Solvers

>>> from sympy import *
>>> x, y, z = symbols('x y z')
>>> init_printing(use_unicode=True)

A Note about Equations

Recall from the gotchas section of this tutorial that symbolic equations in SymPy are not represented by = or ==, but by Eq.

>>> Eq(x, y)
x = y

However, there is an even easier way. In SymPy, any expression not in an Eq is automatically assumed to equal 0 by the solving functions. Since \(a = b\) if and only if \(a - b = 0\), this means that instead of using x == y, you can just use x - y. For example

>>> solveset(Eq(x**2, 1), x)
{-1, 1}
>>> solveset(Eq(x**2 - 1, 0), x)
{-1, 1}
>>> solveset(x**2 - 1, x)
{-1, 1}

This is particularly useful if the equation you wish to solve is already equal to 0. Instead of typing solveset(Eq(expr, 0), x), you can just use solveset(expr, x).

Solving Equations Algebraically

The main function for solving algebraic equations is solveset. The syntax for solveset is solveset(equation, variable=None, domain=S.Complexes) Where equations may be in the form of Eq instances or expressions that are assumed to be equal to zero.

Please note that there is another function called solve which can also be used to solve equations. The syntax is solve(equations, variables) However, it is recommended to use solveset instead.

When solving a single equation, the output of solveset is a FiniteSet or an Interval or ImageSet of the solutions.

>>> solveset(x**2 - x, x)
{0, 1}
>>> solveset(x - x, x, domain=S.Reals)
ℝ
>>> solveset(sin(x) - 1, x, domain=S.Reals)
⎧        π │      ⎫
⎨2⋅n⋅π + ─ │ n ∊ ℤ⎬
⎩        2 │      ⎭

If there are no solutions, an EmptySet is returned and if it is not able to find solutions then a ConditionSet is returned.

>>> solveset(exp(x), x)     # No solution exists
∅
>>> solveset(cos(x) - x, x)  # Not able to find solution
{x │ x ∊ ℂ ∧ (-x + cos(x) = 0)}

In the solveset module, the linear system of equations is solved using linsolve. In future we would be able to use linsolve directly from solveset. Following is an example of the syntax of linsolve.

• List of Equations Form:

  >>> linsolve([x + y + z - 1, x + y + 2*z - 3 ], (x, y, z))
  {(-y - 1, y, 2)}
  

• Augmented Matrix Form:

  >>> linsolve(Matrix(([1, 1, 1, 1], [1, 1, 2, 3])), (x, y, z))
  {(-y - 1, y, 2)}
  

• A*x = b Form

  >>> M = Matrix(((1, 1, 1, 1), (1, 1, 2, 3)))
  >>> system = A, b = M[:, :-1], M[:, -1]
  >>> linsolve(system, x, y, z)
  {(-y - 1, y, 2)}
  

Note

The order of solution corresponds the order of given symbols.

In the solveset module, the non linear system of equations is solved using nonlinsolve. Following are examples of nonlinsolve.

1. When only real solution is present:

   >>> a, b, c, d = symbols('a, b, c, d', real=True)
   >>> nonlinsolve([a**2 + a, a - b], [a, b])
   {(-1, -1), (0, 0)}
   >>> nonlinsolve([x*y - 1, x - 2], x, y)
   {(2, 1/2)}
   

2. When only complex solution is present:

   >>> nonlinsolve([x**2 + 1, y**2 + 1], [x, y])
   {(-ⅈ, -ⅈ), (-ⅈ, ⅈ), (ⅈ, -ⅈ), (ⅈ, ⅈ)}
   

3. When both real and complex solution are present:

   >>> from sympy import sqrt
   >>> system = [x**2 - 2*y**2 -2, x*y - 2]
   >>> vars = [x, y]
   >>> nonlinsolve(system, vars)
   {(-2, -1), (2, 1), (-√2⋅ⅈ, √2⋅ⅈ), (√2⋅ⅈ, -√2⋅ⅈ)}
   

   >>> system = [exp(x) - sin(y), 1/y - 3]
   >>> nonlinsolve(system, vars)
   {({2⋅n⋅ⅈ⋅π + log(sin(1/3)) │ n ∊ ℤ}, 1/3)}
   

