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:

:$\mathcal\=s\mathcal\-f(0)$

:$\mathcal\=s^2\mathcal\-sf(0)-f'(0)$

One can prove by induction that

:$\mathcal\=s^n\mathcal\-\sum_^s^f^(0)$

Now we consider the following differential equation:

:$\sum_^a_if^(t)=\phi(t)$

with given initial conditions

:$f^(0)=c_i$

Using the linearity of the Laplace transform it is equivalent to rewrite the equation as

:$\sum_^a_i\mathcal\=\mathcal\$

obtaining

:$\mathcal\\sum_^a_is^i-\sum_^\sum_^a_is^f^(0)=\mathcal\$

Solving the equation for $\mathcal\$ and substituting $f^(0)$ with $c_i$ one obtains

:$\mathcal\=\frac$

The solution for ''f''(''t'') is obtained by applying the inverse Laplace transform to $\mathcal\.$

Note that if the initial conditions are all zero, i.e.

:$f^(0)=c_i=0\quad\forall i\in\$

then the formula simplifies to

:$f(t)=\mathcal^\left\$

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

:$\mathcal\=\frac$

The equation is then equivalent to

:$s^2\mathcal\-sf(0)-f'(0)+4\mathcal\=\mathcal\$

We deduce

:$\mathcal\=\frac$

Now we apply the Laplace inverse transform to get

:$f(t)=\frac\sin(2t)-\frac\cos(2t)$

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

Integral transforms
Differential equations
Differential calculus
Ordinary differential equations