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.

    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.
  - 2nd order Liouville differential equations.
  - `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:`C.Integral()
<sympy.core.C.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.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 collections import defaultdict

from sympy.core import Add, C, S, Mul, Pow, oo
from sympy.core.compatibility import ordered, iterable, is_sequence, set_union
from sympy.utilities.exceptions import SymPyDeprecationWarning
from sympy.core.exprtools import factor_terms, gcd_terms
from sympy.core.function import (Function, Derivative, AppliedUndef, diff,
    expand, expand_mul)
from sympy.core.multidimensional import vectorize
from sympy.core.numbers import Rational, NaN
from sympy.core.relational import Equality, Eq
from sympy.core.symbol import Symbol, Wild, Dummy, symbols
from sympy.core.sympify import sympify

from sympy.functions import cos, exp, im, log, re, sin, tan, sqrt, sign, Piecewise
from sympy.matrices import wronskian
from sympy.polys import Poly, RootOf, terms_gcd, PolynomialError
from sympy.polys.polytools import cancel, degree
from sympy.series import Order
from sympy.simplify import collect, logcombine, powsimp, separatevars, \
    simplify, trigsimp, denom, fraction, posify
from sympy.simplify.simplify import _mexpand
from sympy.solvers import solve

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",
    "nth_linear_constant_coeff_homogeneous",
    "nth_linear_euler_eq_homogeneous",
    "nth_linear_constant_coeff_undetermined_coefficients",
    "nth_linear_constant_coeff_variation_of_parameters",
    "Liouville",
    "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",
    "Liouville_Integral",
    )


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 = C.Dummy('u')
        repu[u] = d.subs(func, new).doit()
        reps[d] = u

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


def dsolve(eq, func=None, hint="default", simplify=True, **kwargs):
    r"""
    Solves any (supported) kind of ordinary differential equation.

    **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.

    **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``.

    Examples
    ========

    >>> from sympy import Function, dsolve, Eq, Derivative, sin, cos
    >>> from sympy.abc import x
    >>> f = Function('f')
    >>> dsolve(Derivative(f(x), x, x) + 9*f(x), f(x))
    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='separable_reduced')
    f(x) == C1/(C2*x - 1)
    >>> dsolve(eq, hint='1st_exact')
    [f(x) == -acos(C1/cos(x)) + 2*pi, f(x) == acos(C1/cos(x))]
    >>> dsolve(eq, hint='almost_linear')
    [f(x) == -acos(-sqrt(C1/cos(x)**2)) + 2*pi, f(x) == -acos(sqrt(C1/cos(x)**2)) + 2*pi,
    f(x) == acos(-sqrt(C1/cos(x)**2)), f(x) == acos(sqrt(C1/cos(x)**2))]
    >>> dsolve(eq, hint='best')
    f(x) == C1/(C2*x - 1)

    """
    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, type='ode', **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, detail:
                failed_hints[hint] = detail
            else:
                retdict[hint] = rv
        func = hints[hint]['func']
        retdict['best'] = min(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):
    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
        rv = odesimp(solvefunc(eq, func, order, match), func, order, hint)
        return rv
    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, **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',
    '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)
    from sympy import expand

    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')
    if isinstance(eq, Equality):
        if eq.rhs != 0:
            return classify_ode(eq.lhs - eq.rhs, func, 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])

    eq = expand(eq)

    # 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 != 1:
            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 (
                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 1577
            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)

            ## 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(C.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(C.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 1525 and issue 1620.
                pass

        # 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)):
                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

    if 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

    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

        # Euler equation case (a_i * x**i for all i)
        def _test_term(coeff, order):
            r"""
            Linear Euler ODEs have the form  K*x**order*diff(y(x),x,order),
            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.
            """
            assert order >= 0
            if coeff == 0:
                return True
            if order == 0:
                if x in coeff.free_symbols:
                    return False
                return f(x) not in coeff.atoms()
            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

    # 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)

@vectorize(0)
def odesimp(eq, func, order, 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,
    ... 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 = Symbol('C1')

    # First, integrate if the hint allows it.
    eq = _handle_Integral(eq, func, order, hint)
    assert isinstance(eq, 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, x, 2*order)

