There are a lot of ways we can apply this. For example, let
y
(
t
) = cos(
bt
). Then
Y
(
s
) =
s/
(
s
2
+
b
2
),
as we already showed. Using this property, we get that
L
[
e
at
cos(
bt
)] =
Y
(
s

a
) =
s

a
(
s

a
)
2
+
b
2
.
Frequency Differentiation
Let
Y
(
s
) be the Laplace transform of
y
(
t
). Then by definition
Y
(
s
) =
∞
0
e

st
y
(
t
)
dt.
82
If we then differential each side with respect to
s
, we get
Y
(
s
) =
d
ds
∞
0
e

st
y
(
t
)
dt
=
∞
0
d
ds
(
e

st
)
y
(
t
)
dt
=
∞
0

e

st
ty
(
t
)
dt
=

L
[
ty
(
t
)]
.
This gives us the useful identity
L
[
ty
(
t
)] =

Y
(
s
)
.
If this process is repeated
n
times, we get the more general identity
L
[
t
n
y
(
t
)] = (

1)
n
Y
(
n
)
(
s
)
.
Note that if we apply this identity with
y
(
t
) = 1, we get the Laplace transform of
t
n
is
L
[
t
n
] = (

1)
n
d
n
ds
n
1
s
=
n
!
s
n
+1
.
Time Differentiation
Let
Y
(
s
) be the Laplace transform of
y
(
t
).
State the Laplace transform of
y
(
t
) in terms of
Y
.
The Laplace transform for
y
(
t
) is
L
[
y
(
t
)] =
∞
0
e

st
y
(
t
)
dt.
Now we do integration by parts with
u
=
e

st
and =
dv
=
y
(
t
)
dt
. Then
du
=

se

st
dt
and
v
=
y
.
Integration by parts the gives us
∞
0
e

st
y
(
t
)
dt
=
e

st
y
(
t
)
t
=
∞
t
=0
+
∞
0
se

st
y
(
t
)
dt
= (0

y
(0)) +
s
∞
0
e

st
y
(
t
)
dt
=
sY
(
s
)

y
(0)
.
This gives us the most useful property of Laplace transforms for solving differential equations:
L
[
y
(
t
)] =
sY
(
s
)

y
(0)
.
Repeating this process a second time, would yield the result for the Laplace transform of the
second derivative:
L
[
y
(
t
)] =
s
L
[
y
(
t
)]

y
(0) =
s
(
sY
(
s
)

y
(0))

y
(0) =
s
2
Y
(
s
)

sy
0

y
0
.
We can use this property to find the Laplace transform for sin(
at
) since we already know the
Laplace transform of
y
(
t
) = cos(
at
):
L
[sin(
at
)] =

1
a
L
d
dt
cos(
at
)
=

1
a
(
sY
(
s
)

cos(0)) =

1
a
s
2
s
2
+
a
2

1
=
a
s
2
+
a
2
Action on a Differential Equation
Consider a simple firstorder nonhomogeneous equation of the form
ay
+
by
=
g
(
t
)
,
y
(0) =
y
0
83
If we take the Laplace transform of each side, we get
a
L
[
y
] +
b
L
[
y
] =
L
[
g
(
t
)]
which gives us
a
[
sY
(
s
)

y
0
] +
bY
(
s
) =
G
(
s
)
⇒
Y
(
s
) =
G
(
s
) +
ay
0
as
+
b
.
Notice that we were able to solve for
Y
(
s
) algebraically after performing the Laplace transform.
This is what makes Laplace transforms so useful:
differential equations in the time domain become
algebraic equations in the frequency domain.
However, there still remains the problem of inverting
Y
(
s
) to recover the solution
y
(
t
).
3.1.5
Further Examples
We have already seen that the Laplace transform for monomials with integer powers is
L
[
t
n
] =
n
!
s
n
+1
.
But what about noninteger powers?
It turns out that there is a well defined way extend the
idea of a factorial to any positive number, which is called the gamma function Γ(
r
). The gamma
function is defined by the integral
Γ(
r
) =
∞
0
x
r

1
e

x
dx.
When evaluated at integers, the gamma function satisfies the relation
Γ(
n
) = (
n

1)!
which can be proven using successive integration by parts. Therefore we can generalize the Laplace
transform for noninteger powers of
t
:
L
[
t
r
] =
Γ(
r
+ 1)
s
r
+1
.
Other than integers, not many exact values for the gamma function are known. One exception is
Γ(3
/
2) = (1
/
2)! =
√
π
2
.