text: r"""
  This module contains :py:meth:`~sympy.solvers.ode.dsolve` and different helper
  functions that it uses.
  
  :py:meth:`~sympy.solvers.ode.dsolve` solves ordinary differential equations.
  See the docstring on the various functions for their uses.  Note that partial
  differential equations support is in ``pde.py``.  Note that hint functions
  have docstrings describing their various methods, but they are intended for
  internal use.  Use ``dsolve(ode, func, hint=hint)`` to solve an ODE using a
  specific hint.  See also the docstring on
  :py:meth:`~sympy.solvers.ode.dsolve`.
  
  **Functions in this module**
  
      These are the user functions in this module:
  
      - :py:meth:`~sympy.solvers.ode.dsolve` - Solves ODEs.
      - :py:meth:`~sympy.solvers.ode.classify_ode` - Classifies ODEs into
        possible hints for :py:meth:`~sympy.solvers.ode.dsolve`.
      - :py:meth:`~sympy.solvers.ode.checkodesol` - Checks if an equation is the
        solution to an ODE.
      - :py:meth:`~sympy.solvers.ode.homogeneous_order` - Returns the
        homogeneous order of an expression.
      - :py:meth:`~sympy.solvers.ode.infinitesimals` - Returns the infinitesimals
        of the Lie group of point transformations of an ODE, such that it is
        invariant.
      - :py:meth:`~sympy.solvers.ode_checkinfsol` - Checks if the given infinitesimals
        are the actual infinitesimals of a first order ODE.
  
      These are the non-solver helper functions that are for internal use.  The
      user should use the various options to
      :py:meth:`~sympy.solvers.ode.dsolve` to obtain the functionality provided
      by these functions:
  
      - :py:meth:`~sympy.solvers.ode.odesimp` - Does all forms of ODE
        simplification.
      - :py:meth:`~sympy.solvers.ode.ode_sol_simplicity` - A key function for
        comparing solutions by simplicity.
      - :py:meth:`~sympy.solvers.ode.constantsimp` - Simplifies arbitrary
        constants.
      - :py:meth:`~sympy.solvers.ode.constant_renumber` - Renumber arbitrary
        constants.
      - :py:meth:`~sympy.solvers.ode._handle_Integral` - Evaluate unevaluated
        Integrals.
  
      See also the docstrings of these functions.
  
  **Currently implemented solver methods**
  
  The following methods are implemented for solving ordinary differential
  equations.  See the docstrings of the various hint functions for more
  information on each (run ``help(ode)``):
  
    - 1st order separable differential equations.
    - 1st order differential equations whose coefficients or `dx` and `dy` are
      functions homogeneous of the same order.
    - 1st order exact differential equations.
    - 1st order linear differential equations.
    - 1st order Bernoulli differential equations.
    - Power series solutions for first order differential equations.
    - Lie Group method of solving first order differential equations.
    - 2nd order Liouville differential equations.
    - Power series solutions for second order differential equations
      at ordinary and regular singular points.
    - `n`\th order linear homogeneous differential equation with constant
      coefficients.
    - `n`\th order linear inhomogeneous differential equation with constant
      coefficients using the method of undetermined coefficients.
    - `n`\th order linear inhomogeneous differential equation with constant
      coefficients using the method of variation of parameters.
  
  **Philosophy behind this module**
  
  This module is designed to make it easy to add new ODE solving methods without
  having to mess with the solving code for other methods.  The idea is that
  there is a :py:meth:`~sympy.solvers.ode.classify_ode` function, which takes in
  an ODE and tells you what hints, if any, will solve the ODE.  It does this
  without attempting to solve the ODE, so it is fast.  Each solving method is a
  hint, and it has its own function, named ``ode_<hint>``.  That function takes
  in the ODE and any match expression gathered by
  :py:meth:`~sympy.solvers.ode.classify_ode` and returns a solved result.  If
  this result has any integrals in it, the hint function will return an
  unevaluated :py:class:`~sympy.integrals.Integral` class.
  :py:meth:`~sympy.solvers.ode.dsolve`, which is the user wrapper function
  around all of this, will then call :py:meth:`~sympy.solvers.ode.odesimp` on
  the result, which, among other things, will attempt to solve the equation for
  the dependent variable (the function we are solving for), simplify the
  arbitrary constants in the expression, and evaluate any integrals, if the hint
  allows it.
  
  **How to add new solution methods**
  
  If you have an ODE that you want :py:meth:`~sympy.solvers.ode.dsolve` to be
  able to solve, try to avoid adding special case code here.  Instead, try
  finding a general method that will solve your ODE, as well as others.  This
  way, the :py:mod:`~sympy.solvers.ode` module will become more robust, and
  unhindered by special case hacks.  WolphramAlpha and Maple's
  DETools[odeadvisor] function are two resources you can use to classify a
  specific ODE.  It is also better for a method to work with an `n`\th order ODE
  instead of only with specific orders, if possible.
  
  To add a new method, there are a few things that you need to do.  First, you
  need a hint name for your method.  Try to name your hint so that it is
  unambiguous with all other methods, including ones that may not be implemented
  yet.  If your method uses integrals, also include a ``hint_Integral`` hint.
  If there is more than one way to solve ODEs with your method, include a hint
  for each one, as well as a ``<hint>_best`` hint.  Your ``ode_<hint>_best()``
  function should choose the best using min with ``ode_sol_simplicity`` as the
  key argument.  See
  :py:meth:`~sympy.solvers.ode.ode_1st_homogeneous_coeff_best`, for example.
  The function that uses your method will be called ``ode_<hint>()``, so the
  hint must only use characters that are allowed in a Python function name
  (alphanumeric characters and the underscore '``_``' character).  Include a
  function for every hint, except for ``_Integral`` hints
  (:py:meth:`~sympy.solvers.ode.dsolve` takes care of those automatically).
  Hint names should be all lowercase, unless a word is commonly capitalized
  (such as Integral or Bernoulli).  If you have a hint that you do not want to
  run with ``all_Integral`` that doesn't have an ``_Integral`` counterpart (such
  as a best hint that would defeat the purpose of ``all_Integral``), you will
  need to remove it manually in the :py:meth:`~sympy.solvers.ode.dsolve` code.
  See also the :py:meth:`~sympy.solvers.ode.classify_ode` docstring for
  guidelines on writing a hint name.
  
  Determine *in general* how the solutions returned by your method compare with
  other methods that can potentially solve the same ODEs.  Then, put your hints
  in the :py:data:`~sympy.solvers.ode.allhints` tuple in the order that they
  should be called.  The ordering of this tuple determines which hints are
  default.  Note that exceptions are ok, because it is easy for the user to
  choose individual hints with :py:meth:`~sympy.solvers.ode.dsolve`.  In
  general, ``_Integral`` variants should go at the end of the list, and
  ``_best`` variants should go before the various hints they apply to.  For
  example, the ``undetermined_coefficients`` hint comes before the
  ``variation_of_parameters`` hint because, even though variation of parameters
  is more general than undetermined coefficients, undetermined coefficients
  generally returns cleaner results for the ODEs that it can solve than
  variation of parameters does, and it does not require integration, so it is
  much faster.
  
  Next, you need to have a match expression or a function that matches the type
  of the ODE, which you should put in :py:meth:`~sympy.solvers.ode.classify_ode`
  (if the match function is more than just a few lines, like
  :py:meth:`~sympy.solvers.ode._undetermined_coefficients_match`, it should go
  outside of :py:meth:`~sympy.solvers.ode.classify_ode`).  It should match the
  ODE without solving for it as much as possible, so that
  :py:meth:`~sympy.solvers.ode.classify_ode` remains fast and is not hindered by
  bugs in solving code.  Be sure to consider corner cases.  For example, if your
  solution method involves dividing by something, make sure you exclude the case
  where that division will be 0.
  
  In most cases, the matching of the ODE will also give you the various parts
  that you need to solve it.  You should put that in a dictionary (``.match()``
  will do this for you), and add that as ``matching_hints['hint'] = matchdict``
  in the relevant part of :py:meth:`~sympy.solvers.ode.classify_ode`.
  :py:meth:`~sympy.solvers.ode.classify_ode` will then send this to
  :py:meth:`~sympy.solvers.ode.dsolve`, which will send it to your function as
  the ``match`` argument.  Your function should be named ``ode_<hint>(eq, func,
  order, match)`.  If you need to send more information, put it in the ``match``
  dictionary.  For example, if you had to substitute in a dummy variable in
  :py:meth:`~sympy.solvers.ode.classify_ode` to match the ODE, you will need to
  pass it to your function using the `match` dict to access it.  You can access
  the independent variable using ``func.args[0]``, and the dependent variable
  (the function you are trying to solve for) as ``func.func``.  If, while trying
  to solve the ODE, you find that you cannot, raise ``NotImplementedError``.
  :py:meth:`~sympy.solvers.ode.dsolve` will catch this error with the ``all``
  meta-hint, rather than causing the whole routine to fail.
  
  Add a docstring to your function that describes the method employed.  Like
  with anything else in SymPy, you will need to add a doctest to the docstring,
  in addition to real tests in ``test_ode.py``.  Try to maintain consistency
  with the other hint functions' docstrings.  Add your method to the list at the
  top of this docstring.  Also, add your method to ``ode.rst`` in the
  ``docs/src`` directory, so that the Sphinx docs will pull its docstring into
  the main SymPy documentation.  Be sure to make the Sphinx documentation by
  running ``make html`` from within the doc directory to verify that the
  docstring formats correctly.
  
  If your solution method involves integrating, use :py:meth:`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.simplify.logcombine` on them (it also helps to write
  the arbitrary constant as ``log(C1)`` instead of ``C1`` in this case).  Also
  consider common ways that you can rearrange your solution to have
  :py:meth:`~sympy.solvers.ode.constantsimp` take better advantage of it.  It is
  better to put simplification in :py:meth:`~sympy.solvers.ode.odesimp` than in
  your method, because it can then be turned off with the simplify flag in
  :py:meth:`~sympy.solvers.ode.dsolve`.  If you have any extraneous
  simplification in your function, be sure to only run it using ``if
  match.get('simplify', True):``, especially if it can be slow or if it can
  reduce the domain of the solution.
  
  Finally, as with every contribution to SymPy, your method will need to be
  tested.  Add a test for each method in ``test_ode.py``.  Follow the
  conventions there, i.e., test the solver using ``dsolve(eq, f(x),
  hint=your_hint)``, and also test the solution using
  :py:meth:`~sympy.solvers.ode.checkodesol` (you can put these in a separate
  tests and skip/XFAIL if it runs too slow/doesn't work).  Be sure to call your
  hint specifically in :py:meth:`~sympy.solvers.ode.dsolve`, that way the test
  won't be broken simply by the introduction of another matching hint.  If your
  method works for higher order (>1) ODEs, you will need to run ``sol =
  constant_renumber(sol, 'C', 1, order)`` for each solution, where ``order`` is
  the order of the ODE.  This is because ``constant_renumber`` renumbers the
  arbitrary constants by printing order, which is platform dependent.  Try to
  test every corner case of your solver, including a range of orders if it is a
  `n`\th order solver, but if your solver is slow, such as if it involves hard
  integration, try to keep the test run time down.
  
  Feel free to refactor existing hints to avoid duplicating code or creating
  inconsistencies.  If you can show that your method exactly duplicates an
  existing method, including in the simplicity and speed of obtaining the
  solutions, then you can remove the old, less general method.  The existing
  code is tested extensively in ``test_ode.py``, so if anything is broken, one
  of those tests will surely fail.
  
  """
  from __future__ import print_function, division
  
  from collections import defaultdict
  from itertools import islice
  
  from sympy.core import Add, C, S, Mul, Pow, oo
  from sympy.core.compatibility import ordered, iterable, is_sequence, xrange
  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, Subs)
  from sympy.core.multidimensional import vectorize
  from sympy.core.numbers import Rational, NaN, zoo
  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.functions.combinatorial.factorials import factorial
  from sympy.matrices import wronskian
  from sympy.polys import Poly, RootOf, terms_gcd, PolynomialError
  from sympy.polys.polytools import cancel, degree, div
  from sympy.series import Order
  from sympy.series.series import series
  from sympy.simplify import collect, logcombine, powsimp, separatevars, \
      simplify, trigsimp, denom, 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",
      "1st_power_series",
      "lie_group",
      "nth_linear_constant_coeff_homogeneous",
      "nth_linear_euler_eq_homogeneous",
      "nth_linear_constant_coeff_undetermined_coefficients",
      "nth_linear_constant_coeff_variation_of_parameters",
      "Liouville",
      "2nd_power_series_ordinary",
      "2nd_power_series_regular",
      "separable_Integral",
      "1st_exact_Integral",
      "1st_linear_Integral",
      "Bernoulli_Integral",
      "1st_homogeneous_coeff_subs_indep_div_dep_Integral",
      "1st_homogeneous_coeff_subs_dep_div_indep_Integral",
      "almost_linear_Integral",
      "linear_coefficients_Integral",
      "separable_reduced_Integral",
      "nth_linear_constant_coeff_variation_of_parameters_Integral",
      "Liouville_Integral",
      )
  
  lie_heuristics = (
      "abaco1_simple",
      "abaco1_product",
      "abaco2_similar",
      "abaco2_unique_unknown",
      "abaco2_unique_general",
      "linear",
      "function_sum",
      "bivariate",
      "chi"
      )
  
  
  def sub_func_doit(eq, func, new):
      r"""
      When replacing the func with something else, we usually want the
      derivative evaluated, so this function helps in making that happen.
  
      To keep subs from having to look through all derivatives, we mask them off
      with dummy variables, do the func sub, and then replace masked-off
      derivatives with their doit values.
  
      Examples
      ========
  
      >>> from sympy import Derivative, symbols, Function
      >>> from sympy.solvers.ode import sub_func_doit
      >>> x, z = symbols('x, z')
      >>> y = Function('y')
  
      >>> sub_func_doit(3*Derivative(y(x), x) - 1, y(x), x)
      2
  
      >>> sub_func_doit(x*Derivative(y(x), x) - y(x)**2 + y(x), y(x),
      ... 1/(x*(z + 1/x)))
      x*(-1/(x**2*(z + 1/x)) + 1/(x**3*(z + 1/x)**2)) + 1/(x*(z + 1/x))
      ...- 1/(x**2*(z + 1/x)**2)
      """
      reps = {}
      repu = {}
      for d in eq.atoms(Derivative):
          u = 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,
      ics= None, xi=None, eta=None, x0=0, n=6, **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.
  
          ``xi`` and ``eta`` are the infinitesimal functions of an ordinary
              differential equation. They are the infinitesimals of the Lie group
              of point transformations for which the differential equation is
              invariant. The user can specify values for the infinitesimals. If
              nothing is specified, ``xi`` and ``eta`` are calculated using
              :py:meth:`~sympy.solvers.ode.infinitesimals` with the help of various
              heuristics.
  
