# Laplace transform applied to differential equations

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.

First consider the following property of the Laplace transform:

${\mathcal {L}}\{f'\}=s{\mathcal {L}}\{f\}-f(0)$ ${\mathcal {L}}\{f''\}=s^{2}{\mathcal {L}}\{f\}-sf(0)-f'(0)$ One can prove by induction that

${\mathcal {L}}\{f^{(n)}\}=s^{n}{\mathcal {L}}\{f\}-\sum _{i=1}^{n}s^{n-i}f^{(i-1)}(0)$ Now we consider the following differential equation:

$\sum _{i=0}^{n}a_{i}f^{(i)}(t)=\phi (t)$ with given initial conditions

$f^{(i)}(0)=c_{i}$ Using the linearity of the Laplace transform it is equivalent to rewrite the equation as

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

${\mathcal {L}}\{f(t)\}\sum _{i=0}^{n}a_{i}s^{i}-\sum _{i=1}^{n}\sum _{j=1}^{i}a_{i}s^{i-j}f^{(j-1)}(0)={\mathcal {L}}\{\phi (t)\}$ Solving the equation for ${\mathcal {L}}\{f(t)\}$ and substituting $f^{(i)}(0)$ with $c_{i}$ one obtains

${\mathcal {L}}\{f(t)\}={\frac {{\mathcal {L}}\{\phi (t)\}+\sum _{i=1}^{n}\sum _{j=1}^{i}a_{i}s^{i-j}c_{j-1}}{\sum _{i=0}^{n}a_{i}s^{i}}}$ The solution for f(t) is obtained by applying the inverse Laplace transform to ${\mathcal {L}}\{f(t)\}.$ Note that if the initial conditions are all zero, i.e.

$f^{(i)}(0)=c_{i}=0\quad \forall i\in \{0,1,2,...\ n\}$ then the formula simplifies to

$f(t)={\mathcal {L}}^{-1}\left\{{{\mathcal {L}}\{\phi (t)\} \over \sum _{i=0}^{n}a_{i}s^{i}}\right\}$ ## 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

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

The equation is then equivalent to

$s^{2}{\mathcal {L}}\{f(t)\}-sf(0)-f'(0)+4{\mathcal {L}}\{f(t)\}={\mathcal {L}}\{\phi (t)\}$

We deduce

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

Now we apply the Laplace inverse transform to get

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

## Bibliography

• 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