# Ordinary Differential Equations/Laplace Transform

## Definition

Let $f(t)$  be a function on $[0,\infty )$ . The Laplace transform of $f$  is defined by the integral

$F(s)={\mathcal {L}}\{f\}(s)=\int _{0}^{\infty }e^{-st}f(t)dt\,.$

The domain of $F(s)$  is all values of $s$  such that the integral exists.

## Properties

### Linearity

Let $f$  and $g$  be functions whose Laplace transforms exist for $s>\alpha$  and let $a$  and $b$  be constants. Then, for $s>\alpha$ ,

${\mathcal {L}}\{af+bg\}=a{\mathcal {L}}\{f\}+b{\mathcal {L}}\{g\}\,,$

which can be proved using the properties of improper integrals.

### Shifting in s

If the Laplace transform ${\mathcal {L}}\{f\}(s)=F(s)$  exists for $s>\alpha$ , then

${\mathcal {L}}\{e^{at}f(t)\}(s)=F(s-a)\,$

for $s>\alpha +a$ .

Proof.

{\begin{aligned}{\mathcal {L}}\{e^{at}f(t)\}(s)&{}=\int _{0}^{\infty }e^{-st}e^{at}f(t)dt\\&{}=\int _{0}^{\infty }e^{-(s-a)t}f(t)dt\\&{}=F(s-a)\,.\end{aligned}}

### Laplace Transform of Higher-Order Derivatives

If $F(s)={\mathcal {L}}\{f(t)\}$ , then ${\mathcal {L}}\{f'(t)\}=sF(s)-f(0)$

Proof:
${\mathcal {L}}\{f'(t)\}=\int _{0}^{\infty }f'(t)e^{-st}dt$
$=\lim _{C\to \infty }\int _{0}^{C}f'(t)e^{-st}dt$
$=\lim _{C\to \infty }\left.e^{-st}f(t)\right|_{0}^{C}-\int _{0}^{C}-sf(t)e^{-st}dt$  (integrating by parts)
$=-f(0)+s\lim _{C\to \infty }\int _{0}^{C}f(t)e^{-st}dt$
$=s{\mathcal {L}}\{f(t)\}-f(0)$
$=sF(s)-f(0)\,$

Using the above and the linearity of Laplace Transforms, it is easy to prove that ${\mathcal {L}}\{f''(t)\}=s^{2}F(s)-sf(0)-f'(0)$

### Derivatives of the Laplace Transform

If ${\mathcal {L}}\{f(t)\}=F(s)$ , then ${\mathcal {L}}\{tf(t)\}=-F'(s)$

## Laplace Transform of Few Simple Functions

1. ${\mathcal {L}}\{1\}={1 \over s}$
2. ${\mathcal {L}}\{e^{at}\}={1 \over s-a}$
3. ${\mathcal {L}}\{\cos \omega t\}={s \over s^{2}+\omega ^{2}}$
4. ${\mathcal {L}}\{\sin \omega t\}={\omega \over s^{2}+\omega ^{2}}$
5. ${\mathcal {L}}\{1\}={1 \over s}$
6. ${\mathcal {L}}\{t^{n}\}={n! \over s^{n+1}}$