Laplace transform applied to differential equations explained

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:

l{L}\{f'\}=sl{L}\{f\}-f(0)

l{L}\{f''\}=s2l{L}\{f\}-sf(0)-f'(0)

One can prove by induction that

l{L}\{f(n)

n
\}=s
i=1

sn-if(i-1)(0)

Now we consider the following differential equation:

n
\sum
i=0
(i)
a
if

(t)=\phi(t)

with given initial conditions

f(i)(0)=ci

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

n
\sum
i=0
(i)
a
il{L}\{f

(t)\}=l{L}\{\phi(t)\}

obtaining

n
l{L}\{f(t)\}\sum
i=0
n
a
i=1
i
\sum
j=1
i-j
a
is

f(j-1)(0)=l{L}\{\phi(t)\}

Solving the equation for

l{L}\{f(t)\}

and substituting

f(i)(0)

with

ci

one obtains
n
l{L}\{f(t)\}=l{L
\{\phi(t)\}+\sum
i=1
i
\sum
j=1
i-j
a
is

cj-1

}

The solution for f(t) is obtained by applying the inverse Laplace transform to

l{L}\{f(t)\}.

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

f(i)(0)=ci=0   \foralli\in\{0,1,2,...n\}

then the formula simplifies to

f(t)=l{L}-1

n
\left\{{l{L}\{\phi(t)\}\over\sum
i=0
i}\right\}
a
is

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

l{L}\{\phi(t)\}=2
s2+4

The equation is then equivalent to

s2l{L}\{f(t)\}-sf(0)-f'(0)+4l{L}\{f(t)\}=l{L}\{\phi(t)\}

We deduce

l{L}\{f(t)\}=2
(s2+4)2

Now we apply the Laplace inverse transform to get

f(t)=1\sin(2t)-
8
t
4

\cos(2t)

Bibliography