Solvers¶
The solvers module in SymPy implements methods for solving equations.
Note
It is recommended to use solveset()
to solve univariate equations,
sympy.solvers.solveset.linsolve()
to solve system of linear equations
instead of solve()
and sympy.solvers.solveset.nonlinsolve()
to
solve system of non linear equations since sooner or later the solveset
will take over solve
either internally or externally.
Algebraic equations¶
Use solve()
to solve algebraic equations. We suppose all equations are equaled to 0,
so solving x**2 == 1 translates into the following code:
>>> from sympy.solvers import solve
>>> from sympy import Symbol
>>> x = Symbol('x')
>>> solve(x**2  1, x)
[1, 1]
The first argument for solve()
is an equation (equaled to zero) and the second argument
is the symbol that we want to solve the equation for.

sympy.solvers.solvers.
solve
(f, *symbols, **flags)[source]¶ Algebraically solves equations and systems of equations.
 Currently supported are:
 polynomial,
 transcendental
 piecewise combinations of the above
 systems of linear and polynomial equations
 sytems containing relational expressions.
Input is formed as:
 f
 a single Expr or Poly that must be zero,
 an Equality
 a Relational expression or boolean
 iterable of one or more of the above
 symbols (object(s) to solve for) specified as
 none given (other nonnumeric objects will be used)
 single symbol
 denested list of symbols e.g. solve(f, x, y)
 ordered iterable of symbols e.g. solve(f, [x, y])
 flags
 ‘dict’=True (default is False)
return list (perhaps empty) of solution mappings
 ‘set’=True (default is False)
return list of symbols and set of tuple(s) of solution(s)
 ‘exclude=[] (default)’
don’t try to solve for any of the free symbols in exclude; if expressions are given, the free symbols in them will be extracted automatically.
 ‘check=True (default)’
If False, don’t do any testing of solutions. This can be useful if one wants to include solutions that make any denominator zero.
 ‘numerical=True (default)’
do a fast numerical check if
f
has only one symbol. ‘minimal=True (default is False)’
a very fast, minimal testing.
 ‘warn=True (default is False)’
show a warning if checksol() could not conclude.
 ‘simplify=True (default)’
simplify all but polynomials of order 3 or greater before returning them and (if check is not False) use the general simplify function on the solutions and the expression obtained when they are substituted into the function which should be zero
 ‘force=True (default is False)’
make positive all symbols without assumptions regarding sign.
 ‘rational=True (default)’
recast Floats as Rational; if this option is not used, the system containing floats may fail to solve because of issues with polys. If rational=None, Floats will be recast as rationals but the answer will be recast as Floats. If the flag is False then nothing will be done to the Floats.
 ‘manual=True (default is False)’
do not use the polys/matrix method to solve a system of equations, solve them one at a time as you might “manually”
 ‘implicit=True (default is False)’
allows solve to return a solution for a pattern in terms of other functions that contain that pattern; this is only needed if the pattern is inside of some invertible function like cos, exp, ....
 ‘particular=True (default is False)’
instructs solve to try to find a particular solution to a linear system with as many zeros as possible; this is very expensive
 ‘quick=True (default is False)’
when using particular=True, use a fast heuristic instead to find a solution with many zeros (instead of using the very slow method guaranteed to find the largest number of zeros possible)
 ‘cubics=True (default)’
return explicit solutions when cubic expressions are encountered
 ‘quartics=True (default)’
return explicit solutions when quartic expressions are encountered
 ‘quintics=True (default)’
return explicit solutions (if possible) when quintic expressions are encountered
Notes
solve() with check=True (default) will run through the symbol tags to elimate unwanted solutions. If no assumptions are included all possible solutions will be returned.
>>> from sympy import Symbol, solve >>> x = Symbol("x") >>> solve(x**2  1) [1, 1]
By using the positive tag only one solution will be returned:
>>> pos = Symbol("pos", positive=True) >>> solve(pos**2  1) [1]
Assumptions aren’t checked when \(solve()\) input involves relationals or bools.
When the solutions are checked, those that make any denominator zero are automatically excluded. If you do not want to exclude such solutions then use the check=False option:
>>> from sympy import sin, limit >>> solve(sin(x)/x) # 0 is excluded [pi]
If check=False then a solution to the numerator being zero is found: x = 0. In this case, this is a spurious solution since sin(x)/x has the well known limit (without dicontinuity) of 1 at x = 0:
>>> solve(sin(x)/x, check=False) [0, pi]
In the following case, however, the limit exists and is equal to the the value of x = 0 that is excluded when check=True:
>>> eq = x**2*(1/x  z**2/x) >>> solve(eq, x) [] >>> solve(eq, x, check=False) [0] >>> limit(eq, x, 0, '') 0 >>> limit(eq, x, 0, '+') 0
Examples
The output varies according to the input and can be seen by example:
>>> from sympy import solve, Poly, Eq, Function, exp >>> from sympy.abc import x, y, z, a, b >>> f = Function('f')
boolean or univariate Relational
>>> solve(x < 3) And(oo < x, x < 3)
to always get a list of solution mappings, use flag dict=True
>>> solve(x  3, dict=True) [{x: 3}] >>> solve([x  3, y  1], dict=True) [{x: 3, y: 1}]
to get a list of symbols and set of solution(s) use flag set=True
>>> solve([x**2  3, y  1], set=True) ([x, y], set([(sqrt(3), 1), (sqrt(3), 1)]))
single expression and single symbol that is in the expression
>>> solve(x  y, x) [y] >>> solve(x  3, x) [3] >>> solve(Eq(x, 3), x) [3] >>> solve(Poly(x  3), x) [3] >>> solve(x**2  y**2, x, set=True) ([x], set([(y,), (y,)])) >>> solve(x**4  1, x, set=True) ([x], set([(1,), (1,), (I,), (I,)]))
single expression with no symbol that is in the expression
>>> solve(3, x) [] >>> solve(x  3, y) []
single expression with no symbol given
In this case, all free symbols will be selected as potential symbols to solve for. If the equation is univariate then a list of solutions is returned; otherwise – as is the case when symbols are given as an iterable of length > 1 – a list of mappings will be returned.
>>> solve(x  3) [3] >>> solve(x**2  y**2) [{x: y}, {x: y}] >>> solve(z**2*x**2  z**2*y**2) [{x: y}, {x: y}, {z: 0}] >>> solve(z**2*x  z**2*y**2) [{x: y**2}, {z: 0}]