    # 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
            global collectterms
            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
            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 costants 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):
        eqi = constantsimp(eqi, x, 2*order)
        eq[i] = constant_renumber(eqi, '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 = reduce(set.union, funcs, set())
            if len(funcs) != 1:
                raise ValueError(
                    'must pass func arg to checkodesol for this case.')
            func = funcs.pop()
    # ========== deprecation handling
    # After the deprecation period this handling section becomes:
    # ----------
    # if not is_unfunc(func) or len(func.args) != 1:
    #     raise ValueError(
    #         "func must be a function of one variable, not %s" % func)
    # ----------
    # assume, during deprecation, that sol and func are reversed
    if isinstance(sol, AppliedUndef) and len(sol.args) == 1:
        if isinstance(func, AppliedUndef) and len(func.args) == 1:
            msg = "If you really do want sol to be just %s, use Eq(%s, 0) " % \
                (sol, sol) + "instead."
        else:
            msg = ""
        SymPyDeprecationWarning(msg, feature="The order of the "
            "arguments sol and func to checkodesol()",
            useinstead="checkodesol(ode, sol, func)", issue=3384,
        ).warn()
        sol, func = func, sol
    elif not (isinstance(func, AppliedUndef) and len(func.args) == 1):
        from sympy.utilities.misc import filldedent
        raise ValueError(filldedent('''
        func (or sol, during deprecation) must be a function
        of one variable. Got sol = %s, func = %s''' % (sol, func)))
    # ========== end of deprecation handling
    if is_sequence(sol, set):
        return type(sol)(map(lambda i: checkodesol(ode, i, order=order,
            solve_for_func=solve_for_func), 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):
                        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]]
    [26, 33]
    >>> min([eq1, eq2], key=lambda i: ode_sol_simplicity(i, f(x)))
    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(C.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))


@vectorize(0)
def constantsimp(expr, independentsymbol, endnumber, startnumber=1,
                 symbolname='C'):
    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 in to 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 is done with limited assistance: terms of
    :py:class:`~sympy.core.add.Add`\s are collected to try join constants and
    powers with exponents that are :py:class:`~sympy.core.add.Add`\s are
    expanded so `e^x (C_1 \cos(x) + C_2 \cos(x))` will simplify to `e^x C_1
    \cos(x)` and `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, x, 3)
    C1*x
    >>> constantsimp(C1 + 2 + x + y, x, 3)
    C1 + x
    >>> constantsimp(C1*C2 + 2 + x + y + C3*x, x, 3)
    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.

    if type(symbolname) is tuple:
        x, endnumber, startnumber, constantsymbols = symbolname
    else:
        constantsymbols = symbols(
            symbolname + '%i:%i' % (startnumber, endnumber + 1))
        x = independentsymbol
    con_set = set(constantsymbols)
    ARGS = None, None, None, (
        x, endnumber, startnumber, constantsymbols)

    if isinstance(expr, Equality):
        # For now, only treat the special case where one side of the equation
        # is a constant
        if expr.lhs in con_set:
            return Eq(expr.lhs, constantsimp(expr.rhs + expr.lhs, *ARGS) - expr.lhs)
            # this could break if expr.lhs is absorbed into another constant,
            # but for now, the only solutions that return Eq's with a constant
            # on one side are first order.  At any rate, it will still be
            # technically correct.  The expression will just have too many
            # constants in it
        elif expr.rhs in con_set:
            return Eq(constantsimp(expr.lhs + expr.rhs, *ARGS) - expr.rhs, expr.rhs)
        else:
            return Eq(constantsimp(expr.lhs, *ARGS), constantsimp(expr.rhs, *ARGS))

    if not hasattr(expr, 'has') or not expr.has(*constantsymbols):
        return expr
    else:
        # ================ pre-processing ================
        def _take(i):
            # return the lowest numbered constant symbol that appears in ``i``
            # else return ``i``
            c = i.free_symbols & con_set
            if c:
                return min(c)
            return i

        if not (expr.has(x) and x in expr.free_symbols):
            return constantsymbols[0]

