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)\}
f(i)(0)
ci
n | ||||
| ||||
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 |
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
|
The equation is then equivalent to
s2l{L}\{f(t)\}-sf(0)-f'(0)+4l{L}\{f(t)\}=l{L}\{\phi(t)\}
We deduce
|
Now we apply the Laplace inverse transform to get
f(t)= | 1 | \sin(2t)- |
8 |
t | |
4 |
\cos(2t)