when an object other than a Symbol is given as a symbol, it is isolated algebraically and an implicit solution may be obtained. This is mostly provided as a convenience to save one from replacing the object with a Symbol and solving for that Symbol. It will only work if the specified object can be replaced with a Symbol using the subs method.
>>> solve(f(x)  x, f(x)) [x] >>> solve(f(x).diff(x)  f(x)  x, f(x).diff(x)) [x + f(x)] >>> solve(f(x).diff(x)  f(x)  x, f(x)) [x + Derivative(f(x), x)] >>> solve(x + exp(x)**2, exp(x), set=True) ([exp(x)], set([(sqrt(x),), (sqrt(x),)]))
>>> from sympy import Indexed, IndexedBase, Tuple, sqrt >>> A = IndexedBase('A') >>> eqs = Tuple(A[1] + A[2]  3, A[1]  A[2] + 1) >>> solve(eqs, eqs.atoms(Indexed)) {A[1]: 1, A[2]: 2}
To solve for a symbol implicitly, use ‘implicit=True’:
>>> solve(x + exp(x), x) [LambertW(1)] >>> solve(x + exp(x), x, implicit=True) [exp(x)]
It is possible to solve for anything that can be targeted with subs:
>>> solve(x + 2 + sqrt(3), x + 2) [sqrt(3)] >>> solve((x + 2 + sqrt(3), x + 4 + y), y, x + 2) {y: 2 + sqrt(3), x + 2: sqrt(3)}
Nothing heroic is done in this implicit solving so you may end up with a symbol still in the solution:
>>> eqs = (x*y + 3*y + sqrt(3), x + 4 + y) >>> solve(eqs, y, x + 2) {y: sqrt(3)/(x + 3), x + 2: (2*x  6 + sqrt(3))/(x + 3)} >>> solve(eqs, y*x, x) {x: y  4, x*y: 3*y  sqrt(3)}
if you attempt to solve for a number remember that the number you have obtained does not necessarily mean that the value is equivalent to the expression obtained:
>>> solve(sqrt(2)  1, 1) [sqrt(2)] >>> solve(x  y + 1, 1) # /!\ 1 is targeted, too [x/(y  1)] >>> [_.subs(z, 1) for _ in solve((x  y + 1).subs(1, z), 1)] [x + y]
To solve for a function within a derivative, use dsolve.
single expression and more than 1 symbol
when there is a linear solution
>>> solve(x  y**2, x, y) [{x: y**2}] >>> solve(x**2  y, x, y) [{y: x**2}]
when undetermined coefficients are identified
that are linear
>>> solve((a + b)*x  b + 2, a, b) {a: 2, b: 2}
that are nonlinear
>>> solve((a + b)*x  b**2 + 2, a, b, set=True) ([a, b], set([(sqrt(2), sqrt(2)), (sqrt(2), sqrt(2))]))
if there is no linear solution then the first successful attempt for a nonlinear solution will be returned
>>> solve(x**2  y**2, x, y) [{x: y}, {x: y}] >>> solve(x**2  y**2/exp(x), x, y) [{x: 2*LambertW(y/2)}] >>> solve(x**2  y**2/exp(x), y, x) [{y: x*sqrt(exp(x))}, {y: x*sqrt(exp(x))}]
iterable of one or more of the above
involving relationals or bools
>>> solve([x < 3, x  2]) Eq(x, 2) >>> solve([x > 3, x  2]) False
when the system is linear
with a solution
>>> solve([x  3], x) {x: 3} >>> solve((x + 5*y  2, 3*x + 6*y  15), x, y) {x: 3, y: 1} >>> solve((x + 5*y  2, 3*x + 6*y  15), x, y, z) {x: 3, y: 1} >>> solve((x + 5*y  2, 3*x + 6*y  z), z, x, y) {x: 5*y + 2, z: 21*y  6}
without a solution
>>> solve([x + 3, x  3]) []
when the system is not linear
>>> solve([x**2 + y 2, y**2  4], x, y, set=True) ([x, y], set([(2, 2), (0, 2), (2, 2)]))
if no symbols are given, all free symbols will be selected and a list of mappings returned
>>> solve([x  2, x**2 + y]) [{x: 2, y: 4}] >>> solve([x  2, x**2 + f(x)], set([f(x), x])) [{x: 2, f(x): 4}]
if any equation doesn’t depend on the symbol(s) given it will be eliminated from the equation set and an answer may be given implicitly in terms of variables that were not of interest
>>> solve([x  y, y  3], x) {x: y}
Disabling Highorder, Explicit Solutions
When solving polynomial expressions, one might not want explicit solutions (which can be quite long). If the expression is univariate, CRootOf instances will be returned instead:
>>> solve(x**3  x + 1) [1/((1/2  sqrt(3)*I/2)*(3*sqrt(69)/2 + 27/2)**(1/3))  (1/2  sqrt(3)*I/2)*(3*sqrt(69)/2 + 27/2)**(1/3)/3, (1/2 + sqrt(3)*I/2)*(3*sqrt(69)/2 + 27/2)**(1/3)/3  1/((1/2 + sqrt(3)*I/2)*(3*sqrt(69)/2 + 27/2)**(1/3)), (3*sqrt(69)/2 + 27/2)**(1/3)/3  1/(3*sqrt(69)/2 + 27/2)**(1/3)] >>> solve(x**3  x + 1, cubics=False) [CRootOf(x**3  x + 1, 0), CRootOf(x**3  x + 1, 1), CRootOf(x**3  x + 1, 2)]
If the expression is multivariate, no solution might be returned:
>>> solve(x**3  x + a, x, cubics=False) []
Sometimes solutions will be obtained even when a flag is False because the expression could be factored. In the following example, the equation can be factored as the product of a linear and a quadratic factor so explicit solutions (which did not require solving a cubic expression) are obtained:
>>> eq = x**3 + 3*x**2 + x  1 >>> solve(eq, cubics=False) [1, 1 + sqrt(2), sqrt(2)  1]
Solving Equations Involving Radicals
Because of SymPy’s use of the principle root (issue #8789), some solutions to radical equations will be missed unless check=False:
>>> from sympy import root >>> eq = root(x**3  3*x**2, 3) + 1  x >>> solve(eq) [] >>> solve(eq, check=False) [1/3]
In the above example there is only a single solution to the equation. Other expressions will yield spurious roots which must be checked manually; roots which give a negative argument to oddpowered radicals will also need special checking:
>>> from sympy import real_root, S >>> eq = root(x, 3)  root(x, 5) + S(1)/7 >>> solve(eq) # this gives 2 solutions but misses a 3rd [CRootOf(7*_p**5  7*_p**3 + 1, 1)**15, CRootOf(7*_p**5  7*_p**3 + 1, 2)**15] >>> sol = solve(eq, check=False) >>> [abs(eq.subs(x,i).n(2)) for i in sol] [0.48, 0.e110, 0.e110, 0.052, 0.052]
The first solution is negative so real_root must be used to see that it satisfies the expression:
>>> abs(real_root(eq.subs(x, sol[0])).n(2)) 0.e110
If the roots of the equation are not real then more care will be necessary to find the roots, especially for higher order equations. Consider the following expression:
>>> expr = root(x, 3)  root(x, 5)
We will construct a known value for this expression at x = 3 by selecting the 1th root for each radical:
>>> expr1 = root(x, 3, 1)  root(x, 5, 1) >>> v = expr1.subs(x, 3)
The solve function is unable to find any exact roots to this equation:
>>> eq = Eq(expr, v); eq1 = Eq(expr1, v) >>> solve(eq, check=False), solve(eq1, check=False) ([], [])
The function unrad, however, can be used to get a form of the equation for which numerical roots can be found:
>>> from sympy.solvers.solvers import unrad >>> from sympy import nroots >>> e, (p, cov) = unrad(eq) >>> pvals = nroots(e) >>> inversion = solve(cov, x)[0] >>> xvals = [inversion.subs(p, i) for i in pvals]
Although eq or eq1 could have been used to find xvals, the solution can only be verified with expr1:
>>> z = expr  v >>> [xi.n(chop=1e9) for xi in xvals if abs(z.subs(x, xi).n()) < 1e9] [] >>> z1 = expr1  v >>> [xi.n(chop=1e9) for xi in xvals if abs(z1.subs(x, xi).n()) < 1e9] [3.0]