        # collect terms to get constants together
        new_expr = terms_gcd(expr, clear=False, deep=True, expand=False)

        if new_expr.is_Mul:
            # don't let C1*exp(x) + C2*exp(2*x) become exp(x)*(C1 + C2*exp(x))
            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
        expr = new_expr
        # don't allow a number to be factored out of an expression
        # that has no denominator
        if expr.is_Mul:
            h, t = expr.as_coeff_Mul()
            if h != 1 and (t.is_Add or denom(t) == 1):
                args = list(Mul.make_args(t))
                for i, a in enumerate(args):
                    if a.is_Add:
                        args[i] = h*a
                        expr = Mul._from_args(args)
                        break
            # let numbers absorb into constants of an Add, perhaps
            # in the base of a power, if all its terms have a constant
            # symbol in them, e.g. sqrt(2)*(C1 + C2*x) -> C1 + C2*x
            if expr.is_Mul:
                d = sift(expr.args, lambda m: m.is_number is True)
                num = d[True]
                other = d[False]
                if num:
                    for o in other:
                        b, e = o.as_base_exp()
                        if b.is_Add and \
                                all(a.args_cnc(cset=True, warn=False)[0] &
                                con_set for a in b.args):
                            expr = sign(Mul(*num))*Mul._from_args(other)
                            break
        if expr.is_Mul:  # check again that it's still a Mul
            i, d = expr.as_independent(x, strict=True)
            newi = _take(i)
            if newi != i:
                expr = newi*d
        elif expr.is_Add:
            i, d = expr.as_independent(x, strict=True)
            expr = _take(i) + d
            if expr.is_Add:
                terms = {}
                for ai in expr.args:
                    i, d = ai.as_independent(x, strict=True, as_Add=False)
                    terms.setdefault(d, []).append(i)
                expr = Add(*[k*Add(*v) for k, v in terms.items()])
        # handle powers like exp(C0 + g(x)) -> C0*exp(g(x))
        pows = [p for p in expr.atoms(C.Function, C.Pow) if
                (p.is_Pow or p.func is exp) and
                p.exp.is_Add and
                p.exp.as_independent(x, strict=True)[1]]
        if pows:
            reps = []
            for p in pows:
                b, e = p.as_base_exp()
                ei, ed = e.as_independent(x, strict=True)
                e = _take(ei)
                if e != ei or e in constantsymbols:
                    reps.append((p, e*b**ed))
            expr = expr.xreplace(dict(reps))
            # a C1*C2 may have been introduced and the code below won't
            # handle that so handle it now: once to handle the C1*C2
            # and once to handle any C0*f(x) + C0*f(x)
            for _ in range(2):
                muls = [m for m in expr.atoms(Mul) if m.has(*constantsymbols)]
                reps = []
                for m in muls:
                    i, d = m.as_independent(x, strict=True)
                    newi = _take(i)
                    if newi != i:
                        reps.append((m, _take(i)*d))
                expr = expr.xreplace(dict(reps))
        # ================ end of pre-processing ================
        newargs = []
        hasconst = False
        isPowExp = False
        reeval = False
        for i in expr.args:
            if i not in constantsymbols:
                newargs.append(i)
            else:
                newconst = i
                hasconst = True
                if expr.is_Pow and i == expr.exp:
                    isPowExp = True

        for i in range(len(newargs)):
            isimp = constantsimp(newargs[i], *ARGS)
            if isimp in constantsymbols:
                reeval = True
                hasconst = True
                newconst = isimp
                if expr.is_Pow and i == 1:
                    isPowExp = True
            newargs[i] = isimp
        if hasconst:
            newargs = [i for i in newargs if i.has(x)]
            if isPowExp:
                newargs = newargs + [newconst]  # Order matters in this case
            else:
                newargs = [newconst] + newargs
        if expr.is_Pow and len(newargs) == 1:
            newargs.append(S.One)
        if expr.is_Function:
            if (len(newargs) == 0 or hasconst and len(newargs) == 1):
                return newconst
            else:
                newfuncargs = [constantsimp(t, *ARGS) for t in expr.args]
                return expr.func(*newfuncargs)
        else:
            newexpr = expr.func(*newargs)
            if reeval:
                return constantsimp(newexpr, *ARGS)
            else:
                return newexpr


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.

    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)(map(lambda i: constant_renumber(i, symbolname=symbolname,
            startnumber=startnumber, endnumber=endnumber), expr))
    global newstartnumber
    newstartnumber = 1

    def _constant_renumber(expr, symbolname, startnumber, endnumber):
        r"""
        We need to have an internal recursive function so that
        newstartnumber maintains its values throughout recursive calls.

