Laplace transform and inverse Laplace transform: laplace ilaplace invlaplace

### 5.57.2 Laplace transform and inverse Laplace transform: laplace ilaplace invlaplace

Denoting by L the Laplace transform, you get the following:

L(y)(x) | = |
∫ |
+∞
0
| e−x uy(u)du
L−1(g)(x) | = |
1
2iπ
| ∫ |
C
| ez xg(z)dz

where C is a closed contour enclosing the poles of g.

The laplace command finds the Laplace transform of a function.

- laplace takes one mandatory argument and two optional arguments:
  - expr, an expression involving a variable.
  - Optionally, x, the variable name (by default x).
  - Optionally, s, a variable for the output (by default x).
- laplace(expr ⟨ x, ⟩) returns the Laplace transform of expr.

#### Examples.

- Input:
  laplace(sin(x))
  Output:

  1
  x2+1

- Input:
  laplace(sin(t),t)
  Output:

  1
  t2+1

- Input:
  laplace(sin(t),t,s)
  Output:

  1
  s2+1

The ilaplace command finds the Laplace transform of a function.
invlaplace is a synonym for ilaplace.

- ilaplace takes one mandatory argument and two optional arguments:
  - expr, an expression involving a variable.
  - Optionally, x, the variable name (by default x).
  - Optionally, s, a variable for the output (by default x).
- ilaplace(expr ⟨ x, ⟩) returns the inverse Laplace transform of expr.

The Laplace transform has the following properties:

L(y′)(x) | = | −y(0)+xL(y)(x)
L(y″)(x) | = | −y′(0)+xL(y′)(x)
| = | −y′(0)−xy(0)+x2L(y)(x)

These properties make the Laplace transform and inverse Laplace transform useful for solving linear differential equations with constant coefficients. For example, suppose you have

| y′′ +p y′ +q y = f(x) |  |  |  |  |  |  |  |  |
| y(0)=a, y′(0)=b |  |  |  |  |  |  |  |  |

then

L(f)(x) | = | L(y″+py′+qy)(x)
| = | −y′(0)−x y(0)+x2 L(y)(x)−p y(0)+p x L(y)(x))+q L(y)(x)
| = | (x2+p x+q) L(y)(x)−y′(0)−(x+p) y(0)

Therefore, if a=y(0) and b=y′(0), you get

L(f)(x)=(x2+p x+q)L(y)(x)−(x+p) a−b

and the solution of the differential equation is:

y(x)= L−1((L(f)(x)+(x+p) a +b)/(x2+p x+q))

#### Example.

Solve:

y′′ −6 y′+9 y = x e3 x, y(0)=c_0, y′(0)=c_1

Here, p=−6, q=9.
Input:

laplace(x*exp(3*x))

Output:

1
x2−6 x+9

Input:

ilaplace((1/(x^2-6*x+9)+(x-6)*c_0+c_1)/(x^2-6*x+9))

Output:

1
6
|  | ⎛ ⎝ | x3−18 x c0+6 x c1+6 c0 | ⎞ ⎠ | e3 x

Note that this equation could be solved directly.
Input:

desolve(y’’-6*y’+9*y=x*exp(3*x),y)

Output:

e3 x | ⎛ ⎝ | c0 x+c1 | ⎞ ⎠ | + |
1
6
| x3 e3 x

You also can use the addtable command Laplacians of unspecified functions (see Section 5.26.2).