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Chapter 4
k
=30
k
=30
k
=30
. Inth
iscase
,
r
=
−
5
±
√
25
−
30 =
−
5
±
√
5
i
. A general solution has the form
y
=(
C
1
cos
√
5
t
+
C
2
sin
√
5
t
)
e
−
5
t
.F
o
rcon
s
tan
t
s
C
1
and
C
2
, we have the system
y
(0) =
(
C
1
cos
√
5
t
+
C
2
sin
√
5
t
)
e
−
5
t
±
±
t
=0
=
C
1
=1
,
y
0
(0) =
²
(
√
5
C
2
−
5
C
1
)cos
√
5
t
−
(
√
5
C
1
+5
C
2
)sin
√
5
t
³
e
−
5
t
±
±
t
=0
=
√
5
C
2
−
5
C
1
=0
⇒
C
1
=1
,
C
2
=
√
5
,
and so
y
(
t
)=
h
cos
√
5
t
+
√
5sin
√
5
t
i
e
−
5
t
=
√
6
e
−
5
t
sin
´
√
5
t
+
φ
µ
,
where
φ
= arctan(1
/
√
5)
≈
0
.
421 .
Graphs of the solutions for
k
= 20, 25, and 30 are shown in Figures B.23–B.25 in the answers
in the text.
7.
The motion of this massspring system is governed by equation (12) on page 213 in the text.
With
m
=1
/
8,
b
=2
,and
k
= 16 this equation becomes
1
8
y
0
+2
y
0
+16
y
=0
,
(4.12)
and the initial conditions are
y
(0) =
−
3
/
4,
y
0
(0) =
−
2. Since
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 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations

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