        """
        constantsymbols = [Symbol(
            symbolname + "%d" % t) for t in range(startnumber,
        endnumber + 1)]
        global newstartnumber

        if isinstance(expr, Equality):
            return Eq(
                _constant_renumber(
                    expr.lhs, symbolname, startnumber, endnumber),
                _constant_renumber(
                    expr.rhs, symbolname, startnumber, endnumber))

        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:
            # Renumbering happens here
            newconst = Symbol(symbolname + str(newstartnumber))
            newstartnumber += 1
            return newconst
        else:
            from sympy.core.containers import Tuple
            if expr.is_Function or expr.is_Pow or isinstance(expr, Tuple):
                return expr.func(
                    *[_constant_renumber(x, symbolname, startnumber,
                endnumber) for x in expr.args])
            else:
                sortedargs = list(expr.args)
                # 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]
                sortedargs.sort(
                    key=lambda arg: default_sort_key(arg.subs(C_1)))
                return expr.func(
                    *[_constant_renumber(x, symbolname, startnumber,
                    endnumber) for x in sortedargs])

    return _constant_renumber(expr, symbolname, startnumber, endnumber)


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')
    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 = Symbol('C1')
    # 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 = C.Integral(d, x) + C.Integral((e - (C.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:`~ode_sol_simplicity`.

    See the :py:meth:`~ode_1st_homogeneous_coeff_subs_indep_div_dep` and
    :py:meth:`~ode_1st_homogeneous_coeff_subs_dep_div_indep` docstrings for
    more information on these hints.  Note that there is no
    ``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:
        sol1 = odesimp(
            sol1, func, order, "1st_homogeneous_coeff_subs_indep_div_dep")
        sol2 = odesimp(
            sol2, func, order, "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:`~ode_1st_homogeneous_coeff_best` and
    :py:meth:`~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 = Symbol('C1')
    xarg = match.get('xarg', 0)
    yarg = match.get('yarg', 0)
    int = C.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:`~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 = Symbol('C1')
    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 = C.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

    """
    from sympy.simplify.simplify import separatevars

    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 = dum.next()
            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 = Symbol('C1')
    t = exp(C.Integral(r[r['b']]/r[r['a']], x))
    tt = C.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 = Symbol('C1')
    t = exp((1 - r[r['n']])*C.Integral(r[r['b']]/r[r['a']], x))
    tt = (r[r['n']] - 1)*C.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 = Symbol('C1')
    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 = Symbol('C1')
    C2 = Symbol('C2')
    int = C.Integral(exp(C.Integral(r['g'], y)), (y, None, f(x)))
    sol = Eq(int + C1*C.Integral(exp(-C.Integral(r['h'], x)), x) + C2, 0)
    return sol


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
        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

    # A generator of constants
    constants = numbered_symbols(prefix='C', cls=Symbol, start=1)

    # 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 xrange(chareq.degree())]

    # 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.next()
                assert multiplicity == 1
                collectterms = [(0, root, 0)] + collectterms
            elif root.is_real:
                gsol += ln(x)**i*(x**root)*constants.next()
                collectterms = [(i, root, 0)] + collectterms
            else:
                reroot = re(root)
                imroot = im(root)
                gsol += ln(x)**i*(x**reroot)*(
                    constants.next()*sin(abs(imroot)*ln(x))
                    + constants.next()*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_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')
    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
         fi.nargs == 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')
    [f(x) == -x - sqrt(C1 + 7*x**2) - 1, 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')
    [f(x) == (-sqrt(C1*x**2 + 1) + 1)/x, 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_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
    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

    # A generator of constants
    constants = numbered_symbols(prefix='C', cls=Symbol, start=1)

    # 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 xrange(chareq.degree()) ]

    # 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.
    global collectterms
    collectterms = []
    for root, multiplicity in charroots.items():
        for i in range(multiplicity):
            if isinstance(root, RootOf):
                gsol += exp(root*x)*constants.next()
                assert multiplicity == 1
                collectterms = [(0, root, 0)] + collectterms
            else:
                reroot = re(root)
                imroot = im(root)
                gsol += x**i*exp(reroot*x)*(constants.next()*sin(abs(imroot)*x)
                + constants.next()*cos(imroot*x))
                # This ordering is important
                collectterms = [(i, reroot, imroot)] + collectterms
    if returns == 'sol':
        return Eq(f(x), gsol)
    elif returns in ('list' 'both'):
        # 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(x**i*exp(reroot*x))
            else:
                if x**i*exp(reroot*x)*sin(abs(imroot)*x) in gensols:
                    gensols.append(x**i*exp(reroot*x)*cos(imroot*x))
                else:
                    gensols.append(x**i*exp(reroot*x)*sin(abs(imroot)*x))
        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_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 nessesary 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 = coeffs.next()
        coefflist.append(c)
        trialfunc += c*i

    eqs = sub_func_doit(eq, f(x), trialfunc)

    coeffsdict = dict(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(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}

