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\}-sf(0)-f'(0)

{\mathcal {L}}\{f^{(n)}\}=s^{n}{\mathcal {L}}\{f\}-\Sigma _{i=1}^{n}s^{n-i}f^{(i-1)}(0).

Consider the following differential equation:

\sum _{i=0}^{n}a_{i}f^{(i)}(t)=\phi (t).

This equation is equivalent to

\sum _{i=0}^{n}a_{i}{\mathcal {L}}\{f^{(i)}(t)\}={\mathcal {L}}\{\phi (t)\}

which is equivalent to

${\mathcal {L}}\{f(t)\}={{\mathcal {L}}\{\phi (t)\}+\sum _{i=0}^{n}a_{i}\sum _{j=1}^{i}s^{i-j}f^{(j-1)}(0) \over \sum _{i=0}^{n}a_{i}s^{i}}.$

Note that the $f^{(k)}(0)$ are initial conditions.

The solution for f(t) will be given by applying the Laplace inverse transform to ${\mathcal {L}}\{f(t)\}.$

An example

We want to solve

f^{''}(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^{'}(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