The use of Laplace transform makes it much easier to solve linear differential equations.

First consider the folowing relations :

Suppose we want to solve the given differential equation :

this equation is equivalent to :

which is equivalent to :

note that the are initial conditions.

Then all we need to find f(t) is to apply the Laplace inverse transform to

An example

We want to solve :

with initial conditions f(0)=0 and f'(0)=0

we note :

and we get :

so this is equivalent to :

we deduce :

So we apply the Laplace inverse transform and get

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