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Exercises 4.5
(b)
For
t>
3
π/
2,
g
(
t
)
≡
0, and so the given equation becomes homogeneous. Thus, a general
solution,
y
2
(
t
), is given by (4.1), i.e.,
y
2
(
t
)=
y
h
(
t
)=(
c
1
cos 2
t
+
c
2
sin 2
t
)
e
−
t
.
(c)
We want to satisfy the conditions
y
1
(3
2)
=
y
2
(3
2)
,
y
0
1
(3
2)
=
y
0
2
(3
2)
.
Evaluating
y
1
,
y
2
, and their derivatives at
t
=3
2, we solve the system
2
e
−
3
2
+2 =
−
c
1
e
−
3
2
,
0=(
c
1
−
2
c
2
)
e
3
2
⇒
c
1
=
−
2
(
e
3
2
+1
)
,
c
2
=
−
(
e
3
2
)
.
43.
Recall that the motion of a massspring system is governed by the equation
my
0
+
by
0
+
ky
=
g
(
t
)
,
where
m
is the mass,
b
is the damping coeﬃcient,
k
is the spring constant, and
g
(
t
)isthe
external force. Thus, we have an initial value problem
y
0
+4
y
0
+3
y
=5s
in
t,
y
(0) =
1
2
,y
0
(0) = 0
.
The roots of the auxiliary equation,
r
2
r
+3=0,are
r
=
−
3,
−
1, and a general solution
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This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at University of California, Berkeley.
 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations

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