      **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))]
      """
      given_hint = hint  # hint given by the user
  
      # See the docstring of _desolve for more details.
      hints = _desolve(eq, func=func,
          hint=hint, simplify=True, xi=xi, eta=eta, type='ode', ics=ics,
          x0=x0, n=n, **kwargs)
  
      eq = hints.pop('eq', eq)
      all_ = hints.pop('all', False)
      if all_:
          retdict = {}
          failed_hints = {}
          gethints = classify_ode(eq, dict=True)
          orderedhints = gethints['ordered_hints']
          for hint in hints:
              try:
                  rv = _helper_simplify(eq, hint, hints[hint], simplify)
              except NotImplementedError as detail:
                  failed_hints[hint] = detail
              else:
                  retdict[hint] = rv
          func = hints[hint]['func']
  
          retdict['best'] = min(list(retdict.values()), key=lambda x:
              ode_sol_simplicity(x, func, trysolving=not simplify))
          if given_hint == 'best':
              return retdict['best']
          for i in orderedhints:
              if retdict['best'] == retdict.get(i, None):
                  retdict['best_hint'] = i
                  break
          retdict['default'] = gethints['default']
          retdict['order'] = gethints['order']
          retdict.update(failed_hints)
          return retdict
  
      else:
          # The key 'hint' stores the hint needed to be solved for.
          hint = hints['hint']
          return _helper_simplify(eq, hint, hints, simplify)
  
  def _helper_simplify(eq, hint, match, simplify=True, **kwargs):
      r"""
      Helper function of dsolve that calls the respective
      :py:mod:`~sympy.solvers.ode` functions to solve for the ordinary
      differential equations. This minimises the computation in calling
      :py:meth:`~sympy.solvers.deutils._desolve` multiple times.
      """
      r = match
      if hint.endswith('_Integral'):
          solvefunc = globals()['ode_' + hint[:-len('_Integral')]]
      else:
          solvefunc = globals()['ode_' + hint]
      func = r['func']
      order = r['order']
      match = r[hint]
  
      if simplify:
          # odesimp() will attempt to integrate, if necessary, apply constantsimp(),
          # attempt to solve for func, and apply any other hint specific
          # simplifications
          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, ics=None, **kwargs):
      r"""
      Returns a tuple of possible :py:meth:`~sympy.solvers.ode.dsolve`
      classifications for an ODE.
  
      The tuple is ordered so that first item is the classification that
      :py:meth:`~sympy.solvers.ode.dsolve` uses to solve the ODE by default.  In
      general, classifications at the near the beginning of the list will
      produce better solutions faster than those near the end, thought there are
      always exceptions.  To make :py:meth:`~sympy.solvers.ode.dsolve` use a
      different classification, use ``dsolve(ODE, func,
      hint=<classification>)``.  See also the
      :py:meth:`~sympy.solvers.ode.dsolve` docstring for different meta-hints
      you can use.
  
      If ``dict`` is true, :py:meth:`~sympy.solvers.ode.classify_ode` will
      return a dictionary of ``hint:match`` expression terms. This is intended
      for internal use by :py:meth:`~sympy.solvers.ode.dsolve`.  Note that
      because dictionaries are ordered arbitrarily, this will most likely not be
      in the same order as the tuple.
  
      You can get help on different hints by executing
      ``help(ode.ode_hintname)``, where ``hintname`` is the name of the hint
      without ``_Integral``.
  
      See :py:data:`~sympy.solvers.ode.allhints` or the
      :py:mod:`~sympy.solvers.ode` docstring for a list of all supported hints
      that can be returned from :py:meth:`~sympy.solvers.ode.classify_ode`.
  
      Notes
      =====
  
      These are remarks on hint names.
  
      ``_Integral``
  
          If a classification has ``_Integral`` at the end, it will return the
          expression with an unevaluated :py:class:`~sympy.integrals.Integral`
          class in it.  Note that a hint may do this anyway if
          :py:meth:`~sympy.core.expr.Expr.integrate` cannot do the integral,
          though just using an ``_Integral`` will do so much faster.  Indeed, an
          ``_Integral`` hint will always be faster than its corresponding hint
          without ``_Integral`` because
          :py:meth:`~sympy.core.expr.Expr.integrate` is an expensive routine.
          If :py:meth:`~sympy.solvers.ode.dsolve` hangs, it is probably because
          :py:meth:`~sympy.core.expr.Expr.integrate` is hanging on a tough or
          impossible integral.  Try using an ``_Integral`` hint or
          ``all_Integral`` to get it return something.
  
          Note that some hints do not have ``_Integral`` counterparts.  This is
          because :py:meth:`~sympy.solvers.ode.integrate` is not used in solving
          the ODE for those method. For example, `n`\th order linear homogeneous
          ODEs with constant coefficients do not require integration to solve,
          so there is no ``nth_linear_homogeneous_constant_coeff_Integrate``
          hint. You can easily evaluate any unevaluated
          :py:class:`~sympy.integrals.Integral`\s in an expression by doing
          ``expr.doit()``.
  
      Ordinals
  
          Some hints contain an ordinal such as ``1st_linear``.  This is to help
          differentiate them from other hints, as well as from other methods
          that may not be implemented yet. If a hint has ``nth`` in it, such as
          the ``nth_linear`` hints, this means that the method used to applies
          to ODEs of any order.
  
      ``indep`` and ``dep``
  
          Some hints contain the words ``indep`` or ``dep``.  These reference
          the independent variable and the dependent function, respectively. For
          example, if an ODE is in terms of `f(x)`, then ``indep`` will refer to
          `x` and ``dep`` will refer to `f`.
  
      ``subs``
  
          If a hints has the word ``subs`` in it, it means the the ODE is solved
          by substituting the expression given after the word ``subs`` for a
          single dummy variable.  This is usually in terms of ``indep`` and
          ``dep`` as above.  The substituted expression will be written only in
          characters allowed for names of Python objects, meaning operators will
          be spelled out.  For example, ``indep``/``dep`` will be written as
          ``indep_div_dep``.
  
      ``coeff``
  
          The word ``coeff`` in a hint refers to the coefficients of something
          in the ODE, usually of the derivative terms.  See the docstring for
          the individual methods for more info (``help(ode)``).  This is
          contrast to ``coefficients``, as in ``undetermined_coefficients``,
          which refers to the common name of a method.
  
      ``_best``
  
          Methods that have more than one fundamental way to solve will have a
          hint for each sub-method and a ``_best`` meta-classification. This
          will evaluate all hints and return the best, using the same
          considerations as the normal ``best`` meta-hint.
  
  
      Examples
      ========
  
      >>> from sympy import Function, classify_ode, Eq
      >>> from sympy.abc import x
      >>> f = Function('f')
      >>> classify_ode(Eq(f(x).diff(x), 0), f(x))
      ('separable', '1st_linear', '1st_homogeneous_coeff_best',
      '1st_homogeneous_coeff_subs_indep_div_dep',
      '1st_homogeneous_coeff_subs_dep_div_indep',
      '1st_power_series', 'lie_group',
      'nth_linear_constant_coeff_homogeneous',
      'separable_Integral', '1st_linear_Integral',
      '1st_homogeneous_coeff_subs_indep_div_dep_Integral',
      '1st_homogeneous_coeff_subs_dep_div_indep_Integral')
      >>> classify_ode(f(x).diff(x, 2) + 3*f(x).diff(x) + 2*f(x) - 4)
      ('nth_linear_constant_coeff_undetermined_coefficients',
      'nth_linear_constant_coeff_variation_of_parameters',
      'nth_linear_constant_coeff_variation_of_parameters_Integral')
  
      """
      prep = kwargs.pop('prep', True)
      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')
      xi = kwargs.get('xi')
      eta = kwargs.get('eta')
      terms = kwargs.get('n')
  
      if isinstance(eq, Equality):
          if eq.rhs != 0:
              return classify_ode(eq.lhs - eq.rhs, func, ics=ics, xi=xi,
                  n=terms, eta=eta, prep=False)
          eq = eq.lhs
      order = ode_order(eq, f(x))
      # hint:matchdict or hint:(tuple of matchdicts)
      # Also will contain "default":<default hint> and "order":order items.
      matching_hints = {"order": order}
  
      if not order:
          if dict:
              matching_hints["default"] = None
              return matching_hints
          else:
              return ()
  
      df = f(x).diff(x)
      a = Wild('a', exclude=[f(x)])
      b = Wild('b', exclude=[f(x)])
      c = Wild('c', exclude=[f(x)])
      d = Wild('d', exclude=[df, f(x).diff(x, 2)])
      e = Wild('e', exclude=[df])
      k = Wild('k', exclude=[df])
      n = Wild('n', exclude=[f(x)])
      c1 = Wild('c1', exclude=[x])
      a2 = Wild('a2', exclude=[x, f(x), df])
      b2 = Wild('b2', exclude=[x, f(x), df])
      c2 = Wild('c2', exclude=[x, f(x), df])
      d2 = Wild('d2', exclude=[x, f(x), df])
      a3 = Wild('a3', exclude=[f(x), df, f(x).diff(x, 2)])
      b3 = Wild('b3', exclude=[f(x), df, f(x).diff(x, 2)])
      c3 = Wild('c3', exclude=[f(x), df, f(x).diff(x, 2)])
      r3 = {'xi': xi, 'eta': eta}  # Used for the lie_group hint
      boundary = {}  # Used to extract initial conditions
      C0 = Symbol("C0")
      eq = expand(eq)
  
      # Preprocessing to get the initial conditions out
      if ics is not None:
          for funcarg in ics:
              # Separating derivatives
              if isinstance(funcarg, Subs):
                  deriv = funcarg.expr
                  old = funcarg.variables[0]
                  new = funcarg.point[0]
                  if isinstance(deriv, Derivative) and isinstance(deriv.args[0],
                      AppliedUndef) and deriv.args[0].func == f and old == x and not new.has(x):
                      dorder = ode_order(deriv, x)
                      temp = 'f' + str(dorder)
                      boundary.update({temp: new, temp + 'val': ics[funcarg]})
                  else:
                      raise ValueError("Enter valid boundary conditions for Derivatives")
  
  
              # Separating functions
              elif isinstance(funcarg, AppliedUndef):
                  if funcarg.func == f and len(funcarg.args) == 1 and \
                      not funcarg.args[0].has(x):
                      boundary.update({'f0': funcarg.args[0], 'f0val': ics[funcarg]})
                  else:
                      raise ValueError("Enter valid boundary conditions for Function")
  
              else:
                  raise ValueError("Enter boundary conditions of the form ics "
                      " = {f(point}: value, f(point).diff(point, order).subs(arg, point) "
                      ":value")
  
      # Precondition to try remove f(x) from highest order derivative
      reduced_eq = None
      if eq.is_Add:
          deriv_coef = eq.coeff(f(x).diff(x, order))
          if deriv_coef != 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)
  
              # FIRST ORDER POWER SERIES WHICH NEEDS INITIAL CONDITIONS
              # TODO: Hint first order series should match only if d/e is analytic.
              # For now, only d/e and (d/e).diff(arg) is checked for existence at
              # at a given point.
              # This is currently done internally in ode_1st_power_series.
              point = boundary.get('f0', 0)
              value = boundary.get('f0val', C0)
              check = cancel(r[d]/r[e])
              check1 = check.subs({x: point, y: value})
              if not check1.has(oo) and not check1.has(zoo) and \
                  not check1.has(NaN) and not check1.has(-oo):
                  check2 = (check1.diff(x)).subs({x: point, y: value})
                  if not check2.has(oo) and not check2.has(zoo) and \
                      not check2.has(NaN) and not check2.has(-oo):
                      rseries = r.copy()
                      rseries.update({'terms': terms, 'f0': point, 'f0val': value})
                      matching_hints["1st_power_series"] = rseries
  
              r3.update(r)
              ## Exact Differential Equation: P(x, y) + Q(x, y)*y' = 0 where
              # dP/dy == dQ/dx
              try:
                  if r[d] != 0:
                      numerator = simplify(r[d].diff(y) - r[e].diff(x))
                      # The following few conditions try to convert a non-exact
                      # differential equation into an exact one.
                      # References : Differential equations with applications
                      # and historical notes - George E. Simmons
  
                      if numerator:
                          # If (dP/dy - dQ/dx) / Q = f(x)
                          # then exp(integral(f(x))*equation becomes exact
                          factor = simplify(numerator/r[e])
                          variables = factor.free_symbols
                          if len(variables) == 1 and x == variables.pop():
                              factor = exp(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
  
          # Any first order ODE can be ideally solved by the Lie Group
          # method
          matching_hints["lie_group"] = r3
  
          # This match is used for several cases below; we now collect on
          # f(x) so the matching works.
          r = collect(reduced_eq, df, exact=True).match(d + e*df)
          if r:
              # Using r[d] and r[e] without any modification for hints
              # linear-coefficients and separable-reduced.
              num, den = r[d], r[e]  # ODE = d/e + df
              r['d'] = d
              r['e'] = e
              r['y'] = y
              r[d] = num.subs(f(x), y)
              r[e] = den.subs(f(x), y)
  
              ## Separable Case: y' == P(y)*Q(x)
              r[d] = separatevars(r[d])
              r[e] = separatevars(r[e])
              # m1[coeff]*m1[x]*m1[y] + m2[coeff]*m2[x]*m2[y]*y'
              m1 = separatevars(r[d], dict=True, symbols=(x, y))
              m2 = separatevars(r[e], dict=True, symbols=(x, y))
              if m1 and m2:
                  r1 = {'m1': m1, 'm2': m2, 'y': y}
                  matching_hints["separable"] = r1
                  matching_hints["separable_Integral"] = r1
  
              ## First order equation with homogeneous coefficients:
              # dy/dx == F(y/x) or dy/dx == F(x/y)
              ordera = homogeneous_order(r[d], x, y)
              if ordera is not None:
                  orderb = homogeneous_order(r[e], x, y)
                  if ordera == orderb:
                      # u1=y/x and u2=x/y
                      u1 = Dummy('u1')
                      u2 = Dummy('u2')
                      s = "1st_homogeneous_coeff_subs"
                      s1 = s + "_dep_div_indep"
                      s2 = s + "_indep_div_dep"
                      if simplify((r[d] + u1*r[e]).subs({x: 1, y: u1})) != 0:
                          matching_hints[s1] = r
                          matching_hints[s1 + "_Integral"] = r
                      if simplify((r[e] + u2*r[d]).subs({x: u2, y: 1})) != 0:
                          matching_hints[s2] = r
                          matching_hints[s2 + "_Integral"] = r
                      if s1 in matching_hints and s2 in matching_hints:
                          matching_hints["1st_homogeneous_coeff_best"] = r
  
