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

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 is 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

```
>>> solve(Eq(x**2, 1), x)
[-1, 1]
>>> solve(Eq(x**2 - 1, 0), x)
[-1, 1]
>>> solve(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 `solve(Eq(expr, 0), x)`, you can just use
`solve(expr, x)`.

The main function for solving algebraic equations, as we saw above, is
`solve`. The syntax is `solve(equations, variables)`, where, as we saw
above, `equations` may be in the form of `Eq` instances or expressions
that are assumed to be equal to zero.

When solving a single equation, the output of `solve` is a list of the
solutions.

```
>>> solve(x**2 - x, x)
[0, 1]
```

If no solutions are found, an empty list is returned, or
`NotImplementedError` is raised.

```
>>> solve(exp(x), x)
[]
```

Note

If `solve` returns `[]` or raises `NotImplementedError`, it doesn’t
mean that the equation has no solutions. It just means that it couldn’t
find any. Often this means that the solutions cannot be represented
symbolically. For example, the equation \(x = \cos(x)\) has a solution, but
it cannot be represented symbolically using standard functions.

```
>>> solve(x - cos(x), x)
Traceback (most recent call last):
...
NotImplementedError: multiple generators [x, exp(I*x)]
No algorithms are implemented to solve equation exp(I*x)
```

In fact, `solve` makes *no guarantees whatsoever* about the completeness
of the solutions it finds. Much of `solve` is heuristics, which may find
some solutions to an equation or system of equations, but not all of them.

`solve` can also solve systems of equations. Pass a list of equations and a
list of variables to solve for.

```
>>> solve([x - y + 2, x + y - 3], [x, y])
{x: 1/2, y: 5/2}
>>> solve([x*y - 7, x + y - 6], [x, y])
⎡⎛ ___ ___ ⎞ ⎛ ___ ___ ⎞⎤
⎣⎝- ╲╱ 2 + 3, ╲╱ 2 + 3⎠, ⎝╲╱ 2 + 3, - ╲╱ 2 + 3⎠⎦
```

Note

The type of the output of `solve` when solving systems of equations
varies depending on the type of the input. If you want a consistent
interface, pass `dict=True`.

```
>>> solve([x - y + 2, x + y - 3], [x, y], dict=True)
[{x: 1/2, y: 5/2}]
>>> solve([x*y - 7, x + y - 6], [x, y], dict=True)
⎡⎧ ___ ___ ⎫ ⎧ ___ ___ ⎫⎤
⎢⎨x: - ╲╱ 2 + 3, y: ╲╱ 2 + 3⎬, ⎨x: ╲╱ 2 + 3, y: - ╲╱ 2 + 3⎬⎥
⎣⎩ ⎭ ⎩ ⎭⎦
```

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

```
>>> solve(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.

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))), f(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.