Homework 5 solutions
page 186
In these 8 problems, the roots associated to the forcing term are indicated in the answer in
those cases where the answer was yes.
1)
No
, the
t

1
is not a polynomial
2)
Yes
, the roots
±
4
i
are repeated 4 times (polynomial of degree 3)
3)
Yes
, 3
t
=
e
ln 3
t
, so the root is ln 3 and it is not repeated
4)
Yes
, sin(
x
)
/e
4
x
=
e

4
x
sin
x
, so the roots are

4
±
i
, and they are not repeated.
5)
No
, the secant cannot be written in the required form.
6)
Yes
, 4
x
sin
2
x
+ 4
x
cos
2
x
= 4
x
= 4
xe
0
x
, so the root is 0 and it is repeated twice (the
polynomial is degree 1)
7)
No
, the diﬀerential equation does not have constant coeﬃcients
8)
Yes
, The roots are 4
±
25
i
, and since the polynomial has degree 100, the roots are
repeated 101 times.
12) 2
x
0
+
x
= 3
t
2
, The right hand side corresponds to the root zero repeated 3 times (the
polynomial is of degree 2). The left hand side has one root,
r
=

1
/
2. There is no overlap,
so we have
x
p
=
C
1
+
C
2
t
+
C
3
t
2
. Plug this in to the equation and we get
2
x
0
p
+
x
p
= 2(
C
2
+ 2
C
3
t
) + (
C
1
+
C
2
t
+
C
3
t
2
) = (2
C
2
+
C
1
) + (4
C
3
+
C
2
)
t
+
C
3
t
2
= 3
t
2
,
leading to
2
C
2
+
C
1
= 0
,
4
C
3
+
C
2
= 0
, C
3
= 3
.
Solving we get
C
3
= 3
, C
2
=

12
, C
1
= 24
so
x
p
= 24

12
t
+ 3
t
2
.
Of course we have
x
h
=
C
4
e

t/
2
, so
x
=
x
h
+
x
p
=
C
4
e

t/
2
+ 8

4
t
+
t
2
.
16)
θ
00

θ
= (
D
2

1)
θ
=
t
sin
t
. The roots associated to the forcing term
t
sin
t
are
±
i
and
they are repeated twice since the polynomial
t
has degree 1. The left hand side has roots
±
1. Thus
θ
=
C
1
e
t
+
C
2
e

t
+ (
C
3
+
C
4
t
) sin
t
+ (
C
5
+
C
6
t
) cos
t
, and since
θ
h
=
C
1
e
t
+
C
2
e

t
,
we should take
θ
p
= (
C
3
+
C
4
t
) sin
t
+ (
C
5
+
C
6
t
) cos
t
. Plugging in we have
θ
0
p
= (
C
4

C
5

C
6
t
) sin
t
+ (
C
3
+
C
4
t
+
C
6
) cos
t
and
θ
00
p
= (

C
3

C
4
t

2
C
6
) sin
t
+ (2
C
4

C
5

C
6
t
) cos
t
so
θ
00
p

θ
p
= (

2
C
3

2
C
4
t

2
C
6
) sin
t
+ (2
C
4

2
C
5

2
C
6
t
) cos
t
=
t
sin
t.
Thus we must solve

2
C
4
= 1
,

2
C
3

2
C
6
= 0
, C
6
= 0
,
and 2
C
4

2
C
5
= 0
.
That leads to
C
3
= 0
, C
4
=

1
2
, C
5
=

1
2
,
and
C
6
= 0
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θ
p
=

1
2
(
t
sin
t
+ cos
t
)
.
We also know that
θ
=
θ
h
+
θ
p
=
C
1
e
t
+
C
2
e

t

1
2
(
t
sin
t
+ cos
t
)
.
27) The roots from the forcing are
±
3
i
repeated 4 times, from the left hand side we also
have
±
3
i
. Thus the total multiplicity of
±
3
i
is 5 – corresponding to a polynomial of de
gree 4.
y
h
=
C
1
cos 3
t
+
C
2
sin 3
t
, and we do not include terms already in
y
h
in
y
p
, so
y
p
= (
C
3
t
+
C
4
t
2
+
C
5
t
3
+
C
6
t
4
) cos 3
t
+ (
C
7
t
+
C
8
t
2
+
C
9
t
3
+
C
10
t
4
) sin 3
t
.
28) The root corresponding to the forcing term 1 repeated 5 times. The roots from the left
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 Spring '08
 Dienes
 Work, Boundary value problem, 2m, 2 m, 2  k

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