              ## Linear coefficients of the form
              # y'+ F((a*x + b*y + c)/(a'*x + b'y + c')) = 0
              # that can be reduced to homogeneous form.
              F = num/den
              params = _linear_coeff_match(F, func)
              if params:
                  xarg, yarg = params
                  u = Dummy('u')
                  t = Dummy('t')
                  # Dummy substitution for df and f(x).
                  dummy_eq = reduced_eq.subs(((df, t), (f(x), u)))
                  reps = ((x, x + xarg), (u, u + yarg), (t, df), (u, f(x)))
                  dummy_eq = simplify(dummy_eq.subs(reps))
                  # get the re-cast values for e and d
                  r2 = collect(expand(dummy_eq), [df, f(x)]).match(e*df + d)
                  if r2:
                      orderd = homogeneous_order(r2[d], x, f(x))
                      if orderd is not None:
                          ordere = homogeneous_order(r2[e], x, f(x))
                          if orderd == ordere:
                              # Match arguments are passed in such a way that it
                              # is coherent with the already existing homogeneous
                              # functions.
                              r2[d] = r2[d].subs(f(x), y)
                              r2[e] = r2[e].subs(f(x), y)
                              r2.update({'xarg': xarg, 'yarg': yarg,
                                  'd': d, 'e': e, 'y': y})
                              matching_hints["linear_coefficients"] = r2
                              matching_hints["linear_coefficients_Integral"] = r2
  
              ## Equation of the form y' + (y/x)*H(x^n*y) = 0
              # that can be reduced to separable form
  
              factor = simplify(x/f(x)*num/den)
  
              # Try representing factor in terms of x^n*y
              # where n is lowest power of x in factor;
              # first remove terms like sqrt(2)*3 from factor.atoms(Mul)
              u = None
              for mul in ordered(factor.atoms(Mul)):
                  if mul.has(x):
                      _, u = mul.as_independent(x, f(x))
                      break
              if u and u.has(f(x)):
                  t = Dummy('t')
                  r2 = {'t': t}
                  xpart, ypart = u.as_independent(f(x))
                  test = factor.subs(((u, t), (1/u, 1/t)))
                  free = test.free_symbols
                  if len(free) == 1 and free.pop() == t:
                      r2.update({'power': xpart.as_base_exp()[1], 'u': test})
                      matching_hints["separable_reduced"] = r2
                      matching_hints["separable_reduced_Integral"] = r2
  
          ## Almost-linear equation of the form f(x)*g(y)*y' + k(x)*l(y) + m(x) = 0
          r = collect(eq, [df, f(x)]).match(e*df + d)
          if r:
              r2 = r.copy()
              r2[c] = S.Zero
              if r2[d].is_Add:
                  # Separate the terms having f(x) to r[d] and
                  # remaining to r[c]
                  no_f, r2[d] = r2[d].as_independent(f(x))
                  r2[c] += no_f
              factor = simplify(r2[d].diff(f(x))/r[e])
              if factor and not factor.has(f(x)):
                  r2[d] = factor_terms(r2[d])
                  u = r2[d].as_independent(f(x), as_Add=False)[1]
                  r2.update({'a': e, 'b': d, 'c': c, 'u': u})
                  r2[d] /= u
                  r2[e] /= u.diff(f(x))
                  matching_hints["almost_linear"] = r2
                  matching_hints["almost_linear_Integral"] = r2
  
  
      elif order == 2:
          # Liouville ODE in the form
          # f(x).diff(x, 2) + g(f(x))*(f(x).diff(x))**2 + h(x)*f(x).diff(x)
          # See Goldstein and Braun, "Advanced Methods for the Solution of
          # Differential Equations", pg. 98
  
          s = d*f(x).diff(x, 2) + e*df**2 + k*df
          r = reduced_eq.match(s)
          if r and r[d] != 0:
              y = Dummy('y')
              g = simplify(r[e]/r[d]).subs(f(x), y)
              h = simplify(r[k]/r[d])
              if h.has(f(x)) or g.has(x):
                  pass
              else:
                  r = {'g': g, 'h': h, 'y': y}
                  matching_hints["Liouville"] = r
                  matching_hints["Liouville_Integral"] = r
  
          # Homogeneous second order differential equation of the form
          # a3*f(x).diff(x, 2) + b3*f(x).diff(x) + c3, where
          # for simplicity, a3, b3 and c3 are assumed to be polynomials.
          # It has a definite power series solution at point x0 if, b3/a3 and c3/a3
          # are analytic at x0.
          deq = a3*(f(x).diff(x, 2)) + b3*df + c3*f(x)
          r = collect(reduced_eq,
              [f(x).diff(x, 2), f(x).diff(x), f(x)]).match(deq)
          ordinary = False
          if r and r[a3] != 0:
              if all([r[key].is_polynomial() for key in r]):
                  p = cancel(r[b3]/r[a3])  # Used below
                  q = cancel(r[c3]/r[a3])  # Used below
                  point = kwargs.get('x0', 0)
                  check = p.subs(x, point)
                  if not check.has(oo) and not check.has(NaN) and \
                      not check.has(zoo) and not check.has(-oo):
                      check = q.subs(x, point)
                      if not check.has(oo) and not check.has(NaN) and \
                          not check.has(zoo) and not check.has(-oo):
                          ordinary = True
                          r.update({'a3': a3, 'b3': b3, 'c3': c3, 'x0': point, 'terms': terms})
                          matching_hints["2nd_power_series_ordinary"] = r
  
                  # Checking if the differential equation has a regular singular point
                  # at x0. It has a regular singular point at x0, if (b3/a3)*(x - x0)
                  # and (c3/a3)*((x - x0)**2) are analytic at x0.
                  if not ordinary:
                      p = cancel((x - point)*p)
                      check = p.subs(x, point)
                      if not check.has(oo) and not check.has(NaN) and \
                          not check.has(zoo) and not check.has(-oo):
                          q = cancel(((x - point)**2)*q)
                          check = q.subs(x, point)
                          if not check.has(oo) and not check.has(NaN) and \
                              not check.has(zoo) and not check.has(-oo):
                              coeff_dict = {'p': p, 'q': q, 'x0': point, 'terms': terms}
                              matching_hints["2nd_power_series_regular"] = coeff_dict
  
  
      if order > 0:
          # nth order linear ODE
          # a_n(x)y^(n) + ... + a_1(x)y' + a_0(x)y = F(x) = b
  
          r = _nth_linear_match(reduced_eq, func, order)
  
          # Constant coefficient case (a_i is constant for all i)
          if r and not any(r[i].has(x) for i in r if i >= 0):
              # Inhomogeneous case: F(x) is not identically 0
              if r[-1]:
                  undetcoeff = _undetermined_coefficients_match(r[-1], x)
                  s = "nth_linear_constant_coeff_variation_of_parameters"
                  matching_hints[s] = r
                  matching_hints[s + "_Integral"] = r
                  if undetcoeff['test']:
                      r['trialset'] = undetcoeff['trialset']
                      matching_hints["nth_linear_constant_coeff_undetermined_"
                          "coefficients"] = r
              # Homogeneous case: F(x) is identically 0
              else:
                  matching_hints["nth_linear_constant_coeff_homogeneous"] = r
  
          # 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)([checkodesol(ode, i, order=order, solve_for_func=solve_for_func) for i in sol])
  
      if not isinstance(sol, Equality):
          sol = Eq(func, sol)
      x = func.args[0]
      s = True
      testnum = 0
      if order == 'auto':
          order = ode_order(ode, func)
      if solve_for_func and not (
              sol.lhs == func and not sol.rhs.has(func)) and not (
              sol.rhs == func and not sol.lhs.has(func)):
          try:
              solved = solve(sol, func)
              if not solved:
                  raise NotImplementedError
          except NotImplementedError:
              pass
          else:
              if len(solved) == 1:
                  result = checkodesol(ode, Eq(func, solved[0]),
                      order=order, solve_for_func=False)
              else:
                  result = checkodesol(ode, [Eq(func, t) for t in solved],
                  order=order, solve_for_func=False)
  
              return result
  
      while s:
          if testnum == 0:
              # First pass, try substituting a solved solution directly into the
              # ODE. This has the highest chance of succeeding.
              ode_diff = ode.lhs - ode.rhs
  
              if sol.lhs == func:
                  s = sub_func_doit(ode_diff, func, sol.rhs)
              elif sol.rhs == func:
                  s = sub_func_doit(ode_diff, func, sol.lhs)
              else:
                  testnum += 1
                  continue
              ss = simplify(s)
              if ss:
                  # with the new numer_denom in power.py, if we do a simple
                  # expansion then testnum == 0 verifies all solutions.
                  s = ss.expand(force=True)
              else:
                  s = 0
              testnum += 1
          elif testnum == 1:
              # Second pass. If we cannot substitute f, try seeing if the nth
              # derivative is equal, this will only work for odes that are exact,
              # by definition.
              s = simplify(
                  trigsimp(diff(sol.lhs, x, order) - diff(sol.rhs, x, order)) -
                  trigsimp(ode.lhs) + trigsimp(ode.rhs))
              # s2 = simplify(
              #     diff(sol.lhs, x, order) - diff(sol.rhs, x, order) - \
              #     ode.lhs + ode.rhs)
              testnum += 1
          elif testnum == 2:
              # Third pass. Try solving for df/dx and substituting that into the
              # ODE. Thanks to Chris Smith for suggesting this method.  Many of
              # the comments below are his too.
              # The method:
              # - Take each of 1..n derivatives of the solution.
              # - Solve each nth derivative for d^(n)f/dx^(n)
              #   (the differential of that order)
              # - Back substitute into the ODE in decreasing order
              #   (i.e., n, n-1, ...)
              # - Check the result for zero equivalence
              if sol.lhs == func and not sol.rhs.has(func):
                  diffsols = {0: sol.rhs}
              elif sol.rhs == func and not sol.lhs.has(func):
                  diffsols = {0: sol.lhs}
              else:
                  diffsols = {}
              sol = sol.lhs - sol.rhs
              for i in range(1, order + 1):
                  # Differentiation is a linear operator, so there should always
                  # be 1 solution. Nonetheless, we test just to make sure.
                  # We only need to solve once.  After that, we automatically
                  # have the solution to the differential in the order we want.
                  if i == 1:
                      ds = sol.diff(x)
                      try:
                          sdf = solve(ds, func.diff(x, i))
                          if not sdf:
                              raise NotImplementedError
                      except NotImplementedError:
                          testnum += 1
                          break
                      else:
                          diffsols[i] = sdf[0]
                  else:
                      # This is what the solution says df/dx should be.
                      diffsols[i] = diffsols[i - 1].diff(x)
  
              # Make sure the above didn't fail.
              if testnum > 2:
                  continue
              else:
                  # Substitute it into ODE to check for self consistency.
                  lhs, rhs = ode.lhs, ode.rhs
                  for i in range(order, -1, -1):
                      if i == 0 and 0 not in diffsols:
                          # We can only substitute f(x) if the solution was
                          # solved for f(x).
                          break
                      lhs = sub_func_doit(lhs, func.diff(x, i), diffsols[i])
                      rhs = sub_func_doit(rhs, func.diff(x, i), diffsols[i])
                      ode_or_bool = Eq(lhs, rhs)
                      ode_or_bool = simplify(ode_or_bool)
  
                      if isinstance(ode_or_bool, bool):
                          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 of constants is done with limited assistance:
  
      1. terms of :py:class:`~sympy.core.add.Add`\s are collected to try join
         constants so `e^x (C_1 \cos(x) + C_2 \cos(x))` will simplify to `e^x
         C_1 \cos(x)`;
  
      2. powers with exponents that are :py:class:`~sympy.core.add.Add`\s are
         expanded so `e^{C_1 + x}` will be simplified to `C_1 e^x`.
  
      Use :py:meth:`~sympy.solvers.ode.constant_renumber` to renumber constants
      after simplification or else arbitrary numbers on constants may appear,
      e.g. `C_1 + C_3 x`.
  
      In rare cases, a single constant can be "simplified" into two constants.
      Every differential equation solution should have as many arbitrary
      constants as the order of the differential equation.  The result here will
      be technically correct, but it may, for example, have `C_1` and `C_2` in
      an expression, when `C_1` is actually equal to `C_2`.  Use your discretion
      in such situations, and also take advantage of the ability to use hints in
      :py:meth:`~sympy.solvers.ode.dsolve`.
  
      Examples
      ========
  
      >>> from sympy import symbols
      >>> from sympy.solvers.ode import constantsimp
      >>> C1, C2, C3, x, y = symbols('C1,C2,C3,x,y')
      >>> constantsimp(2*C1*x, 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, key=str)
              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)(
              [constant_renumber(i, symbolname=symbolname, startnumber=startnumber, endnumber=endnumber)
                  for i in 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:`~sympy.solvers.ode.ode_sol_simplicity`.
  
      See the
      :py:meth:`~sympy.solvers.ode.ode_1st_homogeneous_coeff_subs_indep_div_dep`
      and
      :py:meth:`~sympy.solvers.ode.ode_1st_homogeneous_coeff_subs_dep_div_indep`
      docstrings for more information on these hints.  Note that there is no
      ``ode_1st_homogeneous_coeff_best_Integral`` hint.
  
      Examples
      ========
  
      >>> from sympy import Function, dsolve, pprint
      >>> from sympy.abc import x
      >>> f = Function('f')
      >>> pprint(dsolve(2*x*f(x) + (x**2 + f(x)**2)*f(x).diff(x), f(x),
      ... hint='1st_homogeneous_coeff_best', simplify=False))
                               /    2    \
                               | 3*x     |
                            log|----- + 1|
                               | 2       |
                               \f (x)    /
      log(f(x)) = log(C1) - --------------
                                  3
  
      References
      ==========
  
      - http://en.wikipedia.org/wiki/Homogeneous_differential_equation
      - M. Tenenbaum & H. Pollard, "Ordinary Differential Equations",
        Dover 1963, pp. 59
  
      # indirect doctest
  
      """
      # There are two substitutions that solve the equation, u1=y/x and u2=x/y
      # They produce different integrals, so try them both and see which
      # one is easier.
      sol1 = ode_1st_homogeneous_coeff_subs_indep_div_dep(eq,
      func, order, match)
      sol2 = ode_1st_homogeneous_coeff_subs_dep_div_indep(eq,
      func, order, match)
      simplify = match.get('simplify', True)
      if simplify:
          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:`~sympy.solvers.ode.ode_1st_homogeneous_coeff_best` and
      :py:meth:`~sympy.solvers.ode.ode_1st_homogeneous_coeff_subs_indep_div_dep`.
  