    """
    from sympy import S
    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.
            """
            # I was using the below match, but it doesn't always put all of the
            # coefficient in c.  c.f. 2**x*6*exp(x)*log(2)
            # The below code is probably cleaner anyway.
#            c = Wild('c', exclude=[x])
#            t = Wild('t')
#            r = expr.match(c*t)
            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/encyclopedia/VariationOfParameters.html
    - 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.

    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 1563, 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*C.Integral(wronskian(filter(lambda x: x != i,
        gensols), 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.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.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 = Symbol('C1')
    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(C.Integral(r['m2']['coeff']*r['m2'][r['y']]/r['m1'][r['y']],
        (r['y'], None, u)), C.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 for the given first order differential equation.
    As of now, it simply checks, by substituting the infinitesimals in the
    partial differential equation.
    (eta.diff(x) + (eta.diff(y) - xi.diff(x))*h -
    (xi.diff(y))*h**2 - xi*(h.diff(x)) - eta*(h.diff(y))) = 0
    where eta, and xi are the infinitesimals and h(x,y) = 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 zero.

    """
    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')
            match = collect(expand(eq), df).match(a*df + b)
            h = -match[b]/match[a]
            y = Symbol('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))
                else:
                    soltup.append((True, 0))
            return soltup

def infinitesimals(eq, func=None, order=None, **kwargs):
    r"""
    The functions xi and eta, are called the infinitesimals which help in
    the process of finding a new co-ordinate system, in which a differential
    equation can be simplified. They are tangents to the coordinate curves, in
    which the differential equation is simplified.

    Consider the transformation (x, y) --> (X, Y) such that X and Y are
    f(x, y, lambda) and g(x, y, lambda) and such that the
    differential equation remains invariant. Xi and eta are the tangents to
    the transformed coordinates X and Y, when lambda is the identity.

    The infinitesimals can be found by solving the following Partial Differential
    Equation.

        >>> 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

    Once the infinitesimals are found, the following partial differential
    equations would help us find the transformed coordinates (X, Y)

    1. X.diff(x)*xi + X.diff(y)*eta = 0
    2. Y.diff(x)*xi + Y.diff(y)*eta = 1


    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},
    {eta(x, f(x)): f(x), xi(x, f(x)): 0},
    {eta(x, f(x)): 0, xi(x, f(x)): x**(-2)},
    {eta(x, f(x)): x**2*f(x) + f(x), xi(x, f(x)): 1}]


    References
    ==========

    - Solving differential equations by Symmetry Groups,
      John Starrett, pp. 1 - pp. 14


    """
    from sympy.integrals.integrals import integrate

    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')
            match = kwargs.get('match',
                collect(expand(eq), df).match(a*df + b))
            h = -simplify(match[b]/match[a])
            y = Symbol('y')
            h = h.subs(func, y)
            hsyms = h.free_symbols
            xi = Function('xi')(x, func)
            eta = Function('eta')(x, func)
            hx = h.diff(x)
            hy = h.diff(y)
            xieta = []
            # This is the PDE that has to be solved using various
            # heuristics. The purpose is to "intelligently" guess
            # the functions xi and eta. All the heuristics are cited
            # from the paper "Computer Algebra Solving of first order
            # ODE's Using Symmetry Methods" unless otherwise specified.
            # Here dy/dx = h
            # [eta.diff(x) + (eta.diff(y) - xi.diff(x))*h
            # - xi.diff(y)*h**2 - xi*(h.diff(x)) - eta*(h.diff(y))) = 0]
            # The first heuristic uses the following four sets of
            # assumptions on xi and eta
            # 1. [xi = 0, eta = f(x)]
            # 2. [xi = 0, eta = f(y)]
            # 3. [xi = f(x), eta = 0]
            # 4. [xi = f(y), eta = 0]
            # Assuming xi = 0 and eta to be a function of x, the PDE
            # reduces to eta.diff(x) - eta*(h.diff(y)) = 0
            # If h.diff(y) is a function of x, then this can usually
            # be integrated easily.
            hysym = hy.free_symbols
            if y not in hysym:
                try:
                    fx = exp(integrate(hy, x))
                except NotImplementedError:
                    pass
                else:
                    inf = {xi: 0, eta: fx.subs(y, func)}
                    if inf not in xieta:
                        xieta.append(inf)