4. When the system is positive-dimensional system (has infinitely many solutions):

   >>> nonlinsolve([x*y, x*y - x], [x, y])
   {(0, y)}
   

   >>> system = [a**2 + a*c, a - b]
   >>> nonlinsolve(system, [a, b])
   {(0, 0), (-c, -c)}
   

Note

1. The order of solution corresponds the order of given symbols.

2. Currently nonlinsolve doesn’t return solution in form of LambertW (if there is solution present in the form of LambertW).

solve can be used for such cases:

>>> solve([x**2 - y**2/exp(x)], [x, y], dict=True)
⎡⎧         ____⎫  ⎧        ____⎫⎤
⎢⎨        ╱  x ⎬  ⎨       ╱  x ⎬⎥
⎣⎩y: -x⋅╲╱  ℯ  ⎭, ⎩y: x⋅╲╱  ℯ  ⎭⎦
>>> solve(x**2 - y**2/exp(x), x, dict=True)
⎡⎧      ⎛-y ⎞⎫  ⎧      ⎛y⎞⎫⎤
⎢⎨x: 2⋅W⎜───⎟⎬, ⎨x: 2⋅W⎜─⎟⎬⎥
⎣⎩      ⎝ 2 ⎠⎭  ⎩      ⎝2⎠⎭⎦

3. Currently nonlinsolve is not properly capable of solving the system of equations having trigonometric functions.

solve can be used for such cases (but does not give all solution):

>>> solve([sin(x + y), cos(x - y)], [x, y])
⎡⎛-3⋅π   3⋅π⎞  ⎛-π   π⎞  ⎛π  3⋅π⎞  ⎛3⋅π  π⎞⎤
⎢⎜─────, ───⎟, ⎜───, ─⎟, ⎜─, ───⎟, ⎜───, ─⎟⎥
⎣⎝  4     4 ⎠  ⎝ 4   4⎠  ⎝4   4 ⎠  ⎝ 4   4⎠⎦

solveset reports each solution only once. To get the solutions of a polynomial including multiplicity use roots.

>>> solveset(x**3 - 6*x**2 + 9*x, x)
{0, 3}
>>> roots(x**3 - 6*x**2 + 9*x, x)
{0: 1, 3: 2}

The output {0: 1, 3: 2} of roots means that 0 is a root of multiplicity 1 and 3 is a root of multiplicity 2.

Note

Currently solveset is not capable of solving the following types of equations:

• Equations solvable by LambertW (Transcendental equation solver).

solve can be used for such cases:

>>> solve(x*exp(x) - 1, x )
[W(1)]

Solving Differential Equations

To solve differential equations, use dsolve. First, create an undefined function by passing cls=Function to the symbols function.

>>> f, g = symbols('f g', cls=Function)

f and g are now undefined functions. We can call f(x), and it will represent an unknown function.

>>> f(x)
f(x)

Derivatives of f(x) are unevaluated.

>>> f(x).diff(x)
d
──(f(x))
dx

(see the Derivatives section for more on derivatives).

To represent the differential equation \(f''(x) - 2f'(x) + f(x) = \sin(x)\), we would thus use

>>> diffeq = Eq(f(x).diff(x, x) - 2*f(x).diff(x) + f(x), sin(x))
>>> diffeq
                      2
         d           d
f(x) - 2⋅──(f(x)) + ───(f(x)) = sin(x)
         dx           2
                    dx

To solve the ODE, pass it and the function to solve for to dsolve.

>>> dsolve(diffeq, f(x))
                    x   cos(x)
f(x) = (C₁ + C₂⋅x)⋅ℯ  + ──────
                          2

dsolve returns an instance of Eq. This is because in general, solutions to differential equations cannot be solved explicitly for the function.

>>> dsolve(f(x).diff(x)*(1 - sin(f(x))) - 1, f(x))
x - f(x) - cos(f(x)) = C₁

The arbitrary constants in the solutions from dsolve are symbols of the form C1, C2, C3, and so on.