      Examples
      ========
  
      >>> from sympy import Function, dsolve
      >>> from sympy.abc import x
      >>> f = Function('f')
      >>> pprint(dsolve(2*x*f(x) + (x**2 + f(x)**2)*f(x).diff(x), f(x),
      ... hint='1st_homogeneous_coeff_subs_dep_div_indep', simplify=False))
                            /          3   \
                            |3*f(x)   f (x)|
                         log|------ + -----|
                            |  x         3 |
                            \           x  /
      log(x) = log(C1) - -------------------
                                  3
  
      References
      ==========
  
      - http://en.wikipedia.org/wiki/Homogeneous_differential_equation
      - M. Tenenbaum & H. Pollard, "Ordinary Differential Equations",
        Dover 1963, pp. 59
  
      # indirect doctest
  
      """
      x = func.args[0]
      f = func.func
      u = Dummy('u')
      u1 = Dummy('u1')  # u1 == f(x)/x
      r = match  # d+e*diff(f(x),x)
      C1 = 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:`~sympy.solvers.ode.ode_1st_homogeneous_coeff_subs_dep_div_indep`.
  
      Examples
      ========
  
      >>> from sympy import Function, pprint, dsolve
      >>> from sympy.abc import x
      >>> f = Function('f')
      >>> pprint(dsolve(2*x*f(x) + (x**2 + f(x)**2)*f(x).diff(x), f(x),
      ... hint='1st_homogeneous_coeff_subs_indep_div_dep',
      ... simplify=False))
                               /    2    \
                               | 3*x     |
                            log|----- + 1|
                               | 2       |
                               \f (x)    /
      log(f(x)) = log(C1) - --------------
                                  3
  
      References
      ==========
  
      - http://en.wikipedia.org/wiki/Homogeneous_differential_equation
      - M. Tenenbaum & H. Pollard, "Ordinary Differential Equations",
        Dover 1963, pp. 59
  
      # indirect doctest
  
      """
      x = func.args[0]
      f = func.func
      u = Dummy('u')
      u2 = Dummy('u2')  # u2 == x/f(x)
      r = match  # d+e*diff(f(x),x)
      C1 = 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 = next(dum)
              eq = eq.subs(i, dummyvar)
              symset.remove(i)
              newsyms.add(dummyvar)
      symset.update(newsyms)
  
      if not eq.free_symbols & symset:
          return None
  
      # assuming order of a nested function can only be equal to zero
      if isinstance(eq, Function):
          return None if homogeneous_order(
              eq.args[0], *tuple(symset)) != 0 else S.Zero
  
      # make the replacement of x with x*t and see if t can be factored out
      t = Dummy('t', positive=True)  # It is sufficient that t > 0
      eqs = separatevars(eq.subs([(i, t*i) for i in symset]), [t], dict=True)[t]
      if eqs is S.One:
          return S.Zero  # there was no term with only t
      i, d = eqs.as_independent(t, as_Add=False)
      b, e = d.as_base_exp()
      if b == t:
          return e
  
  
  def ode_1st_linear(eq, func, order, match):
      r"""
      Solves 1st order linear differential equations.
  
      These are differential equations of the form
  
      .. math:: dy/dx + P(x) y = Q(x)\text{.}
  
      These kinds of differential equations can be solved in a general way.  The
      integrating factor `e^{\int P(x) \,dx}` will turn the equation into a
      separable equation.  The general solution is::
  
          >>> from sympy import Function, dsolve, Eq, pprint, diff, sin
          >>> from sympy.abc import x
          >>> f, P, Q = map(Function, ['f', 'P', 'Q'])
          >>> genform = Eq(f(x).diff(x) + P(x)*f(x), Q(x))
          >>> pprint(genform)
                      d
          P(x)*f(x) + --(f(x)) = Q(x)
                      dx
          >>> pprint(dsolve(genform, f(x), hint='1st_linear_Integral'))
                 /       /                   \
                 |      |                    |
                 |      |         /          |     /
                 |      |        |           |    |
                 |      |        | P(x) dx   |  - | P(x) dx
                 |      |        |           |    |
                 |      |       /            |   /
          f(x) = |C1 +  | Q(x)*e           dx|*e
                 |      |                    |
                 \     /                     /
  
  
      Examples
      ========
  
      >>> f = Function('f')
      >>> pprint(dsolve(Eq(x*diff(f(x), x) - f(x), x**2*sin(x)),
      ... f(x), '1st_linear'))
      f(x) = x*(C1 - cos(x))
  
      References
      ==========
  
      - http://en.wikipedia.org/wiki/Linear_differential_equation#First_order_equation
      - M. Tenenbaum & H. Pollard, "Ordinary Differential Equations",
        Dover 1963, pp. 92
  
      # indirect doctest
  
      """
      x = func.args[0]
      f = func.func
      r = match  # a*diff(f(x),x) + b*f(x) + c
      C1 = 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 ode_2nd_power_series_ordinary(eq, func, order, match):
      r"""
      Gives a power series solution to a second order homogeneous differential
      equation with polynomial coefficients at an ordinary point. A homogenous
      differential equation is of the form
  
      .. math :: P(x)\frac{d^2y}{dx^2} + Q(x)\frac{dy}{dx} + R(x) = 0
  
      For simplicity it is assumed that `P(x)`, `Q(x)` and `R(x)` are polynomials,
      it is sufficient that `\frac{Q(x)}{P(x)}` and `\frac{R(x)}{P(x)}` exists at
      `x0`. A recurrence relation is obtained by substituting `y` as `\sum_{n=0}^\infty anx^n`,
      in the differential equation, and equating the nth term. Using this relation
      various terms can be generated.
  
  
      Examples
      ========
      >>> from sympy import dsolve, Function, pprint
      >>> from sympy.abc import x, y
      >>> f = Function("f")
      >>> eq = f(x).diff(x, 2) + f(x)
      >>> pprint(dsolve(eq, hint='2nd_power_series_ordinary'))
                  /   2    \      / 4    2    \
                  |  x     |      |x    x     |    / 6\
      f(x) = C1*x*|- -- + 1| + C0*|-- - -- + 1| + O\x /
                  \  6     /      \24   2     /
  
  
      References
      ==========
      - http://tutorial.math.lamar.edu/Classes/DE/SeriesSolutions.aspx
      - George E. Simmons, "Differential Equations with Applications and
        Historical Notes", p.p 176 - 184
  
      """
      x = func.args[0]
      f = func.func
      C0, C1 = symbols("C0 C1")
      n = Dummy("n")
      s = Wild("s")
      k = Wild("k", exclude=[x])
      x0 = match.get('x0')
      terms = match.get('terms', 5)
      p = match[match['a3']]
      q = match[match['b3']]
      r = match[match['c3']]
      seriesdict = {}
      recurr = Function("r")
  
      # Generating the recurrence relation which works this way
      # a] For the second order term the summation begins at n = 2. The coefficient
      # p is multiplied with an*(n - 1)*(n - 2)*x**n-2 and a substitution is made such that
      # the exponent of x becomes n.
      # For example, if p is x, then the second degree recurrence term is
      # an*(n - 1)*(n - 2)*x**n-1, substituting (n - 1) as n, it transforms to
      # an+1*n*(n - 1)*x**n.
      # A similar process is done with the first order and zeroth order term.
  
      coefflist = [(recurr(n), r), (n*recurr(n), q), (n*(n - 1)*recurr(n), p)]
      for index, coeff in enumerate(coefflist):
          if coeff[1]:
              f2 = powsimp(expand((coeff[1]*(x - x0)**(n - index)).subs(x, x + x0)))
              if f2.is_Add:
                  addargs = f2.args
              else:
                  addargs = [f2]
              for arg in addargs:
                  powm = arg.match(s*x**k)
                  term = coeff[0]*powm[s]
                  if not powm[k].is_Symbol:
                      term = term.subs(n, n - powm[k].as_independent(n)[0])
                  startind = powm[k].subs(n, index)
                  # Seeing if the startterm can be reduced further.
                  # If it vanishes for n lesser than startind, it is
                  # equal to summation from n.
                  if startind:
                      for i in reversed(range(startind)):
                          if not term.subs(n, i):
                              seriesdict[term] = i
                          else:
                              seriesdict[term] = i + 1
                              break
                  else:
                      seriesdict[term] = S(0)
  
      # Stripping of terms so that the sum starts with the same number.
      teq = S(0)
      suminit = seriesdict.values()
      rkeys = seriesdict.keys()
      req = Add(*rkeys)
      if any(suminit):
          maxval = max(suminit)
          for term in seriesdict:
              val = seriesdict[term]
              if val != maxval:
                  for i in range(val, maxval):
                      teq += term.subs(n, val)
  
      finaldict = {}
      if teq:
          fargs = teq.atoms(AppliedUndef)
          if len(fargs) == 1:
              finaldict[fargs.pop()] = 0
          else:
              maxf = max(fargs, key = lambda x: x.args[0])
              sol = solve(teq, maxf)
              if isinstance(sol, list):
                  sol = sol[0]
              finaldict[maxf] = sol
  
      # Finding the recurrence relation in terms of the largest term.
      fargs = req.atoms(AppliedUndef)
      maxf = max(fargs, key = lambda x: x.args[0])
      minf = min(fargs, key = lambda x: x.args[0])
      if minf.args[0].is_Symbol:
          startiter = 0
      else:
          startiter = -minf.args[0].as_independent(n)[0]
      lhs = maxf
      rhs =  solve(req, maxf)
      if isinstance(rhs, list):
          rhs = rhs[0]
  
      # Checking how many values are already present
      tcounter = len([t for t in finaldict.values() if t])
  
      for count in range(tcounter, terms - 3):  # Assuming c0 and c1 to be arbitrary
      #while tcounter < terms - 2:  # Assuming c0 and c1 to be arbitrary
          check = rhs.subs(n, startiter)
          nlhs = lhs.subs(n, startiter)
          nrhs = check.subs(finaldict)
          finaldict[nlhs] = nrhs
          startiter += 1
  
      # Post processing
      series = C0 + C1*(x - x0)
      for term in finaldict:
          if finaldict[term]:
              fact = term.args[0]
              series += (finaldict[term].subs([(recurr(0), C0), (recurr(1), C1)])*(
                  x - x0)**fact)
      series = collect(expand_mul(series), [C0, C1]) + Order(x**terms)
      return Eq(f(x), series)
  
  
  
  def ode_2nd_power_series_regular(eq, func, order, match):
      r"""
      Gives a power series solution to a second order homogeneous differential
      equation with polynomial coefficients at a regular point. A second order
      homogenous differential equation is of the form
  
      .. math :: P(x)\frac{d^2y}{dx^2} + Q(x)\frac{dy}{dx} + R(x) = 0
  
      A point is said to regular singular at `x0` if `x - x0\frac{Q(x)}{P(x)}`
      and `(x - x0)^{2}\frac{R(x)}{P(x)}` are analytic at `x0`. For simplicity
      `P(x)`, `Q(x)` and `R(x)` are assumed to be polynomials. The algorithm for
      finding the power series solutions is:
  
      1.  Try expressing `(x - x0)P(x)` and `((x - x0)^{2})Q(x)` as power series
          solutions about x0. Find `p0` and `q0` which are the constants of the
          power series expansions.
      2.  Solve the indicial equation `f(m) = m(m - 1) + m*p0 + q0`, to obtain the
          roots `m1` and `m2` of the indicial equation.
      3.  If `m1 - m2` is a non integer there exists two series solutions. If
          `m1 = m2`, there exists only one solution. If `m1 - m2` is an integer,
          then the existence of one solution is confirmed. The other solution may
          or may not exist.
  
      The power series solution is of the form `x^{m}\sum_{n=0}^\infty anx^n`. The
      coefficients are determined by the following recurrence relation.
      `an = -\frac{\sum_{k=0}^{n-1} q_{n-k} + (m + k)p_{n-k}}{f(m + n)}`. For the case
      in which `m1 - m2` is an integer, it can be seen from the recurrence relation
      that for the lower root `m`, when `n` equals the difference of both the
      roots, the denominator becomes zero. So if the numerator is not equal to zero,
      a second series solution exists.
  
  
      Examples
      ========
      >>> from sympy import dsolve, Function, pprint
      >>> from sympy.abc import x, y
      >>> f = Function("f")
      >>> eq = x*(f(x).diff(x, 2)) + 2*(f(x).diff(x)) + x*f(x)
      >>> pprint(dsolve(eq))
                /    6    4    2    \
                |   x    x    x     |
             C1*|- --- + -- - -- + 1|      /  4    2    \
                \  720   24   2     /      | x    x     |    / 6\
      f(x) = ------------------------ + C0*|--- - -- + 1| + O\x /
                        x                  \120   6     /
  
  
      References
      ==========
      - George E. Simmons, "Differential Equations with Applications and
        Historical Notes", p.p 176 - 184
  
      """
      x = func.args[0]
      f = func.func
      C0, C1 = symbols("C0 C1")
      n = Dummy("n")
      m = Dummy("m")  # for solving the indicial equation
      s = Wild("s")
      k = Wild("k", exclude=[x])
      x0 = match.get('x0')
      terms = match.get('terms', 5)
      p = match['p']
      q = match['q']
  
      # Generating the indicial equation
      indicial = []
      for term in [p, q]:
          if not term.has(x):
              indicial.append(term)
          else:
              term = series(term, n=1, x0=x0)
              if isinstance(term, Order):
                  indicial.append(S(0))
              else:
                  for arg in term.args:
                      if not arg.has(x):
                          indicial.append(arg)
                          break
  
      p0, q0 = indicial
      sollist = solve(m*(m - 1) + m*p0 + q0, m)
      if sollist and isinstance(sollist, list) and all(
          [sol.is_real for sol in sollist]):
          serdict1 = {}
          serdict2 = {}
          if len(sollist) == 1:
              # Only one series solution exists in this case.
              m1 = m2 = sollist.pop()
              if terms-m1-1 <= 0:
                return Eq(f(x), Order(terms))
              serdict1 = _frobenius(terms-m1-1, m1, p0, q0, p, q, x0, x, C0)
  
          else:
              m1 = sollist[0]
              m2 = sollist[1]
              if m1 < m2:
                  m1, m2 = m2, m1
              # Irrespective of whether m1 - m2 is an integer or not, one
              # Frobenius series solution exists.
              serdict1 = _frobenius(terms-m1-1, m1, p0, q0, p, q, x0, x, C0)
              if not (m1 - m2).is_integer:
                  # Second frobenius series solution exists.
                  serdict2 = _frobenius(terms-m2-1, m2, p0, q0, p, q, x0, x, C1)
              else:
                  # Check if second frobenius series solution exists.
                  serdict2 = _frobenius(terms-m2-1, m2, p0, q0, p, q, x0, x, C1, check=m1)
  
          if serdict1:
              finalseries1 = C0
              for key in serdict1:
                  power = int(key.name[1:])
                  finalseries1 += serdict1[key]*(x - x0)**power
              finalseries1 = (x - x0)**m1*finalseries1
              finalseries2 = S(0)
              if serdict2:
                  for key in serdict2:
                      power = int(key.name[1:])
                      finalseries2 += serdict2[key]*(x - x0)**power
                  finalseries2 += C1
                  finalseries2 = (x - x0)**m2*finalseries2
              return Eq(f(x), collect(finalseries1 + finalseries2,
                  [C0, C1]) + Order(x**terms))
  
  def _frobenius(n, m, p0, q0, p, q, x0, x, c, check=None):
      r"""
      Returns a dict with keys as coefficients and values as their values in terms of C0
      """
      n = int(n)
      # In cases where m1 - m2 is not an integer
      m2 = check
  
      d = Dummy("d")
      numsyms = numbered_symbols("C", start=0)
      numsyms = [next(numsyms) for i in range(n + 1)]
      C0 = Symbol("C0")
      serlist = []
      for ser in [p, q]:
          # Order term not present
          if ser.is_polynomial(x) and Poly(ser, x).degree() <= n:
              if x0:
                  ser = ser.subs(x, x + x0)
              dict_ = Poly(ser, x).as_dict()
          # Order term present
          else:
              tseries = series(ser, x=x0, n=n+1)
              # Removing order
              dict_ = Poly(list(ordered(tseries.args))[: -1], x).as_dict()
          # Fill in with zeros, if coefficients are zero.
          for i in range(n + 1):
              if (i, ) not in dict_:
                  dict_[(i,)] = S(0)
          serlist.append(dict_)
  
      pseries = serlist[0]
      qseries = serlist[1]
      indicial = d*(d - 1) + d*p0 + q0
      frobdict = {}
      for i in range(1, n + 1):
          num = c*(m*pseries[(i,)] + qseries[(i,)])
          for j in range(1, i):
              sym = Symbol("C" + str(j))
              num += frobdict[sym]*((m + j)*pseries[(i - j,)] + qseries[(i - j,)])
  
