# Ordinary Differential Equations/Laplace Transform

## Definition

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

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

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

## Properties

### Linearity

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

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

which can be proved using the properties of improper integrals.

### Shifting in s

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

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

for ${\displaystyle s>\alpha +a}$.

Proof.

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

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

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

### Derivatives of the Laplace Transform

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

## Laplace Transform of Few Simple Functions

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