Laplace transform applied to differential equations

The use of Laplace transform makes it much easier to solve linear differential equations with given initial conditions.

First consider the following relations:

$\mathcal{L}\{f'\} = s \mathcal{L}\{f\} - f(0)$

$\mathcal{L}\{f''\} = s^2 \mathcal{L}\{f\} - s f(0) - f'(0)$

$\mathcal{L}\{f^{(n)}\} = s^n \mathcal{L}\{f\} - \Sigma_{i = 1}^{n}s^{n - i}f^{(i - 1)}(0)$

Suppose we want to solve the given differential equation:

$\sum^n_{i=0}a_if^{(i)}(t)=\phi(t)$

This equation is equivalent to

$\sum^n_{i=0}a_i\mathcal{L}\{f^{(i)}(t)\}=\mathcal{L}\{\phi(t)\}$

which is equivalent to

$\mathcal{L}\{f(t)\}={\mathcal{L}\{\phi(t)\}+\sum^n_{i=0}a_i\sum^i_{j=1}s^{i-j}f^{(j-1)}(0) \over \sum^n_{i=0}a_is^i}$

note that the $f (k) (0)$ are initial conditions.

Then all we need to get f(t) is to apply the Laplace inverse transform to $\mathcal{L}\{f(t)\}$

An example

We want to solve

$f^{(2)}(t)+4f(t)=\sin(2t) \,\!$

with initial conditions f(0) = 0 and f ′(0)=0.

We note that

$\phi(t)=\sin(2t) \,\!$

and we get

$\mathcal{L}\{\phi(t)\}=\frac{2}{s^2+4}$

So this is equivalent to

$s^2\mathcal{L}\{f(t)\}-sf(0)-f^{(1)}(0)+4\mathcal{L}\{f(t)\}=\mathcal{L}\{\phi(t)\}$

We deduce

$\mathcal{L}\{f(t)\}=\frac{2}{(s^2+4)^2}$

So we apply the Laplace inverse transform and get

$f(t)=\frac{1}{8}\sin(2t)-\frac{t}{4}\cos(2t)$

A. D. Polyanin, Handbook of Linear Partial Differential Equations for Engineers and Scientists, Chapman & Hall/CRC Press, Boca Raton, 2002. ISBN 1-58488-299-9