          # Checking for cases when m1 - m2 is an integer. If num equals zero
          # then a second Frobenius series solution cannot be found. If num is not zero
          # then set constant as zero and proceed.
          if m2 is not None and i == m2 - m:
              if num:
                  return False
              else:
                  frobdict[numsyms[i]] = S(0)
          else:
              frobdict[numsyms[i]] = -num/(indicial.subs(d, m+i))
  
      return frobdict
  
  def _nth_linear_match(eq, func, order):
      r"""
      Matches a differential equation to the linear form:
  
      .. math:: a_n(x) y^{(n)} + \cdots + a_1(x)y' + a_0(x) y + B(x) = 0
  
      Returns a dict of order:coeff terms, where order is the order of the
      derivative on each term, and coeff is the coefficient of that derivative.
      The key ``-1`` holds the function `B(x)`. Returns ``None`` if the ODE is
      not linear.  This function assumes that ``func`` has already been checked
      to be good.
  
      Examples
      ========
  
      >>> from sympy import Function, cos, sin
      >>> from sympy.abc import x
      >>> from sympy.solvers.ode import _nth_linear_match
      >>> f = Function('f')
      >>> _nth_linear_match(f(x).diff(x, 3) + 2*f(x).diff(x) +
      ... x*f(x).diff(x, 2) + cos(x)*f(x).diff(x) + x - f(x) -
      ... sin(x), f(x), 3)
      {-1: x - sin(x), 0: -1, 1: cos(x) + 2, 2: x, 3: 1}
      >>> _nth_linear_match(f(x).diff(x, 3) + 2*f(x).diff(x) +
      ... x*f(x).diff(x, 2) + cos(x)*f(x).diff(x) + x - f(x) -
      ... sin(f(x)), f(x), 3) == None
      True
  
      """
      x = func.args[0]
      one_x = set([x])
      terms = dict([(i, S.Zero) for i in range(-1, order + 1)])
      for i in Add.make_args(eq):
          if not i.has(func):
              terms[-1] += i
          else:
              c, f = i.as_independent(func)
              if not ((isinstance(f, Derivative) and set(f.variables) == one_x) \
                      or f == func):
                  return None
              else:
                  terms[len(f.args[1:])] += c
      return terms
  
  
  def ode_nth_linear_euler_eq_homogeneous(eq, func, order, match, returns='sol'):
      r"""
      Solves an `n`\th order linear homogeneous variable-coefficient
      Cauchy-Euler equidimensional ordinary differential equation.
  
      This is an equation with form `0 = a_0 f(x) + a_1 x f'(x) + a_2 x^2 f''(x)
      \cdots`.
  
      These equations can be solved in a general manner, by substituting
      solutions of the form `f(x) = x^r`, and deriving a characteristic equation
      for `r`.  When there are repeated roots, we include extra terms of the
      form `C_{r k} \ln^k(x) x^r`, where `C_{r k}` is an arbitrary integration
      constant, `r` is a root of the characteristic equation, and `k` ranges
      over the multiplicity of `r`.  In the cases where the roots are complex,
      solutions of the form `C_1 x^a \sin(b \log(x)) + C_2 x^a \cos(b \log(x))`
      are returned, based on expansions with Eulers formula.  The general
      solution is the sum of the terms found.  If SymPy cannot find exact roots
      to the characteristic equation, a
      :py:class:`~sympy.polys.rootoftools.RootOf` instance will be returned
      instead.
  
      >>> from sympy import Function, dsolve, Eq
      >>> from sympy.abc import x
      >>> f = Function('f')
      >>> dsolve(4*x**2*f(x).diff(x, 2) + f(x), f(x),
      ... hint='nth_linear_euler_eq_homogeneous')
      ... # doctest: +NORMALIZE_WHITESPACE
          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) * next(constants)
                  assert multiplicity == 1
                  collectterms = [(0, root, 0)] + collectterms
              elif root.is_real:
                  gsol += ln(x)**i*(x**root) * next(constants)
                  collectterms = [(i, root, 0)] + collectterms
              else:
                  reroot = re(root)
                  imroot = im(root)
                  gsol += ln(x)**i * (x**reroot) * (
                      next(constants) * sin(abs(imroot)*ln(x))
                      + next(constants) * 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_1st_power_series(eq, func, order, match):
      r"""
      The power series solution is a method which gives the Taylor series expansion
      to the solution of a differential equation.
  
      For a first order differential equation `\frac{dy}{dx} = h(x, y)`, a power
      series solution exists at a point `x = x0` if `h(x, y)` is analytic at `x0`.
      The solution is given by
  
      .. math:: f(x0) + \sum_{n = 1}^{\infty} \frac{f^n(x)(x0)(x - x0)^n}{n!}
  
  
      The following algorithm is followed, till the required number of terms are
      generated.
  
      1. F_1 = `h(x, y)`
      2. F_n+1 = \frac{\partial F_n}{\partial x} + \frac{\partial F_n}{\partial y}F_1
  
      Examples
      ========
      >>> from sympy import Function, Derivative, pprint, exp
      >>> from sympy.solvers.ode import dsolve
      >>> from sympy.abc import x
      >>> f = Function('f')
      >>> eq = exp(x)*(f(x).diff(x)) - f(x)
      >>> pprint(dsolve(eq, hint='1st_power_series'))
                             3       4       5
                         C0*x    C0*x    C0*x     / 6\
      f(x) = C0 + C0*x - ----- + ----- + ----- + O\x /
                           6       24      60
  
  
      References
      ==========
  
      - Travis W. Walker, Analytic power series technique for solving first-order
        differential equations, p.p 17, 18
  
      """
      x = func.args[0]
      y = match['y']
      f = func.func
      h = -match[match['d']]/match[match['e']]
      C0 = Symbol("C0")
      point = match.get('f0')
      value = match.get('f0val')
      terms = match.get('terms')
  
      # First term
      F = h
      if not h:
          return Eq(f(x), value)
  
      # Initialisation
      series = value
      if terms > 1:
          hc = h.subs({x: point, y: value})
          if hc.has(oo) or hc.has(NaN) or hc.has(zoo):
              # Derivative does not exist, not analytic
              return Eq(f(x), oo)
          elif hc:
              series += hc*(x - point)
  
      for factcount in range(2, terms):
          Fnew = F.diff(x) + F.diff(y)*h
          Fnewc = Fnew.subs({x: point, y: value})
          # Same logic as above
          if Fnewc.has(oo) or Fnewc.has(NaN) or Fnewc.has(-oo) or Fnewc.has(zoo):
              return Eq(f(x), oo)
          series += Fnewc*((x - point)**factcount)/factorial(factcount)
          F = Fnew
      series += Order(x**terms)
      return Eq(f(x), series)
  
  
  def ode_nth_linear_constant_coeff_homogeneous(eq, func, order, match,
          returns='sol'):
      r"""
      Solves an `n`\th order linear homogeneous differential equation with
      constant coefficients.
  
      This is an equation of the form
  
      .. math:: a_n f^{(n)}(x) + a_{n-1} f^{(n-1)}(x) + \cdots + a_1 f'(x)
                  + a_0 f(x) = 0\text{.}
  
      These equations can be solved in a general manner, by taking the roots of
      the characteristic equation `a_n m^n + a_{n-1} m^{n-1} + \cdots + a_1 m +
      a_0 = 0`.  The solution will then be the sum of `C_n x^i e^{r x}` terms,
      for each where `C_n` is an arbitrary constant, `r` is a root of the
      characteristic equation and `i` is one of each from 0 to the multiplicity
      of the root - 1 (for example, a root 3 of multiplicity 2 would create the
      terms `C_1 e^{3 x} + C_2 x e^{3 x}`).  The exponential is usually expanded
      for complex roots using Euler's equation `e{I x} = \cos(x) + I \sin(x)`.
      Complex roots always come in conjugate pairs in polynomials with real
      coefficients, so the two roots will be represented (after simplifying the
      constants) as `e^{a x} \left(C_1 \cos(b x) + C_2 \sin(b x)\right)`.
  
      If SymPy cannot find exact roots to the characteristic equation, a
      :py:class:`~sympy.polys.rootoftools.RootOf` instance will be return
      instead.
  
      >>> from sympy import Function, dsolve, Eq
      >>> from sympy.abc import x
      >>> f = Function('f')
      >>> dsolve(f(x).diff(x, 5) + 10*f(x).diff(x) - 2*f(x), f(x),
      ... hint='nth_linear_constant_coeff_homogeneous')
      ... # doctest: +NORMALIZE_WHITESPACE
      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 range(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) * next(constants)
                  assert multiplicity == 1
                  collectterms = [(0, root, 0)] + collectterms
              else:
                  reroot = re(root)
                  imroot = im(root)
                  gsol += x**i*exp(reroot*x) * (next(constants) * sin(abs(imroot) * x) + \
                      next(constants) * 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 = next(coeffs)
          coefflist.append(c)
          trialfunc += c*i
  
      eqs = sub_func_doit(eq, f(x), trialfunc)
  
      coeffsdict = dict(list(zip(trialset, [0]*(len(trialset) + 1))))
  
      eqs = expand_mul(eqs)
  
      for i in Add.make_args(eqs):
          s = separatevars(i, dict=True, symbols=[x])
          coeffsdict[s[x]] += s['coeff']
  
      coeffvals = solve(list(coeffsdict.values()), coefflist)
  
      if not coeffvals:
          raise NotImplementedError(
              "Could not solve `%s` using the "
              "method of undetermined coefficients "
              "(unable to solve for coefficients)." % eq)
  
      psol = trialfunc.subs(coeffvals)
  
      return Eq(f(x), gsol.rhs + psol)
  
  
  def _undetermined_coefficients_match(expr, x):
      r"""
      Returns a trial function match if undetermined coefficients can be applied
      to ``expr``, and ``None`` otherwise.
  
      A trial expression can be found for an expression for use with the method
      of undetermined coefficients if the expression is an
      additive/multiplicative combination of constants, polynomials in `x` (the
      independent variable of expr), `\sin(a x + b)`, `\cos(a x + b)`, and
      `e^{a x}` terms (in other words, it has a finite number of linearly
      independent derivatives).
  
      Note that you may still need to multiply each term returned here by
      sufficient `x` to make it linearly independent with the solutions to the
      homogeneous equation.
  
      This is intended for internal use by ``undetermined_coefficients`` hints.
  
      SymPy currently has no way to convert `\sin^n(x) \cos^m(y)` into a sum of
      only `\sin(a x)` and `\cos(b x)` terms, so these are not implemented.  So,
      for example, you will need to manually convert `\sin^2(x)` into `[1 +
      \cos(2 x)]/2` to properly apply the method of undetermined coefficients on
      it.
  
      Examples
      ========
  
      >>> from sympy import log, exp
      >>> from sympy.solvers.ode import _undetermined_coefficients_match
      >>> from sympy.abc import x
      >>> _undetermined_coefficients_match(9*x*exp(x) + exp(-x), x)
      {'test': True, 'trialset': set([x*exp(x), exp(-x), exp(x)])}
      >>> _undetermined_coefficients_match(log(x), x)
      {'test': False}
  
      """
      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([sol for sol in gensols if sol != i], x)*r[-1]/wr, x)*i/r[order]
          negoneterm *= -1
  
      if r.get('simplify', True):
          psol = simplify(psol)
          psol = trigsimp(psol, deep=True)
      return Eq(f(x), gsol.rhs + psol)
  
  
  def ode_separable(eq, func, order, match):
      r"""
      Solves separable 1st order differential equations.
  
      This is any differential equation that can be written as `P(y)
      \tfrac{dy}{dx} = Q(x)`.  The solution can then just be found by
      rearranging terms and integrating: `\int P(y) \,dy = \int Q(x) \,dx`.
      This hint uses :py:meth:`sympy.simplify.simplify.separatevars` as its back
      end, so if a separable equation is not caught by this solver, it is most
      likely the fault of that function.
      :py:meth:`~sympy.simplify.simplify.separatevars` is
      smart enough to do most expansion and factoring necessary to convert a
      separable equation `F(x, y)` into the proper form `P(x)\cdot{}Q(y)`.  The
      general solution is::
  
          >>> from sympy import Function, dsolve, Eq, pprint
          >>> from sympy.abc import x
          >>> a, b, c, d, f = map(Function, ['a', 'b', 'c', 'd', 'f'])
          >>> genform = Eq(a(x)*b(f(x))*f(x).diff(x), c(x)*d(f(x)))
          >>> pprint(genform)
                       d
          a(x)*b(f(x))*--(f(x)) = c(x)*d(f(x))
                       dx
          >>> pprint(dsolve(genform, f(x), hint='separable_Integral'))
               f(x)
             /                  /
            |                  |
            |  b(y)            | c(x)
            |  ---- dy = C1 +  | ---- dx
            |  d(y)            | a(x)
            |                  |
           /                  /
  
      Examples
      ========
  
      >>> from sympy import Function, dsolve, Eq
      >>> from sympy.abc import x
      >>> f = Function('f')
      >>> pprint(dsolve(Eq(f(x)*f(x).diff(x) + x, 3*x*f(x)**2), f(x),
      ... hint='separable', simplify=False))
         /   2       \         2
      log\3*f (x) - 1/        x
      ---------------- = C1 + --
             6                2
  
      References
      ==========
  
      - M. Tenenbaum & H. Pollard, "Ordinary Differential Equations",
        Dover 1963, pp. 52
  
      # indirect doctest
  
      """
      x = func.args[0]
      f = func.func
      C1 = 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 of the given first order differential equation.
      This method is specific to the Lie Group Solver of ODEs.
  