            # Assuming xi = 0 and eta to be a function of y, the PDE
            # reduces to eta.diff(y)*h - eta*(h.diff(y)) = 0
            # If h.diff(y)/h is a function of y, then this can usually
            # be integrated easily.
            factor = hy/h
            facsym = factor.free_symbols
            if x not in facsym:
                try:
                    fy = exp(integrate(factor, y))
                except NotImplementedError:
                    pass
                else:
                    inf = {xi: 0, eta: fy.subs(y, func)}
                    if inf not in xieta:
                        xieta.append(inf)

            # Assuming eta = 0 and xi to be a function of x, the PDE
            # reduces to - (xi.diff(x))*h - xi*(h.diff(x))
            # If -h.diff(x)/h is a function of x, then this can usually
            # be integrated easily.
            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.subs(y, func), eta: 0}
                    if inf not in xieta:
                        xieta.append(inf)

            # Assuming eta = 0 and xi to be a function of y, the PDE
            # reduces to - xi.diff(y)*h**2 - xi*(h.diff(x))
            # If -h.diff(x)/h**2 is a function of y, then this can usually
            # be integrated easily.
            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: 0}
                    if inf not in xieta:
                        xieta.append(inf)

            # The second heuristic uses the following four sets of
            # assumptions on xi and eta
            # 1. [xi = 0, eta = f(x)*g(y)]
            # 2. [xi = 0, eta = f(y)*g(x)]
            # 3. [xi = f(x)*g(y), eta = 0]
            # 4. [xi = f(y)*g(x), eta = 0]
            # g is a function built by extracting algebraic factors from the
            # numerator and denominator of the ODE to be solved and f is an
            # unknown function to be determined by solving auxiliary ODE's
            # after substituting g in the PDE.
            # If the auxilliary ODE obtained is separable in f, it is assumed
            # that this heuristic works.
            argx = []  # Extracting factors containing x only
            argy = []  # Extracting factors containing y only
            Fx = Function('F')(x)
            Fy = Function('F')(y)
            c = Wild('c')
            d = Wild('d', exclude=[Fx.diff(x), Fy.diff(y)])
            gcd = gcd_terms(h)
            if gcd.is_Pow:
                base, power = gcd.as_base_exp()
                if gcd.has(y) and not gcd.has(x) and \
                    gcd.is_algebraic_expr(y):
                    argy.append(base**power)
                if gcd.has(x) and not gcd.has(y) and \
                    gcd.is_algebraic_expr(x):
                    argx.append(base**power)
            elif gcd.is_Mul:
                factors = gcd.args
                for arg in factors:
                    if arg.has(y) and not arg.has(x) and \
                        arg.is_algebraic_expr(y):
                        if argy:
                            argy.extend([arg*factory for factory in argy])
                        argy.append(arg)
                    if arg.has(x) and not arg.has(y) and \
                        arg.is_algebraic_expr(x):
                        if argx:
                            argx.extend([arg*factorx for factorx in argx])
                        argx.append(arg)

            if argy:
                for arg in argy:
                    # Assuming xi = 0, and eta to be f(x)*g(y), the PDE reduces to
                    # (f(x).diff(x)*g(y) + f(x)*g(y).diff(y)*h - f(x)*g(y)*(hy) = 0)
                    eq = (Fx.diff(x))*arg + Fx*h*(arg.diff(y)) - Fx*arg*(hy)
                    # This means y can be successfully eliminated from eq.
                    # The same logic applies for the four assumptions below
                    var = separatevars(eq, [x, y], dict=True)
                    if var:
                        coeffx = var[x]
                        match = coeffx.match(c*Fx.diff(x) + d)
                        if match:
                            match = (-match[d]/match[c]).subs(Fx, y)
                            # This heuristic is assumed to work if the auxilliary ODE
                            # obtained is separable in Fx and x. The same logic
                            # is used for the cases below.
                            red = separatevars(match, [x, y], dict=True)
                            if red:
                                inty = integrate(1/(red['coeff']*red[y]), y)
                                intx = integrate(red[x], x)
                                try:
                                    msol = solve(Eq(inty, intx), y)
                                except NotImplementedError:
                                    pass
                                else:
                                    for sol in msol:
                                        if not sol is S.NaN:
                                            inf = {xi: 0, eta: (sol*arg).subs(y, func)}
                                            if inf not in xieta:
                                                xieta.append(inf)

