Laplace Transforms

The Laplace transform converts a differential equation into an algebraic equation, making it easier to solve.

Definition

$\mathcal{L}\{f(t)\} = F(s) = \int_0^\infty e^{-st}f(t)\,dt$

Essential transform pairs

$f(t)$	$F(s)$
$1$	$1/s$
$t^n$	$n!/s^{n+1}$
$e^{at}$	$1/(s-a)$
$\sin(\omega t)$	$\omega/(s^{2}+\omega^{2})$
$\cos(\omega t)$	$s/(s^{2}+\omega^{2})$

Key properties

Linearity: $\mathcal{L}\{af+bg\} = aF + bG$
Derivative: $\mathcal{L}\{f'\} = sF(s) - f(0)$
Second derivative: $\mathcal{L}\{f''\} = s^{2}F(s) - sf(0) - f'(0)$

Solving an ODE — Example: $y'' + 4y = 0$ , $y(0)=1$ , $y'(0)=0$

Transform: $s^{2}Y - s + 4Y = 0$
Solve: $Y = \dfrac{s}{s^{2}+4}$
Inverse: $y = \cos(2t)$

Partial fractions are used to invert complex $F(s)$ expressions back to $f(t)$ .

Physics link: Laplace transforms are the standard tool for analysing linear circuits and control systems, where initial conditions appear naturally in the transformed equation.

Common pitfall: Laplace transforms absorb initial conditions at the very first step, via $\mathcal{L}\{y'\} = sY - y(0)$ . Bolting the conditions on at the end, ODE-style, double-counts or drops them.

Placeholder: Laplace Transforms

Practise this lesson

Differential Equations and Vector Calculus