      As of now, it simply checks, by substituting the infinitesimals in the
      partial differential equation.
  
  
      .. math:: \frac{\partial \eta}{\partial x} + \left(\frac{\partial \eta}{\partial y}
                  - \frac{\partial \xi}{\partial x}\right)*h
                  - \frac{\partial \xi}{\partial y}*h^{2}
                  - \xi\frac{\partial h}{\partial x} - \eta\frac{\partial h}{\partial y} = 0
  
  
      where `\eta`, and `\xi` are the infinitesimals and `h(x,y) = \frac{dy}{dx}`
  
      The infinitesimals should be given in the form of a list of dicts
      ``[{xi(x, y): inf, eta(x, y): inf}]``, corresponding to the
      output of the function infinitesimals. It returns a list
      of values of the form ``[(True/False, sol)]`` where ``sol`` is the value
      obtained after substituting the infinitesimals in the PDE. If it
      is ``True``, then ``sol`` would be 0.
  
      """
      if isinstance(eq, Equality):
          eq = eq.lhs - eq.rhs
      if not func:
          eq, func = _preprocess(eq)
      variables = func.args
      if len(variables) != 1:
          raise ValueError("ODE's have only one independent variable")
      else:
          x = variables[0]
          if not order:
              order = ode_order(eq, func)
          if order != 1:
              raise NotImplementedError("Lie groups solver has been implemented "
              "only for first order differential equations")
          else:
              df = func.diff(x)
              a = Wild('a', exclude = [df])
              b = Wild('b', exclude = [df])
              match = collect(expand(eq), df).match(a*df + b)
  
              if match:
                  h = -simplify(match[b]/match[a])
              else:
                  try:
                      sol = solve(eq, df)
                  except NotImplementedError:
                      raise NotImplementedError("Infinitesimals for the "
                          "first order ODE could not be found")
                  else:
                      h = sol[0]  # Find infinitesimals for one solution
  
              y = Dummy('y')
              h = h.subs(func, y)
              xi = Function('xi')(x, y)
              eta = Function('eta')(x, y)
              dxi = Function('xi')(x, func)
              deta = Function('eta')(x, func)
              pde = (eta.diff(x) + (eta.diff(y) - xi.diff(x))*h -
                  (xi.diff(y))*h**2 - xi*(h.diff(x)) - eta*(h.diff(y)))
              soltup = []
              for sol in infinitesimals:
                  tsol = {xi: S(sol[dxi]).subs(func, y),
                      eta: S(sol[deta]).subs(func, y)}
                  sol = simplify(pde.subs(tsol).doit())
                  if sol:
                      soltup.append((False, sol.subs(y, func)))
                  else:
                      soltup.append((True, 0))
              return soltup
  
  def ode_lie_group(eq, func, order, match):
      r"""
      This hint implements the Lie group method of solving first order differential
      equations. The aim is to convert the given differential equation from the
      given coordinate given system into another coordinate system where it becomes
      invariant under the one-parameter Lie group of translations. The converted ODE is
      quadrature and can be solved easily. It makes use of the
      :py:meth:`sympy.solvers.ode.infinitesimals` function which returns the
      infinitesimals of the transformation.
  
      The coordinates `r` and `s` can be found by solving the following Partial
      Differential Equations.
  
      .. math :: \xi\frac{\partial r}{\partial x} + \eta\frac{\partial r}{\partial y}
                    = 0
  
      .. math :: \xi\frac{\partial s}{\partial x} + \eta\frac{\partial s}{\partial y}
                    = 1
  
      The differential equation becomes separable in the new coordinate system
  
      .. math :: \frac{ds}{dr} = \frac{\frac{\partial s}{\partial x} +
                   h(x, y)\frac{\partial s}{\partial y}}{
                   \frac{\partial r}{\partial x} + h(x, y)\frac{\partial r}{\partial y}}
  
      After finding the solution by integration, it is then converted back to the original
      coordinate system by subsituting `r` and `s` in terms of `x` and `y` again.
  
      Examples
      ========
  
      >>> from sympy import Function, dsolve, Eq, exp, pprint
      >>> from sympy.abc import x
      >>> f = Function('f')
      >>> pprint(dsolve(f(x).diff(x) + 2*x*f(x) - x*exp(-x**2), f(x),
      ... hint='lie_group'))
             /      2\    2
             |     x |  -x
      f(x) = |C1 + --|*e
             \     2 /
  
  
      References
      ==========
  
      - Solving differential equations by Symmetry Groups,
        John Starrett, pp. 1 - pp. 14
  
      """
      from sympy.integrals.integrals import integrate
      from sympy.solvers.pde import pdsolve
  
      heuristics = lie_heuristics
      inf = {}
      f = func.func
      x = func.args[0]
      df = func.diff(x)
      xi = Function("xi")
      eta = Function("eta")
      a = Wild('a', exclude = [df])
      b = Wild('b', exclude = [df])
      xis = match.pop('xi')
      etas = match.pop('eta')
  
      if match:
          h = -simplify(match[match['d']]/match[match['e']])
          y = match['y']
      else:
          try:
              sol = solve(eq, df)
          except NotImplementedError:
              raise NotImplementedError("Unable to solve the differential equation " +
                  str(eq) + " by the lie group method")
          else:
              y = Dummy("y")
              h = sol[0].subs(func, y)
  
      if xis is not None and etas is not None:
          inf = [{xi(x, f(x)): S(xis), eta(x, f(x)): S(etas)}]
  
          if not checkinfsol(eq, inf, func=f(x), order=1)[0][0]:
              raise ValueError("The given infinitesimals xi and eta"
                  " are not the infinitesimals to the given equation")
          else:
              heuristics = ["user_defined"]
  
      match = {'h': h, 'y': y}
  
      # This is done so that if:
      # a] solve raises a NotImplementedError.
      # b] any heuristic raises a ValueError
      # another heuristic can be used.
      tempsol = []  # Used by solve below
      for heuristic in heuristics:
          try:
              if not inf:
                  inf = infinitesimals(eq, hint=heuristic, func=func, order=1, match=match)
          except ValueError:
              continue
          else:
              for infsim in inf:
                  xiinf = (infsim[xi(x, func)]).subs(func, y)
                  etainf = (infsim[eta(x, func)]).subs(func, y)
                  # This condition creates recursion while using pdsolve.
                  # Since the first step while solving a PDE of form
                  # a*(f(x, y).diff(x)) + b*(f(x, y).diff(y)) + c = 0
                  # is to solve the ODE dy/dx = b/a
                  if simplify(etainf/xiinf) == h:
                      continue
                  rpde = f(x, y).diff(x)*xiinf + f(x, y).diff(y)*etainf
                  r = pdsolve(rpde, func=f(x, y)).rhs
                  s = pdsolve(rpde - 1, func=f(x, y)).rhs
                  newcoord = [_lie_group_remove(coord) for coord in [r, s]]
                  r = Dummy("r")
                  s = Dummy("s")
                  C1 = Symbol("C1")
                  rcoord = newcoord[0]
                  scoord = newcoord[-1]
                  try:
                      sol = solve([r - rcoord, s - scoord], x, y, dict=True)
                  except NotImplementedError:
                      continue
                  else:
                      sol = sol[0]
                      xsub = sol[x]
                      ysub = sol[y]
                      num = simplify(scoord.diff(x) + scoord.diff(y)*h)
                      denom = simplify(rcoord.diff(x) + rcoord.diff(y)*h)
                      if num and denom:
                          diffeq = simplify((num/denom).subs([(x, xsub), (y, ysub)]))
                          sep = separatevars(diffeq, symbols=[r, s], dict=True)
                          if sep:
                              # Trying to separate, r and s coordinates
                              deq = integrate((1/sep[s]), s) + C1 - integrate(sep['coeff']*sep[r], r)
                              # Substituting and reverting back to original coordinates
                              deq = deq.subs([(r, rcoord), (s, scoord)])
                              try:
                                  sdeq = solve(deq, y)
                              except NotImplementedError:
                                  tempsol.append(deq)
                              else:
                                  if len(sdeq) == 1:
                                      return Eq(f(x), sdeq.pop())
                                  else:
                                      return [Eq(f(x), sol) for sol in sdeq]
  
  
                      elif denom: # (ds/dr) is zero which means s is constant
                          return Eq(f(x), solve(scoord - C1, y)[0])
  
                      elif num: # (dr/ds) is zero which means r is constant
                          return Eq(f(x), solve(rcoord - C1, y)[0])
  
      # If nothing works, return solution as it is, without solving for y
      if tempsol:
          if len(tempsol) == 1:
              return Eq(tempsol.pop().subs(y, f(x)), 0)
          else:
              return [Eq(sol.subs(y, f(x)), 0) for sol in tempsol]
  
      raise NotImplementedError("The given ODE " + str(eq) + " cannot be solved by"
          + " the lie group method")
  
  
  def _lie_group_remove(coords):
      r"""
      This function is strictly meant for internal use by the Lie group ODE solving
      method. It replaces arbitrary functions returned by pdsolve with either 0 or 1 or the
      args of the arbitrary function.
  
      The algorithm used is:
      1] If coords is an instance of an Undefined Function, then the args are returned
      2] If the arbitrary function is present in an Add object, it is replaced by zero.
      3] If the arbitrary function is present in an Mul object, it is replaced by one.
      4] If coords has no Undefined Function, it is returned as it is.
  
      Examples
      ========
      >>> from sympy.solvers.ode import _lie_group_remove
      >>> from sympy import Function
      >>> from sympy.abc import x, y
      >>> F = Function("F")
      >>> eq = x**2*y
      >>> _lie_group_remove(eq)
      x**2*y
      >>> eq = F(x**2*y)
      >>> _lie_group_remove(eq)
      x**2*y
      >>> eq = y**2*x + F(x**3)
      >>> _lie_group_remove(eq)
      x*y**2
      >>> eq = (F(x**3) + y)*x**4
      >>> _lie_group_remove(eq)
      x**4*y
  
      """
      if isinstance(coords, AppliedUndef):
          return coords.args[0]
      elif coords.is_Add:
          subfunc = coords.atoms(AppliedUndef)
          if subfunc:
              for func in subfunc:
                  coords = coords.subs(func, 0)
          return coords
      elif coords.is_Pow:
          base, expr = coords.as_base_exp()
          base = _lie_group_remove(base)
          expr = _lie_group_remove(expr)
          return base**expr
      elif coords.is_Mul:
          mulargs = []
          coordargs = coords.args
          for arg in coordargs:
              if not isinstance(coords, AppliedUndef):
                  mulargs.append(_lie_group_remove(arg))
          return Mul(*mulargs)
      return coords
  
  def infinitesimals(eq, func=None, order=None, hint='default', match=None):
      r"""
      The infinitesimal functions of an ordinary differential equation, `\xi(x,y)`
      and `\eta(x,y)`, are the infinitesimals of the Lie group of point transformations
      for which the differential equation is invariant. So, the ODE `y'=f(x,y)`
      would admit a Lie group `x^*=X(x,y;\varepsilon)=x+\varepsilon\xi(x,y)`,
      `y^*=Y(x,y;\varepsilon)=y+\varepsilon\eta(x,y)` such that `(y^*)'=f(x^*, y^*)`.
      A change of coordinates, to `r(x,y)` and `s(x,y)`, can be performed so this Lie group
      becomes the translation group, `r^*=r` and `s^*=s+\varepsilon`.
      They are tangents to the coordinate curves of the new system.
  
      Consider the transformation `(x, y) \to (X, Y)` such that the
      differential equation remains invariant. `\xi` and `\eta` are the tangents to
      the transformed coordinates `X` and `Y`, at `\varepsilon=0`.
  
      .. math:: \left(\frac{\partial X(x,y;\varepsilon)}{\partial\varepsilon
                  }\right)|_{\varepsilon=0} = \xi,
                \left(\frac{\partial Y(x,y;\varepsilon)}{\partial\varepsilon
                  }\right)|_{\varepsilon=0} = \eta,
  
      The infinitesimals can be found by solving the following PDE:
  
          >>> from sympy import Function, diff, Eq, pprint
          >>> from sympy.abc import x, y
          >>> xi, eta, h = map(Function, ['xi', 'eta', 'h'])
          >>> h = h(x, y)  # dy/dx = h
          >>> eta = eta(x, y)
          >>> xi = xi(x, y)
          >>> genform = Eq(eta.diff(x) + (eta.diff(y) - xi.diff(x))*h
          ... - (xi.diff(y))*h**2 - xi*(h.diff(x)) - eta*(h.diff(y)), 0)
          >>> pprint(genform)
          /d               d           \                     d              2       d
          |--(eta(x, y)) - --(xi(x, y))|*h(x, y) - eta(x, y)*--(h(x, y)) - h (x, y)*--(x
          \dy              dx          /                     dy                     dy
          <BLANKLINE>
                              d             d
          i(x, y)) - xi(x, y)*--(h(x, y)) + --(eta(x, y)) = 0
                              dx            dx
  
      Solving the above mentioned PDE is not trivial, and can be solved only by
      making intelligent assumptions for `\xi` and `\eta` (heuristics). Once an
      infinitesimal is found, the attempt to find more heuristics stops. This is done to
      optimise the speed of solving the differential equation. If a list of all the
      infinitesimals is needed, ``hint`` should be flagged as ``all``, which gives
      the complete list of infinitesimals. If the infinitesimals for a particular
      heuristic needs to be found, it can be passed as a flag to ``hint``.
  