sympy.solvers.solvers.
solve_linear
(lhs, rhs=0, symbols=[], exclude=[])[source]¶ Return a tuple derived from f = lhs  rhs that is one of the following:
(0, 1) meaning that
f
is independent of the symbols insymbols
that aren’t inexclude
, e.g:>>> from sympy.solvers.solvers import solve_linear >>> from sympy.abc import x, y, z >>> from sympy import cos, sin >>> eq = y*cos(x)**2 + y*sin(x)**2  y # = y*(1  1) = 0 >>> solve_linear(eq) (0, 1) >>> eq = cos(x)**2 + sin(x)**2 # = 1 >>> solve_linear(eq) (0, 1) >>> solve_linear(x, exclude=[x]) (0, 1)
(0, 0) meaning that there is no solution to the equation amongst the symbols given.
(If the first element of the tuple is not zero then the function is guaranteed to be dependent on a symbol insymbols
.)(symbol, solution) where symbol appears linearly in the numerator of
f
, is insymbols
(if given) and is not inexclude
(if given). No simplification is done tof
other than amul=True
expansion, so the solution will correspond strictly to a unique solution.(n, d)
wheren
andd
are the numerator and denominator off
when the numerator was not linear in any symbol of interest;n
will never be a symbol unless a solution for that symbol was found (in which case the second element is the solution, not the denominator).Examples
>>> from sympy.core.power import Pow >>> from sympy.polys.polytools import cancel
The variable
x
appears as a linear variable in each of the following:>>> solve_linear(x + y**2) (x, y**2) >>> solve_linear(1/x  y**2) (x, y**(2))
When not linear in x or y then the numerator and denominator are returned.
>>> solve_linear(x**2/y**2  3) (x**2  3*y**2, y**2)
If the numerator of the expression is a symbol then (0, 0) is returned if the solution for that symbol would have set any denominator to 0:
>>> eq = 1/(1/x  2) >>> eq.as_numer_denom() (x, 2*x + 1) >>> solve_linear(eq) (0, 0)
But automatic rewriting may cause a symbol in the denominator to appear in the numerator so a solution will be returned:
>>> (1/x)**1 x >>> solve_linear((1/x)**1) (x, 0)
Use an unevaluated expression to avoid this:
>>> solve_linear(Pow(1/x, 1, evaluate=False)) (0, 0)
If
x
is allowed to cancel in the following expression, then it appears to be linear inx
, but this sort of cancellation is not done bysolve_linear
so the solution will always satisfy the original expression without causing a division by zero error.>>> eq = x**2*(1/x  z**2/x) >>> solve_linear(cancel(eq)) (x, 0) >>> solve_linear(eq) (x**2*(z**2 + 1), x)
A list of symbols for which a solution is desired may be given:
>>> solve_linear(x + y + z, symbols=[y]) (y, x  z)
A list of symbols to ignore may also be given:
>>> solve_linear(x + y + z, exclude=[x]) (y, x  z)
(A solution for
y
is obtained because it is the first variable from the canonically sorted list of symbols that had a linear solution.)

