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Exercises 4.5
Thus a general solution on (
−
π/
(2
V
)
,π/
(2
V
)) is
y
1
(
t
)=
y
h
(
t
)+
y
p
(
t
C
1
cos
t
+
C
2
sin
t
+
1
1
−
V
2
cos
Vt.
(4.3)
Since
y
(
t
)
≡
0for
t
≤−
(2
V
), the initial conditions for the above solution are
y
1
±
−
π
2
V
²
=
y
0
1
±
−
π
2
V
²
=0
.
From (4.3) we obtain
y
1
±
−
π
2
V
²
=
C
1
cos
±
−
π
2
V
²
+
C
2
sin
±
−
π
2
V
²
y
0
1
±
−
π
2
V
²
=
−
C
1
sin
±
−
π
2
V
²
+
C
2
cos
±
−
π
2
V
²
+
V
1
−
V
2
.
Solving the system yields
C
1
=
V
V
2
−
1
sin
π
2
V
,C
2
=
V
V
2
−
1
cos
π
2
V
,
and
y
1
(
t
V
V
2
−
1
sin
π
2
V
cos
t
+
V
V
2
−
1
cos
π
2
V
sin
t
+
1
1
−
V
2
cos
Vt
=
V
V
2
−
1
sin
±
t
+
π
2
V
²
−
1
V
2
−
1
cos
Vt,
−
π
2
V
<t<
π
2
V
.
For
t>π/
(2
V
) given equation is homogeneous, and its 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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