y(t),v).subs(z(t),w) 

    F2 = r[F2].subs(x(t),u).subs(y(t),v).subs(z(t),w) 

    F3 = r[F3].subs(x(t),u).subs(y(t),v).subs(z(t),w) 

    x_yz = sqrt((C1 - b*v**2 - c*w**2)/a) 

    y_zx = sqrt((C1 - c*w**2 - a*u**2)/b) 

    z_xy = sqrt((C1 - a*u**2 - b*v**2)/c) 

    y_x = dsolve(diff(v(u),u) - ((a*u*F3-c*w*F1)/(c*w*F2-b*v*F3)).subs(w,z_xy).subs(v,v(u))).rhs 

    z_x = dsolve(diff(w(u),u) - ((b*v*F1-a*u*F2)/(c*w*F2-b*v*F3)).subs(v,y_zx).subs(w,w(u))).rhs 

    z_y = dsolve(diff(w(v),v) - ((b*v*F1-a*u*F2)/(a*u*F3-c*w*F1)).subs(u,x_yz).subs(w,w(v))).rhs 

    x_y = dsolve(diff(u(v),v) - ((c*w*F2-b*v*F3)/(a*u*F3-c*w*F1)).subs(w,z_xy).subs(u,u(v))).rhs 

    y_z = dsolve(diff(v(w),w) - ((a*u*F3-c*w*F1)/(b*v*F1-a*u*F2)).subs(u,x_yz).subs(v,v(w))).rhs 

    x_z = dsolve(diff(u(w),w) - ((c*w*F2-b*v*F3)/(b*v*F1-a*u*F2)).subs(v,y_zx).subs(u,u(w))).rhs 

    sol1 = dsolve(diff(u(t),t) - (c*w*F2 - b*v*F3).subs(v,y_x).subs(w,z_x).subs(u,u(t))).rhs 

    sol2 = dsolve(diff(v(t),t) - (a*u*F3 - c*w*F1).subs(u,x_y).subs(w,z_y).subs(v,v(t))).rhs 

    sol3 = dsolve(diff(w(t),t) - (b*v*F1 - a*u*F2).subs(u,x_z).subs(v,y_z).subs(w,w(t))).rhs 

    return [Eq(x(t), sol1), Eq(y(t), sol2), Eq(z(t), sol3)] 

 

def _nonlinear_3eq_order1_type5(x, y, t, eq): 

    r""" 

    .. math:: x' = x (c F_2 - b F_3), \enspace y' = y (a F_3 - c F_1), \enspace z' = z (b F_1 - a F_2) 

 

    where `F_n = F_n (x, y, z, t)` and are arbitrary functions. 

 

    First Integral: 

 

    .. math:: \left|x\right|^{a} \left|y\right|^{b} \left|z\right|^{c} = C_1 

 

    where `C` is an arbitrary constant. If the function `F_n` is independent of `t`, 

    then, by eliminating `t` and `z` from the first two equations of the system, one 

    arrives at a first-order equation. 

 

    References 

    ========== 

    -http://eqworld.ipmnet.ru/en/solutions/sysode/sode0406.pdf 

 

    """ 

    C1 = symbols('C1') 

    u, v, w = symbols('u, v, w') 

    p = Wild('p', exclude=[x(t), y(t), z(t), t]) 

    q = Wild('q', exclude=[x(t), y(t), z(t), t]) 

    s = Wild('s', exclude=[x(t), y(t), z(t), t]) 

    F1, F2, F3 = symbols('F1, F2, F3', cls=Wild) 

    r1 = eq[0].match(diff(x(t),t) - x(t)*(F2 - F3)) 

    r = collect_const(r1[F2]).match(s*F2) 

    r.update(collect_const(r1[F3]).match(q*F3)) 

    if eq[1].has(r[F2]) and not eq[1].has(r[F3]): 

        r[F2], r[F3] = r[F3], r[F2] 

        r[s], r[q] = -r[q], -r[s] 

    r.update((diff(y(t),t) - eq[1]).match(y(t)*(a*r[F3] - r[c]*F1))) 

    a = r[p]; b = r[q]; c = r[s] 

    F1 = r[F1].subs(x(t),u).subs(y(t),v).subs(z(t),w) 

    F2 = r[F2].subs(x(t),u).subs(y(t),v).subs(z(t),w) 

    F3 = r[F3].subs(x(t),u).subs(y(t),v).subs(z(t),w) 

    x_yz = (C1*v**-b*w**-c)**-a 

    y_zx = (C1*w**-c*u**-a)**-b 

    z_xy = (C1*u**-a*v**-b)**-c 

    y_x = dsolve(diff(v(u),u) - ((v*(a*F3-c*F1))/(u*(c*F2-b*F3))).subs(w,z_xy).subs(v,v(u))).rhs 

    z_x = dsolve(diff(w(u),u) - ((w*(b*F1-a*F2))/(u*(c*F2-b*F3))).subs(v,y_zx).subs(w,w(u))).rhs 

    z_y = dsolve(diff(w(v),v) - ((w*(b*F1-a*F2))/(v*(a*F3-c*F1))).subs(u,x_yz).subs(w,w(v))).rhs 

    x_y = dsolve(diff(u(v),v) - ((u*(c*F2-b*F3))/(v*(a*F3-c*F1))).subs(w,z_xy).subs(u,u(v))).rhs 

    y_z = dsolve(diff(v(w),w) - ((v*(a*F3-c*F1))/(w*(b*F1-a*F2))).subs(u,x_yz).subs(v,v(w))).rhs 

    x_z = dsolve(diff(u(w),w) - ((u*(c*F2-b*F3))/(w*(b*F1-a*F2))).subs(v,y_zx).subs(u,u(w))).rhs 

    sol1 = dsolve(diff(u(t),t) - (u*(c*F2-b*F3)).subs(v,y_x).subs(w,z_x).subs(u,u(t))).rhs 

    sol2 = dsolve(diff(v(t),t) - (v*(a*F3-c*F1)).subs(u,x_y).subs(w,z_y).subs(v,v(t))).rhs 

    sol3 = dsolve(diff(w(t),t) - (w*(b*F1-a*F2)).subs(u,x_z).subs(v,y_z).subs(w,w(t))).rhs 

    return [Eq(x(t), sol1), Eq(y(t), sol2), Eq(z(t), sol3)]