sympy.solvers.solvers.
solve_linear_system
(system, *symbols, **flags)[source]¶ Solve system of N linear equations with M variables, which means both under and overdetermined systems are supported. The possible number of solutions is zero, one or infinite. Respectively, this procedure will return None or a dictionary with solutions. In the case of underdetermined systems, all arbitrary parameters are skipped. This may cause a situation in which an empty dictionary is returned. In that case, all symbols can be assigned arbitrary values.
Input to this functions is a Nx(M+1) matrix, which means it has to be in augmented form. If you prefer to enter N equations and M unknowns then use \(solve(Neqs, *Msymbols)\) instead. Note: a local copy of the matrix is made by this routine so the matrix that is passed will not be modified.
The algorithm used here is fractionfree Gaussian elimination, which results, after elimination, in an uppertriangular matrix. Then solutions are found using backsubstitution. This approach is more efficient and compact than the GaussJordan method.
>>> from sympy import Matrix, solve_linear_system >>> from sympy.abc import x, y
Solve the following system:
x + 4 y == 2 2 x + y == 14
>>> system = Matrix(( (1, 4, 2), (2, 1, 14))) >>> solve_linear_system(system, x, y) {x: 6, y: 2}
A degenerate system returns an empty dictionary.
>>> system = Matrix(( (0,0,0), (0,0,0) )) >>> solve_linear_system(system, x, y) {}

