In mathematics, the Laplace transform is a powerful integral transform used to switch a function from the time domain to the s-domain. The Laplace transform can be used in some cases to solve linear differential equations with given initial conditions.
Approach
First consider the following property of the Laplace transform:
:<math>\mathcal{L}\{f'\}=s\mathcal{L}\{f\}-f(0)</math>
:<math>\mathcal{L}\{f\}=s^2\mathcal{L}\{f\}-sf(0)-f'(0)</math>
One can prove by induction that
:<math>\mathcal{L}\{f^{(n)}\}=s^n\mathcal{L}\{f\}-\sum_{i=1}^{n}s^{n-i}f^{(i-1)}(0)</math>
Now we consider the following differential equation:
:<math>\sum_{i=0}^{n}a_if^{(i)}(t)=\phi(t)</math>
with given initial conditions
:<math>f^{(i)}(0)=c_i</math>
Using the linearity of the Laplace transform it is equivalent to rewrite the equation as
:<math>\sum_{i=0}^{n}a_i\mathcal{L}\{f^{(i)}(t)\}=\mathcal{L}\{\phi(t)\}</math>
obtaining
:<math>\mathcal{L}\{f(t)\}\sum_{i=0}^{n}a_is^i-\sum_{i=1}^{n}\sum_{j=1}^{i}a_is^{i-j}f^{(j-1)}(0)=\mathcal{L}\{\phi(t)\}</math>
Solving the equation for <math> \mathcal{L}\{f(t)\}</math> and substituting <math>f^{(i)}(0)</math> with <math>c_i</math> one obtains
:<math>\mathcal{L}\{f(t)\}=\frac{\mathcal{L}\{\phi(t)\}+\sum_{i=1}^{n}\sum_{j=1}^{i}a_is^{i-j}c_{j-1{\sum_{i=0}^{n}a_is^i}</math>
The solution for f(t) is obtained by applying the inverse Laplace transform to <math>\mathcal{L}\{f(t)\}.</math>
Note that if the initial conditions are all zero, i.e.
:<math>f^{(i)}(0)=c_i=0\quad\forall i\in\{0,1,2,...\ n\}</math>
then the formula simplifies to
:<math>f(t)=\mathcal{L}^{-1}\left\