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
{
f
′
}
=
s
L
{
f
}
−
f
(
0
)
{\displaystyle {\mathcal {L}}\{f'\}=s{\mathcal {L}}\{f\}-f(0)}
L
{
f
″
}
=
s
2
L
{
f
}
−
s
f
(
0
)
−
f
′
(
0
)
{\displaystyle {\mathcal {L}}\{f''\}=s^{2}{\mathcal {L}}\{f\}-sf(0)-f'(0)}
One can prove by induction that
L
{
f
(
n
)
}
=
s
n
L
{
f
}
−
∑
i
=
1
n
s
n
−
i
f
(
i
−
1
)
(
0
)
{\displaystyle {\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:
∑
i
=
0
n
a
i
f
(
i
)
(
t
)
=
ϕ
(
t
)
{\displaystyle \sum _{i=0}^{n}a_{i}f^{(i)}(t)=\phi (t)}
with given initial conditions
f
(
i
)
(
0
)
=
c
i
{\displaystyle f^{(i)}(0)=c_{i}}
Using the linearity of the Laplace transform it is equivalent to rewrite the equation as
∑
i
=
0
n
a
i
L
{
f
(
i
)
(
t
)
}
=
L
{
ϕ
(
t
)
}
{\displaystyle \sum _{i=0}^{n}a_{i}{\mathcal {L}}\{f^{(i)}(t)\}={\mathcal {L}}\{\phi (t)\}}
obtaining
L
{
f
(
t
)
}
∑
i
=
0
n
a
i
s
i
−
∑
i
=
1
n
∑
j
=
1
i
a
i
s
i
−
j
f
(
j
−
1
)
(
0
)
=
L
{
ϕ
(
t
)
}
{\displaystyle {\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
L
{
f
(
t
)
}
{\displaystyle {\mathcal {L}}\{f(t)\}}
f
(
i
)
(
0
)
{\displaystyle f^{(i)}(0)}
c
i
{\displaystyle c_{i}}
L
{
f
(
t
)
}
=
L
{
ϕ
(
t
)
}
+
∑
i
=
1
n
∑
j
=
1
i
a
i
s
i
−
j
c
j
−
1
∑
i
=
0
n
a
i
s
i
{\displaystyle {\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
L
{
f
(
t
)
}
.
{\displaystyle {\mathcal {L}}\{f(t)\}.}
Note that if the initial conditions are all zero, i.e.
f
(
i
)
(
0
)
=
c
i
=
0
∀
i
∈
{
0
,
1
,
2
,
.
.
.
n
}
{\displaystyle f^{(i)}(0)=c_{i}=0\quad \forall i\in \{0,1,2,...\ n\}}
then the formula simplifies to
f
(
t
)
=
L
−
1
{
L
{
ϕ
(
t
)
}
∑
i
=
0
n
a
i
s
i
}
{\displaystyle f(t)={\mathcal {L}}^{-1}\left\{{{\mathcal {L}}\{\phi (t)\} \over \sum _{i=0}^{n}a_{i}s^{i}}\right\}}
We want to solve
f
″
(
t
)
+
4
f
(
t
)
=
sin
(
2
t
)
{\displaystyle f''(t)+4f(t)=\sin(2t)}
f (0) = 0 and f′ (0)=0.
We note that
ϕ
(
t
)
=
sin
(
2
t
)
{\displaystyle \phi (t)=\sin(2t)}
and we get
L
{
ϕ
(
t
)
}
=
2
s
2
+
4
{\displaystyle {\mathcal {L}}\{\phi (t)\}={\frac {2}{s^{2}+4}}}
The equation is then equivalent to
s
2
L
{
f
(
t
)
}
−
s
f
(
0
)
−
f
′
(
0
)
+
4
L
{
f
(
t
)
}
=
L
{
ϕ
(
t
)
}
{\displaystyle s^{2}{\mathcal {L}}\{f(t)\}-sf(0)-f'(0)+4{\mathcal {L}}\{f(t)\}={\mathcal {L}}\{\phi (t)\}}
We deduce
L
{
f
(
t
)
}
=
2
(
s
2
+
4
)
2
{\displaystyle {\mathcal {L}}\{f(t)\}={\frac {2}{(s^{2}+4)^{2}}}}
Now we apply the Laplace inverse transform to get
f
(
t
)
=
1
8
sin
(
2
t
)
−
t
4
cos
(
2
t
)
{\displaystyle 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 