sympy.solvers.solvers.
solve_linear_system_LU
(matrix, syms)[source]¶ Solves the augmented matrix system using LUsolve and returns a dictionary in which solutions are keyed to the symbols of syms as ordered.
The matrix must be invertible.
See also
sympy.matrices.LUsolve
Examples
>>> from sympy import Matrix >>> from sympy.abc import x, y, z >>> from sympy.solvers.solvers import solve_linear_system_LU
>>> solve_linear_system_LU(Matrix([ ... [1, 2, 0, 1], ... [3, 2, 2, 1], ... [2, 0, 0, 1]]), [x, y, z]) {x: 1/2, y: 1/4, z: 1/2}

sympy.solvers.solvers.
solve_undetermined_coeffs
(equ, coeffs, sym, **flags)[source]¶ Solve equation of a type p(x; a_1, ..., a_k) == q(x) where both p, q are univariate polynomials and f depends on k parameters. The result of this functions is a dictionary with symbolic values of those parameters with respect to coefficients in q.
This functions accepts both Equations class instances and ordinary SymPy expressions. Specification of parameters and variable is obligatory for efficiency and simplicity reason.
>>> from sympy import Eq >>> from sympy.abc import a, b, c, x >>> from sympy.solvers import solve_undetermined_coeffs
>>> solve_undetermined_coeffs(Eq(2*a*x + a+b, x), [a, b], x) {a: 1/2, b: 1/2}
>>> solve_undetermined_coeffs(Eq(a*c*x + a+b, x), [a, b], x) {a: 1/c, b: 1/c}

sympy.solvers.solvers.
nsolve
(*args, **kwargs)[source]¶ Solve a nonlinear equation system numerically:
nsolve(f, [args,] x0, modules=['mpmath'], **kwargs)
f is a vector function of symbolic expressions representing the system. args are the variables. If there is only one variable, this argument can be omitted. x0 is a starting vector close to a solution.
Use the modules keyword to specify which modules should be used to evaluate the function and the Jacobian matrix. Make sure to use a module that supports matrices. For more information on the syntax, please see the docstring of lambdify.
Overdetermined systems are supported.
>>> from sympy import Symbol, nsolve >>> import sympy >>> import mpmath >>> mpmath.mp.dps = 15 >>> x1 = Symbol('x1') >>> x2 = Symbol('x2') >>> f1 = 3 * x1**2  2 * x2**2  1 >>> f2 = x1**2  2 * x1 + x2**2 + 2 * x2  8 >>> print(nsolve((f1, f2), (x1, x2), (1, 1))) Matrix([[1.19287309935246], [1.27844411169911]])
For onedimensional functions the syntax is simplified:
>>> from sympy import sin, nsolve >>> from sympy.abc import x >>> nsolve(sin(x), x, 2) 3.14159265358979 >>> nsolve(sin(x), 2) 3.14159265358979
To solve with higher precision than the default, use the prec argument.
>>> from sympy import cos >>> nsolve(cos(x)  x, 1) 0.739085133215161 >>> nsolve(cos(x)  x, 1, prec=50) 0.73908513321516064165531208767387340401341175890076 >>> cos(_) 0.73908513321516064165531208767387340401341175890076
mpmath.findroot is used, you can find there more extensive documentation, especially concerning keyword parameters and available solvers. Note, however, that this routine works only with the numerator of the function in the onedimensional case, and for very steep functions near the root this may lead to a failure in the verification of the root. In this case you should use the flag \(verify=False\) and independently verify the solution.
>>> from sympy import cos, cosh >>> from sympy.abc import i >>> f = cos(x)*cosh(x)  1 >>> nsolve(f, 3.14*100) Traceback (most recent call last): ... ValueError: Could not find root within given tolerance. (1.39267e+230 > 2.1684e19) >>> ans = nsolve(f, 3.14*100, verify=False); ans 312.588469032184 >>> f.subs(x, ans).n(2) 2.1e+121 >>> (f/f.diff(x)).subs(x, ans).n(2) 7.4e15
One might safely skip the verification if bounds of the root are known and a bisection method is used:
>>> bounds = lambda i: (3.14*i, 3.14*(i + 1)) >>> nsolve(f, bounds(100), solver='bisect', verify=False) 315.730061685774

