3. a. Consider
y
00

9
y
= 2
te

3
t
sin(
t
) + 6
t
2
e
3
t
. The solution of the homogeneous equation is
y
c
(
t
) =
c
1
e

3
t
+
c
2
e
3
t
.
The guess that one makes for the method of undetermined coefficients is
y
p
(
t
) =
e

3
t
((
A
1
t
+
A
0
) cos(
t
) + (
B
1
t
+
B
0
) sin(
t
)) +
t
(
C
2
t
2
+
C
1
t
+
C
0
)
e
3
t
.
b. Consider
y
00
+ 2
y
0
+ 2
y
= 4
t
2
e

t
+ 3
te

t
cos(
t
). The solution of the homogeneous equation is
y
c
(
t
) =
e

t
(
c
1
cos(
t
) +
c
2
sin(
t
))
.
The guess that one makes for the method of undetermined coefficients is
y
p
(
t
) =
(
A
2
t
2
+
A
1
t
+
A
0
)
e

t
+
te

t
((
B
1
t
+
B
0
) cos(
t
) + (
C
1
t
+
C
0
) sin(
t
))
.
4. a. Consider
y
00

4
y
= 16
e

2
t
,
y
(0) = 1
,
y
0
(0) = 2
.
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Laplace transforms give:
s
2
Y
(
s
)

s

2

4
Y
(
s
) =
16
s
+ 2
or
Y
(
s
) =
s
+ 2
s
2

4
+
16
(
s
+ 2)(
s
2

4)
.
Partial fractions on the second expression give
16
(
s
+ 2)(
s
2

4)
=
A
(
s
+ 2)
2
+
B
s
+ 2
+
C
s

2
or
16 =
A
(
s

2) +
B
(
s
+ 2)(
s

2) +
C
(
s
+ 2)
2
.
From
s
=

2,
A
=

4. From
s
= 2,
C
= 1. From
s
2
,
B
=

C
=

1. It follows that
Y
(
s
) =

4
(
s
+ 2)
2

1
s
+ 2
+
2
s

2
.
The inverse Laplace transform gives
y
(
t
) = 2
e
2
t

e

2
t

4
te

2
t
.
b. Consider
y
00

y
0

12
y
=
t
2
δ
(
t

3)
,
y
(0) = 2
,
y
0
(0) =

6
.
Laplace transforms give:
s
2
Y
(
s
)

2
s
+ 6

sY
(
s
) + 2

12
Y
(
s
) = 9
e

3
s
or
Y
(
s
) =
2(
s

4)
(
s

4)(
s
+ 3)
+
9
e

3
s
(
s

4)(
s
+ 3)
.
Partial fractions on the second expression give
9
(
s

4)(
s
+ 3)
=
A
s

4
+
B
s
+ 3
or
9 =
A
(
s
+ 3) +
B
(
s

4)
.
From
s
= 4,
A
=
9
7
. From
s
=

3,
B
=

9
7
. It follows that
Y
(
s
) =
2
s
+ 3
+
9
7
e

3
s
s

4

e

3
s
s
+ 3
.
The inverse Laplace transform gives
y
(
t
) = 2
e

3
t
+
9
7
u
3
(
t
)
e
4(
t

3)

e

3(
t

3)
.
c. Consider
y
00
+ 2
y
0
+ 5
y
=
5
,
0
≤
t <
4
0
,
t
≥
4
y
(0) = 0
,
y
0
(0) = 4
.
Laplace transforms give:
s
2
Y
(
s
)

4 + 2
sY
(
s
) + 5
Y
(
s
) =
5
s

5
e

4
s
s
or
Y
(
s
) =
4
(
s
+ 1)
2
+ 4
+
5

5
e

4
s
s
((
s
+ 1)
2
+ 4)
.
Partial fractions on the second expression give
5
s
((
s
+ 1)
2
+ 4)
=
A
s
+
B
(
s
+ 1) + 2
C
(
s
+ 1)
2
+ 4
or
5 =
A
((
s
+ 1)
2
+ 4) +
Bs
(
s
+ 1) + 2
Cs.
From
s
= 0,
A
= 1. From
s
2
,
B
=

A
=

1. From
s
=

1, 5 = 4
A

2
C
or
C
=

1
2
. It follows
that
Y
(
s
) =
4
(
s
+ 1)
2
+ 4
+
1
s

s
+ 1
(
s
+ 1)
2
+ 4

1
2
2
(
s
+ 1)
2
+ 4
(1

e

4
s
)
.
The inverse Laplace transform gives
y
(
t
) = 1 +
e

t
3
2
sin(2
t
)

cos(2
t
)
+
u
4
(
t
)
1

e

(
t

4)
cos(2(
t

4))

1
2
sin(2(
t

4))
.
d. Consider
y
00
+ 4
y
= cos(
t
)
,
y
(0) = 1
,
y
0
(0) = 0
.
Laplace transforms give:
s
2
Y
(
s
)

s
+ 4
Y
(
s
) =
s
s
2
+ 1
or
Y
(
s
) =
s
s
2
+ 4
+
s
(
s
2
+ 1)(
s
2
+ 4)
.
Partial fractions on the second expression give
s
(
s
2
+ 1)(
s
2
+ 4)
=
As
+
B
s
2
+ 1
+
Cs
+ 2
D
s
2
+ 4
or
s
= (
As
+
B
)(
s
2
+ 4) + (
Cs
+ 2
D
)(
s
2
+ 1)
.
From
s
=
i
,
i
= 3(
iA
+
B
), so
A
=
1
3
and
B
= 0. From
s
= 2
i
, 2
i
=

3(2
iC
+ 2
D
), so
C
=

1
3
and
D
= 0. It follows that
Y
(
s
) =
s
s
2
+ 4
+
1
3
s
s
2
+ 1

1
3
s
s
2
+ 4
=
2
3
s
s
2
+ 4
+
1
3
s
s
2
+ 1
.
The inverse Laplace transform gives
y
(
t
) =
2
3
cos(2
t
) +
1
3
cos(
t
)
.
e. Consider
y
00
+ 2
y
0

3
y
=
t
2
,
0
≤
t <
3
0
,
t
≥
3
y
(0) = 0
,
y
0
(0) = 0
.
The function on the RHS can be written
f
(
t
) =
t
2

t
2
u
3
(
t
) =
t
2

(
t

3)
2
u
3
(
t
)

6(
t

3)
u
3
(
t
)

9
u
3
(
t
).
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 Fall '08
 staff