      Examples
      ========
  
      >>> from sympy import Function, diff
      >>> from sympy.solvers.ode import infinitesimals
      >>> from sympy.abc import x
      >>> f = Function('f')
      >>> eq = f(x).diff(x) - x**2*f(x)
      >>> infinitesimals(eq)
      [{eta(x, f(x)): exp(x**3/3), xi(x, f(x)): 0}]
  
      References
      ==========
  
      - Solving differential equations by Symmetry Groups,
        John Starrett, pp. 1 - pp. 14
  
      """
  
      if isinstance(eq, Equality):
          eq = eq.lhs - eq.rhs
      if not func:
          eq, func = _preprocess(eq)
      variables = func.args
      if len(variables) != 1:
          raise ValueError("ODE's have only one independent variable")
      else:
          x = variables[0]
          if not order:
              order = ode_order(eq, func)
          if order != 1:
              raise NotImplementedError("Infinitesimals for only "
                  "first order ODE's have been implemented")
          else:
              df = func.diff(x)
              # Matching differential equation of the form a*df + b
              a = Wild('a', exclude = [df])
              b = Wild('b', exclude = [df])
              if match:  # Used by lie_group hint
                  h = match['h']
                  y = match['y']
              else:
                  match = collect(expand(eq), df).match(a*df + b)
                  if match:
                      h = -simplify(match[b]/match[a])
                  else:
                      try:
                          sol = solve(eq, df)
                      except NotImplementedError:
                          raise NotImplementedError("Infinitesimals for the "
                              "first order ODE could not be found")
                      else:
                          h = sol[0]  # Find infinitesimals for one solution
                  y = Dummy("y")
                  h = h.subs(func, y)
  
              u = Dummy("u")
              hx = h.diff(x)
              hy = h.diff(y)
              hinv = ((1/h).subs([(x, u), (y, x)])).subs(u, y)  # Inverse ODE
              match = {'h': h, 'func': func, 'hx': hx, 'hy': hy, 'y': y, 'hinv': hinv}
              if hint == 'all':
                  xieta = []
                  for heuristic in lie_heuristics:
                      function = globals()['lie_heuristic_' + heuristic]
                      inflist = function(match, comp=True)
                      if inflist:
                          xieta.extend([inf for inf in inflist if inf not in xieta])
                  if xieta:
                      return xieta
                  else:
                      raise NotImplementedError("Infinitesimals could not be found for"
                          "the given ODE")
  
              elif hint == 'default':
                  for heuristic in lie_heuristics:
                      function = globals()['lie_heuristic_' + heuristic]
                      xieta = function(match, comp=False)
                      if xieta:
                          return xieta
  
                  raise NotImplementedError("Infinitesimals could not be found for"
                      " the given ODE")
  
              elif hint not in lie_heuristics:
                   raise ValueError("Heuristic not recognized: " + hint)
  
              else:
                   function = globals()['lie_heuristic_' + hint]
                   xieta = function(match, comp=True)
                   if xieta:
                       return xieta
                   else:
                       raise ValueError("Infinitesimals could not be found using the"
                           " given heuristic")
  
  
  def lie_heuristic_abaco1_simple(match, comp=False):
      r"""
      The first heuristic uses the following four sets of
      assumptions on `\xi` and `\eta`
  
      .. math:: \xi = 0, \eta = f(x)
  
      .. math:: \xi = 0, \eta = f(y)
  
      .. math:: \xi = f(x), \eta = 0
  
      .. math:: \xi = f(y), \eta = 0
  
      The success of this heuristic is determined by algebraic factorisation.
      For the first assumption `\xi = 0` and `eta` to be a function of `x`, the PDE
  
      .. math:: \frac{\partial \eta}{\partial x} + (\frac{\partial \eta}{\partial y}
                  - \frac{\partial \xi}{\partial x})*h
                  - \frac{\partial \xi}{\partial y}*h^{2}
                  - \xi*\frac{\partial h}{\partial x} - \eta*\frac{\partial h}{\partial y} = 0
  
      reduces to `f'(x) - f\frac{\partial h}{\partial y} = 0`
      If `\frac{\partial h}{\partial y}` is a function of `x`, then this can usually
      be integrated easily. A similar idea is applied to the other 3 assumptions as well.
  
  
      References
      ==========
  
      - E.S Cheb-Terrab, L.G.S Duarte and L.A,C.P da Mota, Computer Algebra
        Solving of First Order ODEs Using Symmetry Methods, pp. 8
  
  
      """
      from sympy.integrals.integrals import integrate
  
      xieta = []
      y = match['y']
      h = match['h']
      func = match['func']
      x = func.args[0]
      hx = match['hx']
      hy = match['hy']
      xi = Function('xi')(x, func)
      eta = Function('eta')(x, func)
  
      hysym = hy.free_symbols
      if y not in hysym:
          try:
              fx = exp(integrate(hy, x))
          except NotImplementedError:
              pass
          else:
              inf = {xi: S(0), eta: fx}
              if not comp:
                  return [inf]
              if comp and inf not in xieta:
                  xieta.append(inf)
  
      factor = hy/h
      facsym = factor.free_symbols
      if x not in facsym:
          try:
              fy = exp(integrate(factor, y))
          except NotImplementedError:
              pass
          else:
              inf = {xi: S(0), eta: fy.subs(y, func)}
              if not comp:
                  return [inf]
              if comp and inf not in xieta:
                  xieta.append(inf)
  
      factor = -hx/h
      facsym = factor.free_symbols
      if y not in facsym:
          try:
              fx = exp(integrate(factor, x))
          except NotImplementedError:
              pass
          else:
              inf = {xi: fx, eta: S(0)}
              if not comp:
                  return [inf]
              if comp and inf not in xieta:
                  xieta.append(inf)
  
      factor = -hx/(h**2)
      facsym = factor.free_symbols
      if x not in facsym:
          try:
              fy = exp(integrate(factor, y))
          except NotImplementedError:
              pass
          else:
              inf = {xi: fy.subs(y, func), eta: S(0)}
              if not comp:
                  return [inf]
              if comp and inf not in xieta:
                  xieta.append(inf)
  
      if xieta:
          return xieta
  
  def lie_heuristic_abaco1_product(match, comp=False):
      r"""
      The second heuristic uses the following two assumptions on `\xi` and `\eta`
  
      .. math:: \eta = 0, \xi = f(x)*g(y)
  
      .. math:: \eta = f(x)*g(y), \xi = 0
  
      The first assumption of this heuristic holds good if
      `\frac{1}{h^{2}}\frac{\partial^2}{\partial x \partial y}\log(h)` is
      separable in `x` and `y`, then the separated factors containing `x`
      is `f(x)`, and `g(y)` is obtained by
  
      .. math:: exp^{\int f\frac{\partial}{\partial x}\left(\frac{1}{f*h}\right)\,dy}
  
      provided `f\frac{\partial}{\partial x}\left(\frac{1}{f*h}\right)` is a function
      of `y` only.
  
      The second assumption holds good if `\frac{dy}{dx} = h(x, y)` is rewritten as
      `\frac{dy}{dx} = \frac{1}{h(y, x)}` and the same properties of the first assumption
      satisifes. After obtaining `f(x)` and `g(y)`, the coordinates are again
      interchanged, to get `\eta` as `f(x)*g(y)`
  
  
      References
      ==========
      - E.S. Cheb-Terrab, A.D. Roche, Symmetries and First Order
        ODE Patterns, pp. 7 - pp. 8
  
      """
      from sympy.integrals.integrals import integrate
  
      xieta = []
      y = match['y']
      h = match['h']
      hinv = match['hinv']
      func = match['func']
      x = func.args[0]
      xi = Function('xi')(x, func)
      eta = Function('eta')(x, func)
  
  
      inf = separatevars(((log(h).diff(y)).diff(x))/h**2, dict=True, symbols=[x, y])
      if inf and inf['coeff']:
          fx = inf[x]
          gy = simplify(fx*((1/(fx*h)).diff(x)))
          gysyms = gy.free_symbols
          if x not in gysyms:
              gy = exp(integrate(gy, y))
              inf = {eta: S(0), xi: (fx*gy).subs(y, func)}
              if not comp:
                  return [inf]
              if comp and inf not in xieta:
                  xieta.append(inf)
  
      u1 = Dummy("u1")
      inf = separatevars(((log(hinv).diff(y)).diff(x))/hinv**2, dict=True, symbols=[x, y])
      if inf and inf['coeff']:
          fx = inf[x]
          gy = simplify(fx*((1/(fx*hinv)).diff(x)))
          gysyms = gy.free_symbols
          if x not in gysyms:
              gy = exp(integrate(gy, y))
              etaval = fx*gy
              etaval = (etaval.subs([(x, u1), (y, x)])).subs(u1, y)
              inf = {eta: etaval.subs(y, func), xi: S(0)}
              if not comp:
                  return [inf]
              if comp and inf not in xieta:
                  xieta.append(inf)
  
      if xieta:
          return xieta
  
  def lie_heuristic_bivariate(match, comp=False):
      r"""
      The third heuristic assumes the infinitesimals `\xi` and `\eta`
      to be bi-variate polynomials in `x` and `y`. The assumption made here
      for the logic below is that `h` is a rational function in `x` and `y`
      though that may not be necessary for the infinitesimals to be
      bivariate polynomials. The coefficients of the infinitesimals
      are found out by substituting them in the PDE and grouping similar terms
      that are polynomials and since they form a linear system, solve and check
      for non trivial solutions. The degree of the assumed bivariates
      are increased till a certain maximum value.
  
      References
      ==========
      - Lie Groups and Differential Equations
        pp. 327 - pp. 329
  
      """
  
      h = match['h']
      hx = match['hx']
      hy = match['hy']
      func = match['func']
      x = func.args[0]
      y = match['y']
      xi = Function('xi')(x, func)
      eta = Function('eta')(x, func)
  
      if h.is_rational_function():
          # The maximum degree that the infinitesimals can take is
          # calculated by this technique.
          etax, etay, etad, xix, xiy, xid = symbols("etax etay etad xix xiy xid")
          ipde = etax + (etay - xix)*h - xiy*h**2 - xid*hx - etad*hy
          num, denom = cancel(ipde).as_numer_denom()
          deg = Poly(num, x, y).total_degree()
          deta = Function('deta')(x, y)
          dxi = Function('dxi')(x, y)
          ipde = (deta.diff(x) + (deta.diff(y) - dxi.diff(x))*h - (dxi.diff(y))*h**2
              - dxi*hx - deta*hy)
          xieq = Symbol("xi0")
          etaeq = Symbol("eta0")
  
          for i in range(deg + 1):
              if i:
                  xieq += Add(*[
                      Symbol("xi_" + str(power) + "_" + str(i - power))*x**power*y**(i - power)
                      for power in range(i + 1)])
                  etaeq += Add(*[
                      Symbol("eta_" + str(power) + "_" + str(i - power))*x**power*y**(i - power)
                      for power in range(i + 1)])
              pden, denom = (ipde.subs({dxi: xieq, deta: etaeq}).doit()).as_numer_denom()
              pden = expand(pden)
  
              # If the individual terms are monomials, the coefficients
              # are grouped
              if pden.is_polynomial(x, y) and pden.is_Add:
                  polyy = Poly(pden, x, y).as_dict()
              if polyy:
                  symset = xieq.free_symbols.union(etaeq.free_symbols) - set([x, y])
                  soldict = solve(polyy.values(), *symset)
                  if isinstance(soldict, list):
                      soldict = soldict[0]
                  if any(x for x in soldict.values()):
                      xired = xieq.subs(soldict)
                      etared = etaeq.subs(soldict)
                      # Scaling is done by substituting one for the parameters
                      # This can be any number except zero.
                      dict_ = dict((sym, 1) for sym in symset)
                      inf = {eta: etared.subs(dict_).subs(y, func),
                          xi: xired.subs(dict_).subs(y, func)}
                      return [inf]
  
  def lie_heuristic_chi(match, comp=False):
      r"""
      The aim of the fourth heuristic is to find the function `\chi(x, y)`
      that satisifies the PDE `\frac{d\chi}{dx} + h\frac{d\chi}{dx}
      - \frac{\partial h}{\partial y}\chi = 0`.
  
      This assumes `\chi` to be a bivariate polynomial in `x` and `y`. By intution,
      `h` should be a rational function in `x` and `y`. The method used here is
      to substitute a general binomial for `\chi` up to a certain maximum degree
      is reached. The coefficients of the polynomials, are calculated by by collecting
      terms of the same order in `x` and `y`.
  
      After finding `\chi`, the next step is to use `\eta = \xi*h + \chi`, to
      determine `\xi` and `\eta`. This can be done by dividing `\chi` by `h`
      which would give `-\xi` as the quotient and `\eta` as the remainder.
  
  
      References
      ==========
      - E.S Cheb-Terrab, L.G.S Duarte and L.A,C.P da Mota, Computer Algebra
        Solving of First Order ODEs Using Symmetry Methods, pp. 8
  
      """
  
      h = match['h']
      hx = match['hx']
      hy = match['hy']
      func = match['func']
      x = func.args[0]
      y = match['y']
      xi = Function('xi')(x, func)
      eta = Function('eta')(x, func)
  
      if h.is_rational_function():
          schi, schix, schiy = symbols("schi, schix, schiy")
          cpde = schix + h*schiy - hy*schi
          num, denom = cancel(cpde).as_numer_denom()
          deg = Poly(num, x, y).total_degree()
  
          chi = Function('chi')(x, y)
          chix = chi.diff(x)
          chiy = chi.diff(y)
          cpde = chix + h*chiy - hy*chi
          chieq = Symbol("chi")
          for i in range(1, deg + 1):
              chieq += Add(*[
                  Symbol("chi_" + str(power) + "_" + str(i - power))*x**power*y**(i - power)
                  for power in range(i + 1)])
              cnum, cden = cancel(cpde.subs({chi : chieq}).doit()).as_numer_denom()
              cnum = expand(cnum)
              if cnum.is_polynomial(x, y) and cnum.is_Add:
                  cpoly = Poly(cnum, x, y).as_dict()
                  if cpoly:
                      solsyms = chieq.free_symbols - set([x, y])
                      soldict = solve(cpoly.values(), *solsyms)
                      if isinstance(soldict, list):
                          soldict = soldict[0]
                      if any(x for x in soldict.values()):
                          chieq = chieq.subs(soldict)
                          dict_ = dict((sym, 1) for sym in solsyms)
                          chieq = chieq.subs(dict_)
                          # After finding chi, the main aim is to find out
                          # eta, xi by the equation eta = xi*h + chi
                          # One method to set xi, would be rearranging it to
                          # (eta/h) - xi = (chi/h). This would mean dividing
                          # chi by h would give -xi as the quotient and eta
                          # as the remainder. Thanks to Sean Vig for suggesting
                          # this method.
                          xic, etac = div(chieq, h)
                          inf = {eta: etac.subs(y, func), xi: -xic.subs(y, func)}
                          return [inf]
  
  def lie_heuristic_function_sum(match, comp=False):
      r"""
      This heuristic uses the following two assumptions on `\xi` and `\eta`
  
      .. math:: \eta = 0, \xi = f(x) + g(y)
  
      .. math:: \eta = f(x) + g(y), \xi = 0
  
      The first assumption of this heuristic holds good if
  
      .. math:: \frac{\partial}{\partial y}[(h\frac{\partial^{2}}{
                  \partial x^{2}}(h^{-1}))^{-1}]
  
      is separable in `x` and `y`,
  
      1. The separated factors containing `y` is `\frac{\partial g}{\partial y}`.
         From this `g(y)` can be determined.
      2. The separated factors containing `x` is `f''(x)`.
      3. `h\frac{\partial^{2}}{\partial x^{2}}(h^{-1})` equals
         `\frac{f''(x)}{f(x) + g(y)}`. From this `f(x)` can be determined.
  
      The second assumption holds good if `\frac{dy}{dx} = h(x, y)` is rewritten as
      `\frac{dy}{dx} = \frac{1}{h(y, x)}` and the same properties of the first
      assumption satisifes. After obtaining `f(x)` and `g(y)`, the coordinates
      are again interchanged, to get `\eta` as `f(x) + g(y)`.
  