sympy.solvers.solvers.
check_assumptions
(expr, against=None, **assumptions)[source]¶ Checks whether expression \(expr\) satisfies all assumptions.
\(assumptions\) is a dict of assumptions: {‘assumption’: TrueFalse, ...}.
Examples
>>> from sympy import Symbol, pi, I, exp, check_assumptions
>>> check_assumptions(5, integer=True) True >>> check_assumptions(pi, real=True, integer=False) True >>> check_assumptions(pi, real=True, negative=True) False >>> check_assumptions(exp(I*pi/7), real=False) True
>>> x = Symbol('x', real=True, positive=True) >>> check_assumptions(2*x + 1, real=True, positive=True) True >>> check_assumptions(2*x  5, real=True, positive=True) False
To check assumptions of
expr
against another variable or expression, pass the expression or variable asagainst
.>>> check_assumptions(2*x + 1, x) True
\(None\) is returned if check_assumptions() could not conclude.
>>> check_assumptions(2*x  1, real=True, positive=True) >>> z = Symbol('z') >>> check_assumptions(z, real=True)

sympy.solvers.solvers.
checksol
(f, symbol, sol=None, **flags)[source]¶ Checks whether sol is a solution of equation f == 0.
Input can be either a single symbol and corresponding value or a dictionary of symbols and values. When given as a dictionary and flag
simplify=True
, the values in the dictionary will be simplified.f
can be a single equation or an iterable of equations. A solution must satisfy all equations inf
to be considered valid; if a solution does not satisfy any equation, False is returned; if one or more checks are inconclusive (and none are False) then None is returned.Examples
>>> from sympy import symbols >>> from sympy.solvers import checksol >>> x, y = symbols('x,y') >>> checksol(x**4  1, x, 1) True >>> checksol(x**4  1, x, 0) False >>> checksol(x**2 + y**2  5**2, {x: 3, y: 4}) True
To check if an expression is zero using checksol, pass it as
f
and send an empty dictionary forsymbol
:>>> checksol(x**2 + x  x*(x + 1), {}) True
None is returned if checksol() could not conclude.
 flags:
 ‘numerical=True (default)’
 do a fast numerical check if
f
has only one symbol.  ‘minimal=True (default is False)’
 a very fast, minimal testing.
 ‘warn=True (default is False)’
 show a warning if checksol() could not conclude.
 ‘simplify=True (default)’
 simplify solution before substituting into function and simplify the function before trying specific simplifications
 ‘force=True (default is False)’
 make positive all symbols without assumptions regarding sign.
Deutils (Utilities for solving ODE’s and PDE’s)¶

sympy.solvers.deutils.
ode_order
(expr, func)[source]¶ Returns the order of a given differential equation with respect to func.
This function is implemented recursively.
Examples
>>> from sympy import Function >>> from sympy.solvers.deutils import ode_order >>> from sympy.abc import x >>> f, g = map(Function, ['f', 'g']) >>> ode_order(f(x).diff(x, 2) + f(x).diff(x)**2 + ... f(x).diff(x), f(x)) 2 >>> ode_order(f(x).diff(x, 2) + g(x).diff(x, 3), f(x)) 2 >>> ode_order(f(x).diff(x, 2) + g(x).diff(x, 3), g(x)) 3
Recurrence Equations¶

