Math 135, HW2 solutions to selected exercises
Mike Hall
p. 384
5d)
Find a function whose Laplace transform is equal to
1
p
(
p
+1)
.
Using partial fractions decomposition we see that
1
p
(
p
+ 1)
=
1
p

1
p
+ 1
=
L
[1]

L
[
e

x
] =
L
[1

e

x
]
so
f
(
x
) = 1

e

x
is a solution (which is unique, up to some restrictions we don’t care that much about,
by some theorem we take for granted for now).
p. 394
1b)
Find the Laplace transform of
f
(
x
) = (1

x
2
)
e

x
.
By the properties of the Laplace transform, we have
L
[
f
(
x
)](
p
) =
L
[(1

x
2
)
e

x
](
p
) =
L
[1

x
2
](
p
+ 1) =
±
1
s

2
s
3
²³
³
³
³
s
=
p
+1
=
1
p
+ 1

2
(
p
+ 1)
3
.
3d)
Solve:
y
00
+
y
0
= 3
x
2
,
y
(0) = 0
,
y
0
(0) = 1.
Taking Laplace transforms of both sides, we get
(
p
2
L
[
y
]

py
(0)

y
0
(0)) + (
pL
[
y
]

y
(0)) = 3
·
2
p
3
and plugging in for
y
(0) and
y
0
(0), we get
L
[
y
](
p
2
+
p
)

1 =
6
p
3
L
[
y
] =
±
6
p
3
+ 1
²
1
p
(
p
+ 1)
We use partial fractions decomposition to break up the right hand side into recognizable Laplace transforms.
One thing to note is that it’s not always best to combine everything into one big fraction, as it may only
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 Spring '10
 DR.RAHULA
 Laplace, PierreSimon Laplace, partial fractions decomposition

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