      For both assumptions, the constant factors are separated among `g(y)`
      and `f''(x)`, such that `f''(x)` obtained from 3] is the same as that
      obtained from 2]. If not possible, then this heuristic fails.
  
  
      References
      ==========
      - E.S. Cheb-Terrab, A.D. Roche, Symmetries and First Order
        ODE Patterns, pp. 7 - pp. 8
  
      """
      from sympy.integrals.integrals import integrate
      xieta = []
      h = match['h']
      hx = match['hx']
      hy = match['hy']
      func = match['func']
      hinv = match['hinv']
      x = func.args[0]
      y = match['y']
      xi = Function('xi')(x, func)
      eta = Function('eta')(x, func)
  
      for odefac in [h, hinv]:
          factor = odefac*((1/odefac).diff(x, 2))
          sep = separatevars((1/factor).diff(y), dict=True, symbols=[x, y])
          if sep and sep['coeff'] and sep[x].has(x) and sep[y].has(y):
              k = Dummy("k")
              try:
                  gy = k*integrate(sep[y], y)
              except NotImplementedError:
                  pass
              else:
                  fdd = 1/(k*sep[x]*sep['coeff'])
                  fx = simplify(fdd/factor - gy)
                  check = simplify(fx.diff(x, 2) - fdd)
                  if fx:
                      if not check:
                          fx = fx.subs(k, 1)
                          gy = (gy/k)
                      else:
                          sol = solve(check, k)
                          if sol:
                              sol = sol[0]
                              fx = fx.subs(k, sol)
                              gy = (gy/k)*sol
                          else:
                              continue
                      if odefac == hinv:  # Inverse ODE
                          fx = fx.subs(x, y)
                          gy = gy.subs(y, x)
                      etaval = factor_terms(fx + gy)
                      if etaval.is_Mul:
                          etaval = Mul(*[arg for arg in etaval.args if arg.has(x, y)])
                      if odefac == hinv:  # Inverse ODE
                          inf = {eta: etaval.subs(y, func), xi : S(0)}
                      else:
                          inf = {xi: etaval.subs(y, func), eta : S(0)}
                      if not comp:
                          return [inf]
                      else:
                          xieta.append(inf)
  
          if xieta:
              return xieta
  
  def lie_heuristic_abaco2_similar(match, comp=False):
      r"""
      This heuristic uses the following two assumptions on `\xi` and `\eta`
  
      .. math:: \eta = g(x), \xi = f(x)
  
      .. math:: \eta = f(y), \xi = g(y)
  
      For the first assumption,
  
      1. First `\frac{\frac{\partial h}{\partial y}}{\frac{\partial^{2} h}{
         \partial yy}}` is calculated. Let us say this value is A
  
      2. If this is constant, then `h` is matched to the form `A(x) + B(x)e^{
         \frac{y}{C}}` then, `\frac{e^{\int \frac{A(x)}{C} \,dx}}{B(x)}` gives `f(x)`
         and `A(x)*f(x)` gives `g(x)`
  
      3. Otherwise `\frac{\frac{\partial A}{\partial X}}{\frac{\partial A}{
         \partial Y}} = \gamma` is calculated. If
  
         a] `\gamma` is a function of `x` alone
  
         b] `\frac{\gamma\frac{\partial h}{\partial y} - \gamma'(x) - \frac{
         \partial h}{\partial x}}{h + \gamma} = G` is a function of `x` alone.
         then, `e^{\int G \,dx}` gives `f(x)` and `-\gamma*f(x)` gives `g(x)`
  
      The second assumption holds good if `\frac{dy}{dx} = h(x, y)` is rewritten as
      `\frac{dy}{dx} = \frac{1}{h(y, x)}` and the same properties of the first assumption
      satisifes. After obtaining `f(x)` and `g(x)`, the coordinates are again
      interchanged, to get `\xi` as `f(x^*)` and `\eta` as `g(y^*)`
  
      References
      ==========
      - E.S. Cheb-Terrab, A.D. Roche, Symmetries and First Order
        ODE Patterns, pp. 10 - pp. 12
  
      """
  
      from sympy.integrals.integrals import integrate
  
      xieta = []
      h = match['h']
      hx = match['hx']
      hy = match['hy']
      func = match['func']
      hinv = match['hinv']
      x = func.args[0]
      y = match['y']
      xi = Function('xi')(x, func)
      eta = Function('eta')(x, func)
  
      factor = cancel(h.diff(y)/h.diff(y, 2))
      factorx = factor.diff(x)
      factory = factor.diff(y)
      if not factor.has(x) and not factor.has(y):
          A = Wild('A', exclude=[y])
          B = Wild('B', exclude=[y])
          C = Wild('C', exclude=[x, y])
          match = h.match(A + B*exp(y/C))
          try:
              tau = exp(-integrate(match[A]/match[C]), x)/match[B]
          except NotImplementedError:
              pass
          else:
              gx = match[A]*tau
              return [{xi: tau, eta: gx}]
  
      else:
          gamma = cancel(factorx/factory)
          if not gamma.has(y):
              tauint = cancel((gamma*hy - gamma.diff(x) - hx)/(h + gamma))
              if not tauint.has(y):
                  try:
                      tau = exp(integrate(tauint, x))
                  except NotImplementedError:
                      pass
                  else:
                      gx = -tau*gamma
                      return [{xi: tau, eta: gx}]
  
      factor = cancel(hinv.diff(y)/hinv.diff(y, 2))
      factorx = factor.diff(x)
      factory = factor.diff(y)
      if not factor.has(x) and not factor.has(y):
          A = Wild('A', exclude=[y])
          B = Wild('B', exclude=[y])
          C = Wild('C', exclude=[x, y])
          match = h.match(A + B*exp(y/C))
          try:
              tau = exp(-integrate(match[A]/match[C]), x)/match[B]
          except NotImplementedError:
              pass
          else:
              gx = match[A]*tau
              return [{eta: tau.subs(x, func), xi: gx.subs(x, func)}]
  
      else:
          gamma = cancel(factorx/factory)
          if not gamma.has(y):
              tauint = cancel((gamma*hinv.diff(y) - gamma.diff(x) - hinv.diff(x))/(
                  hinv + gamma))
              if not tauint.has(y):
                  try:
                      tau = exp(integrate(tauint, x))
                  except NotImplementedError:
                      pass
                  else:
                      gx = -tau*gamma
                      return [{eta: tau.subs(x, func), xi: gx.subs(x, func)}]
  
  
  def lie_heuristic_abaco2_unique_unknown(match, comp=False):
      r"""
      This heuristic assumes the presence of unknown functions or known functions
      with non-integer powers.
  
      1. A list of all functions and non-integer powers containing x and y
      2. Loop over each element `f` in the list, find `\frac{\frac{\partial f}{\partial x}}{
         \frac{\partial f}{\partial x}} = R`
  
         If it is separable in `x` and `y`, let `X` be the factors containing `x`. Then
  
         a] Check if `\xi = X` and `\eta = -\frac{X}{R}` satisfy the PDE. If yes, then return
            `\xi` and `\eta`
         b] Check if `\xi = \frac{-R}{X}` and `\eta = -\frac{1}{X}` satisfy the PDE.
             If yes, then return `\xi` and `\eta`
  
         If not, check if the following satisfy the ODE
  
         a] `\xi = -R`, `\eta = 1`
         b] `\xi = 1`, `\eta = -\frac{1}{R}`
  
      References
      ==========
      - E.S. Cheb-Terrab, A.D. Roche, Symmetries and First Order
        ODE Patterns, pp. 10 - pp. 12
  
      """
  
      xieta = []
      h = match['h']
      hx = match['hx']
      hy = match['hy']
      func = match['func']
      hinv = match['hinv']
      x = func.args[0]
      y = match['y']
      xi = Function('xi')(x, func)
      eta = Function('eta')(x, func)
  
      funclist = []
      for atom in h.atoms(Pow):
          base, exp = atom.as_base_exp()
          if base.has(x) and base.has(y):
              if not exp.is_Integer:
                  funclist.append(atom)
  
      for function in h.atoms(AppliedUndef):
          syms = function.free_symbols
          if x in syms and y in syms:
              funclist.append(function)
  
      for f in funclist:
          frac = cancel(f.diff(y)/f.diff(x))
          sep = separatevars(frac, dict=True, symbols=[x, y])
          if sep and sep['coeff']:
              xitry1 = sep[x]
              etatry1 = -1/(sep[y]*sep['coeff'])
              pde1 = etatry1.diff(y)*h - xitry1.diff(x)*h - xitry1*hx - etatry1*hy
              if not simplify(pde1):
                  return [{xi: xitry1, eta: etatry1.subs(y, func)}]
              xitry2 = 1/etatry1
              etatry2 = 1/xitry1
              pde2 = etatry2.diff(x) - (xitry2.diff(y))*h**2 - xitry2*hx - etatry2*hy
              if not simplify(expand(pde2)):
                  return [{xi: xitry2.subs(y, func), eta: etatry2}]
  
          else:
              etatry = -1/frac
              pde = etatry.diff(x) + etatry.diff(y)*h - hx - etatry*hy
              if not simplify(pde):
                  return [{xi: S(1), eta: etatry.subs(y, func)}]
              xitry = -frac
              pde = -xitry.diff(x)*h -xitry.diff(y)*h**2 - xitry*hx -hy
              if not simplify(expand(pde)):
                  return [{xi: xitry.subs(y, func), eta: S(1)}]
  
  
  def lie_heuristic_abaco2_unique_general(match, comp=False):
      r"""
      This heuristic finds if infinitesimals of the form `\eta = f(x)`, `\xi = g(y)`
      without making any assumptions on `h`.
  
      The complete sequence of steps is given in the paper mentioned below.
  
      References
      ==========
      - E.S. Cheb-Terrab, A.D. Roche, Symmetries and First Order
        ODE Patterns, pp. 10 - pp. 12
  
      """
      xieta = []
      h = match['h']
      hx = match['hx']
      hy = match['hy']
      func = match['func']
      hinv = match['hinv']
      x = func.args[0]
      y = match['y']
      xi = Function('xi')(x, func)
      eta = Function('eta')(x, func)
  
      C = S(0)
      A = hx.diff(y)
      B = hy.diff(y) + hy**2
      C = hx.diff(x) - hx**2
  
      if not (A and B and C):
          return
  
      Ax = A.diff(x)
      Ay = A.diff(y)
      Axy = Ax.diff(y)
      Axx = Ax.diff(x)
      Ayy = Ay.diff(y)
      D = simplify(2*Axy + hx*Ay - Ax*hy + (hx*hy + 2*A)*A)*A - 3*Ax*Ay
      if not D:
          E1 = simplify(3*Ax**2 + ((hx**2 + 2*C)*A - 2*Axx)*A)
          if E1:
              E2 = simplify((2*Ayy + (2*B - hy**2)*A)*A - 3*Ay**2)
              if not E2:
                  E3 = simplify(
                      E1*((28*Ax + 4*hx*A)*A**3 - E1*(hy*A + Ay)) - E1.diff(x)*8*A**4)
                  if not E3:
                      etaval = cancel((4*A**3*(Ax - hx*A) + E1*(hy*A - Ay))/(S(2)*A*E1))
                      if x not in etaval:
                          try:
                              etaval = exp(integrate(etaval, y))
                          except NotImplementedError:
                              pass
                          else:
                              xival = -4*A**3*etaval/E1
                              if y not in xival:
                                  return [{xi: xival, eta: etaval.subs(y, func)}]
  
      else:
          E1 = simplify((2*Ayy + (2*B - hy**2)*A)*A - 3*Ay**2)
          if E1:
              E2 = simplify(
                  4*A**3*D - D**2 + E1*((2*Axx - (hx**2 + 2*C)*A)*A - 3*Ax**2))
              if not E2:
                  E3 = simplify(
                     -(A*D)*E1.diff(y) + ((E1.diff(x) - hy*D)*A + 3*Ay*D +
                      (A*hx - 3*Ax)*E1)*E1)
                  if not E3:
                      etaval = cancel(((A*hx - Ax)*E1 - (Ay + A*hy)*D)/(S(2)*A*D))
                      if x not in etaval:
                          try:
                              etaval = exp(integrate(etaval, y))
                          except NotImplementedError:
                              pass
                          else:
                              xival = -E1*etaval/D
                              if y not in xival:
                                  return [{xi: xival, eta: etaval.subs(y, func)}]
  
  
  def lie_heuristic_linear(match, comp=False):
      r"""
      This heuristic assumes
  
      1. `\xi = ax + by + c` and
      2. `\eta = fx + gy + h`
  
      After substituting the following assumptions in the determining PDE, it
      reduces to
  
      .. math:: f + (g - a)h - bh^{2} - (ax + by + c)\frac{\partial h}{\partial x}
                   - (fx + gy + c)\frac{\partial h}{\partial y}
  
      Solving the reduced PDE obtained, using the method of characteristics, becomes
      impractical. The method followed is grouping similar terms and solving the system
      of linear equations obtained. The difference between the bivariate heuristic is that
      `h` need not be a rational function in this case.
  
      References
      ==========
      - E.S. Cheb-Terrab, A.D. Roche, Symmetries and First Order
        ODE Patterns, pp. 10 - pp. 12
  
      """
      xieta = []
      h = match['h']
      hx = match['hx']
      hy = match['hy']
      func = match['func']
      hinv = match['hinv']
      x = func.args[0]
      y = match['y']
      xi = Function('xi')(x, func)
      eta = Function('eta')(x, func)
  
      coeffdict = {}
      symbols = numbered_symbols("c", cls=Dummy)
      symlist = [next(symbols) for i in islice(symbols, 6)]
      C0, C1, C2, C3, C4, C5 = symlist
      pde = C3 + (C4 - C0)*h -(C0*x + C1*y + C2)*hx - (C3*x + C4*y + C5)*hy - C1*h**2
      pde, denom = pde.as_numer_denom()
      pde = powsimp(expand(pde))
      if pde.is_Add:
          terms = pde.args
          for term in terms:
              if term.is_Mul:
                  rem = Mul(*[m for m in term.args if not m.has(x, y)])
                  xypart = term/rem
                  if xypart not in coeffdict:
                      coeffdict[xypart] = rem
                  else:
                      coeffdict[xypart] += rem
              else:
                  if term not in coeffdict:
                      coeffdict[term] = S(1)
                  else:
                      coeffdict[term] += S(1)
  
      sollist = coeffdict.values()
      soldict = solve(sollist, symlist)
      if soldict:
          if isinstance(soldict, list):
              soldict = soldict[0]
          subval = soldict.values()
          if any(t for t in subval):
              onedict = dict(zip(symlist, [1]*6))
              xival = C0*x + C1*func + C2
              etaval = C3*x + C4*func + C5
              xival = xival.subs(soldict)
              etaval = etaval.subs(soldict)
              xival = xival.subs(onedict)
              etaval = etaval.subs(onedict)
              return [{xi: xival, eta: etaval}]
  