sympy.solvers.recurr.
rsolve
(f, y, init=None)[source]¶ Solve univariate recurrence with rational coefficients.
Given \(k\)th order linear recurrence \(\operatorname{L} y = f\), or equivalently:
\[a_{k}(n) y(n+k) + a_{k1}(n) y(n+k1) + \cdots + a_{0}(n) y(n) = f(n)\]where \(a_{i}(n)\), for \(i=0, \ldots, k\), are polynomials or rational functions in \(n\), and \(f\) is a hypergeometric function or a sum of a fixed number of pairwise dissimilar hypergeometric terms in \(n\), finds all solutions or returns
None
, if none were found.Initial conditions can be given as a dictionary in two forms:
{ n_0 : v_0, n_1 : v_1, ..., n_m : v_m }
{ y(n_0) : v_0, y(n_1) : v_1, ..., y(n_m) : v_m }
or as a list
L
of values:L = [ v_0, v_1, ..., v_m ]
where
L[i] = v_i
, for \(i=0, \ldots, m\), maps to \(y(n_i)\).See also
Examples
Lets consider the following recurrence:
\[(n  1) y(n + 2)  (n^2 + 3 n  2) y(n + 1) + 2 n (n + 1) y(n) = 0\]>>> from sympy import Function, rsolve >>> from sympy.abc import n >>> y = Function('y')
>>> f = (n  1)*y(n + 2)  (n**2 + 3*n  2)*y(n + 1) + 2*n*(n + 1)*y(n)
>>> rsolve(f, y(n)) 2**n*C0 + C1*factorial(n)
>>> rsolve(f, y(n), { y(0):0, y(1):3 }) 3*2**n  3*factorial(n)

sympy.solvers.recurr.
rsolve_poly
(coeffs, f, n, **hints)[source]¶ Given linear recurrence operator \(\operatorname{L}\) of order \(k\) with polynomial coefficients and inhomogeneous equation \(\operatorname{L} y = f\), where \(f\) is a polynomial, we seek for all polynomial solutions over field \(K\) of characteristic zero.
The algorithm performs two basic steps:
 Compute degree \(N\) of the general polynomial solution.
 Find all polynomials of degree \(N\) or less of \(\operatorname{L} y = f\).
There are two methods for computing the polynomial solutions. If the degree bound is relatively small, i.e. it’s smaller than or equal to the order of the recurrence, then naive method of undetermined coefficients is being used. This gives system of algebraic equations with \(N+1\) unknowns.
In the other case, the algorithm performs transformation of the initial equation to an equivalent one, for which the system of algebraic equations has only \(r\) indeterminates. This method is quite sophisticated (in comparison with the naive one) and was invented together by Abramov, Bronstein and Petkovsek.
It is possible to generalize the algorithm implemented here to the case of linear qdifference and differential equations.
Lets say that we would like to compute \(m\)th Bernoulli polynomial up to a constant. For this we can use \(b(n+1)  b(n) = m n^{m1}\) recurrence, which has solution \(b(n) = B_m + C\). For example:
>>> from sympy import Symbol, rsolve_poly >>> n = Symbol('n', integer=True)
>>> rsolve_poly([1, 1], 4*n**3, n) C0 + n**4  2*n**3 + n**2
References
[R487] S. A. Abramov, M. Bronstein and M. Petkovsek, On polynomial solutions of linear operator equations, in: T. Levelt, ed., Proc. ISSAC ‘95, ACM Press, New York, 1995, 290296. [R488] M. Petkovsek, Hypergeometric solutions of linear recurrences with polynomial coefficients, J. Symbolic Computation, 14 (1992), 243264. [R489]  Petkovsek, H. S. Wilf, D. Zeilberger, A = B, 1996.