                    # Assuming eta = 0, and xi to be f(x)*g(y), the PDE reduces to
                    # (f(x).diff(x)*g(y)*h + f(x)*g(y).diff(y)*h**2 +
                    # f(x)*g(y)*h.diff(x) = 0)
                    eq = ((Fx.diff(x))*arg*h + h**2*Fx*(arg.diff(y)) +
                        Fx*arg*(hx))
                    var = separatevars(eq, [x, y], dict=True)
                    if var:
                        coeffx = var[x]
                        match = coeffx.match(c*Fx.diff(x) + d)
                        if match:
                            match = (-match[d]/match[c]).subs(Fx, y)
                            red = separatevars(match, [x, y], dict=True)
                            if red:
                                inty = integrate(1/(red['coeff']*red[y]), y)
                                intx = integrate(red[x], x)
                                try:
                                    msol = solve(Eq(inty, intx), y)
                                except NotImplementedError:
                                    pass
                                else:
                                    for sol in msol:
                                        if not sol is S.NaN:
                                            inf = {xi: (sol*arg).subs(y, func), eta: 0}
                                            if inf not in xieta:
                                                xieta.append(inf)

            if argx:
                for arg in argx:
                    # Assuming xi = 0, and eta to be g(x)*f(y), the PDE reduces to
                    # (g(x).diff(x)*f(y) + g(x)*f(y).diff(y)*h - g(x)*f(y)*hy = 0)
                    eq = (arg.diff(x))*Fy + arg*h*(Fy.diff(y)) - arg*hy*Fy
                    var = separatevars(eq, [x, y], dict=True)
                    if var:
                        coeffy = var[y]
                        match = coeffy.match(c*Fy.diff(y) + d)
                        if match:
                            match = (-match[d]/match[c]).subs(Fy, x)
                            red = separatevars(match, [x, y], dict=True)
                            if red:
                                intx = integrate(1/(red['coeff']*red[x]), x)
                                inty = integrate(red[y], y)
                                try:
                                    msol = solve(Eq(intx, inty), x)
                                except NotImplementedError:
                                    pass
                                else:
                                    for sol in msol:
                                        if not sol is S.NaN:
                                            inf = {xi: 0, eta: (sol*arg).subs(y, func)}
                                            if inf not in xieta:
                                                xieta.append(inf)

                    # Assuming eta = 0, and xi to be g(x)*f(y), the PDE reduces to
                    # (g(x).diff(x)*f(y)*h + g(x)*f(y).diff(y)*h**2 +
                    # g(x)*f(y)*h.diff(x) = 0)
                    eq = ((arg.diff(x))*h*Fy + h**2*arg*(Fy.diff(y)) +
                        arg*Fy*(h.diff(x)))
                    var = separatevars(eq, [x, y], dict=True)
                    if var:
                        coeffy = var[y]
                        match = coeffy.match(c*Fy.diff(y) + d)
                        if match:
                            match = (-match[d]/match[c]).subs(Fy, x)
                            red = separatevars(match, [x, y], dict=True)
                            if red:
                                intx = integrate(1/(red['coeff']*red[x]), x)
                                inty = integrate(red[y], y)
                                try:
                                    msol = solve(Eq(intx, inty), x)
                                except NotImplementedError:
                                    pass
                                else:
                                    for sol in msol:
                                        if not sol is S.NaN:
                                            inf = {xi: (sol*arg).subs(y, func), eta: 0}
                                            if inf not in xieta:
                                                xieta.append(inf)

            # 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
            # be bivariate polynomials. The coefficients of the infinitesimals
            # are found out by substituting them in the PDE and grouping terms
            # that are monomials in y. The degree of the assumed bivariates
            # are increased till a certain maximum value.

            if h.is_rational_function():
                # The maximum degree that the infinitesimals can take is
                # calculated by this technique.
                deglist = [degree(term) for term in [h, h**2, hx, hy]]
                maxdeg = max(deglist)
                mindeg = min(deglist)
                if mindeg < 0:
                    deg = maxdeg - mindeg
                else:
                    deg = maxdeg

                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)}

                            if inf not in xieta:
                                xieta.append(inf)
                            break

            return xieta
