CHAPTER 9 PROBLEM SOLUTIONS
S
OLUTION TO
P
ROBLEM
9.1.
If we can compute expressions for K and q that are real,
then these quantities exist by construction.
Consider that A, B, K and q must satisfy the
following relationship:
K
cos(
ϖ
t
+ θ
)
=
K
cos(
θ
)
(
29
cos(
ϖ
t
)
+

K
sin(
θ
)
(
29
sin(
ϖ
t
)
≡
A
cos(
ϖ
t
)
+
B
sin(
ϖ
t
)
Therefore
K
cos(
θ
)
=
A
and

K
sin(
θ
)
=
B
.
Consequently,
K
cos(
θ
)
(
29
2
+

K
sin(
θ
)
(
29
2
=
K
2
=
A
2
+
B
2
in which case
K
=
A
2
+
B
2
.
Further,
K
sin(
θ
)
K
cos(
θ
)
=
tan(
θ
)
=

B
A
in which case
θ =
tan

1

B
A
with due regard to quadrant.
S
OLUTION TO
P
ROBLEM
9.2.
For the inductor,
W
L
(
t
)
=
1
2
L
⋅
i
L
2
(
t
)
=
1
2
L V
o
C
L
sin
1
LC
t
2
=
CV
0
2
2
sin
2
1
LC
t
and for the capacitor,
W
C
(
t
)
=
1
2
C
⋅
v
C
2
(
t
)
=
1
2
C V
o
cos
1
LC
t
2
=
CV
0
2
2
cos
2
1
LC
t
.
Hence,
W
C
+
W
L
=
1
2
C
⋅
v
C
2
(
t
)
+
1
2
L
⋅
i
L
2
(
t
)
=
CV
0
2
2
sin
2
1
LC
t
+
cos
2
1
LC
t
=
CV
0
2
2
S
OLUTION TO
P
ROBLEM
9.3.
Since
x
(
t
)
=
(
K
1
+
K
2
t
)
e
α
t
,
x
' (
t
)
=  α ⋅
K
1
e
α
t
+
K
2
e
α
t
 α ⋅
t
⋅
K
2
e
α
t
and
x
' ' (
t
)
= α
2
⋅
K
1
e
α
t
 α ⋅
K
2
e
α
t
 α ⋅
K
2
e
α
t
+ α
2
⋅
t
⋅
K
2
e
α
t
Substituting into the differential equation, we have
α
2
⋅
K
1
e
α
t

2
α ⋅
K
2
e
α
t
+ α
2
⋅
t
⋅
K
2
e
α
t
+
2
α  α ⋅
K
1
e
α
t
+
K
2
e
α
t
 α ⋅
t
⋅
K
2
e
α
t
[
]
+ α
2
K
1
e
α
t
+
K
2
te
α
t
[
]
=
0
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