sympy.solvers.recurr.
rsolve_ratio
(coeffs, f, n, **hints)[source]¶ Given linear recurrence operator \(\operatorname{L}\) of order \(k\) with polynomial coefficients and inhomogeneous equation \(\operatorname{L} y = f\), where \(f\) is a polynomial, we seek for all rational solutions over field \(K\) of characteristic zero.
This procedure accepts only polynomials, however if you are interested in solving recurrence with rational coefficients then use
rsolve
which will preprocess the given equation and run this procedure with polynomial arguments.The algorithm performs two basic steps:
 Compute polynomial \(v(n)\) which can be used as universal denominator of any rational solution of equation \(\operatorname{L} y = f\).
 Construct new linear difference equation by substitution
\(y(n) = u(n)/v(n)\) and solve it for \(u(n)\) finding all its
polynomial solutions. Return
None
if none were found.
Algorithm implemented here is a revised version of the original Abramov’s algorithm, developed in 1989. The new approach is much simpler to implement and has better overall efficiency. This method can be easily adapted to qdifference equations case.
Besides finding rational solutions alone, this functions is an important part of Hyper algorithm were it is used to find particular solution of inhomogeneous part of a recurrence.
See also
References
[R490] S. A. Abramov, Rational solutions of linear difference and qdifference equations with polynomial coefficients, in: T. Levelt, ed., Proc. ISSAC ‘95, ACM Press, New York, 1995, 285289 Examples
>>> from sympy.abc import x >>> from sympy.solvers.recurr import rsolve_ratio >>> rsolve_ratio([2*x**3 + x**2 + 2*x  1, 2*x**3 + x**2  6*x, ...  2*x**3  11*x**2  18*x  9, 2*x**3 + 13*x**2 + 22*x + 8], 0, x) C2*(2*x  3)/(2*(x**2  1))

sympy.solvers.recurr.
rsolve_hyper
(coeffs, f, n, **hints)[source]¶ Given linear recurrence operator \(\operatorname{L}\) of order \(k\) with polynomial coefficients and inhomogeneous equation \(\operatorname{L} y = f\) we seek for all hypergeometric solutions over field \(K\) of characteristic zero.
The inhomogeneous part can be either hypergeometric or a sum of a fixed number of pairwise dissimilar hypergeometric terms.
The algorithm performs three basic steps:
 Group together similar hypergeometric terms in the inhomogeneous part of \(\operatorname{L} y = f\), and find particular solution using Abramov’s algorithm.
 Compute generating set of \(\operatorname{L}\) and find basis in it, so that all solutions are linearly independent.
 Form final solution with the number of arbitrary constants equal to dimension of basis of \(\operatorname{L}\).
Term \(a(n)\) is hypergeometric if it is annihilated by first order linear difference equations with polynomial coefficients or, in simpler words, if consecutive term ratio is a rational function.
The output of this procedure is a linear combination of fixed number of hypergeometric terms. However the underlying method can generate larger class of solutions  D’Alembertian terms.
Note also that this method not only computes the kernel of the inhomogeneous equation, but also reduces in to a basis so that solutions generated by this procedure are linearly independent
References
[R491] M. Petkovsek, Hypergeometric solutions of linear recurrences with polynomial coefficients, J. Symbolic Computation, 14 (1992), 243264. [R492]  Petkovsek, H. S. Wilf, D. Zeilberger, A = B, 1996.
Examples
>>> from sympy.solvers import rsolve_hyper >>> from sympy.abc import x
>>> rsolve_hyper([1, 1, 1], 0, x) C0*(1/2 + sqrt(5)/2)**x + C1*(sqrt(5)/2 + 1/2)**x
>>> rsolve_hyper([1, 1], 1 + x, x) C0 + x*(x + 1)/2
Systems of Polynomial Equations¶

sympy.solvers.polysys.
solve_poly_system
(seq, *gens, **args)[source]¶ Solve a system of polynomial equations.
Examples
>>> from sympy import solve_poly_system >>> from sympy.abc import x, y
>>> solve_poly_system([x*y  2*y, 2*y**2  x**2], x, y) [(0, 0), (2, sqrt(2)), (2, sqrt(2))]

sympy.solvers.polysys.
solve_triangulated
(polys, *gens, **args)[source]¶ Solve a polynomial system using GianniKalkbrenner algorithm.
The algorithm proceeds by computing one Groebner basis in the ground domain and then by iteratively computing polynomial factorizations in appropriately constructed algebraic extensions of the ground domain.
References
1. Patrizia Gianni, Teo Mora, Algebraic Solution of System of Polynomial Equations using Groebner Bases, AAECC5 on Applied Algebra, Algebraic Algorithms and ErrorCorrecting Codes, LNCS 356 247–257, 1989
Examples
>>> from sympy.solvers.polysys import solve_triangulated >>> from sympy.abc import x, y, z
>>> F = [x**2 + y + z  1, x + y**2 + z  1, x + y + z**2  1]
>>> solve_triangulated(F, x, y, z) [(0, 0, 1), (0, 1, 0), (1, 0, 0)]
Diophantine Equations (DEs)¶
See Diophantine