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Chapter 5
⇒
A
=0
,B
=
−
1
⇒
x
h
(
t
)=
−
sin 2
t
and
x
(
t
x
h
(
t
)+
x
p
(
t
c
1
cos
t
+
c
2
sin
t
+
c
3
cos
√
6
t
+
c
4
sin
√
6
t
−
sin 2
t.
From the frst equation in the original system, we have
y
(
t
1
2
(
x
0
+5
x
)
=2
c
1
cos
t
+2
c
2
sin
t
−
1
2
c
3
cos
√
6
t
−
1
2
c
4
sin
√
6
t
−
1
2
sin 2
We determine constants
c
1
and
c
3
using the initial conditions
x
(0) = 0 and
y
(0) = 1.
0=
x
(0) =
c
1
+
c
3
,
1=
y
(0) = 2
c
1
−
c
3
/
2
⇒
c
3
=
−
c
1
,
2
c
1
−
(
−
c
1
)
/
2=1
⇒
c
3
=
−
2
/
5
,
c
1
/
5
.
To fnd
c
2
and
c
4
, compute
x
0
(
t
)and
y
0
(
t
), evaluate these ±unctions at
t
= 0, and use the other
two initial conditions,
x
0
(0) =
y
0
(0) = 0. This yields
x
0
(0) =
c
2
+
√
6
c
4
−
2
,
y
0
(0) = 2
c
2
−
√
6
c
4
/
2
−
1
⇒
c
4
=
√
6
/
5
,
c
2
=4
/
5
.
There±ore, the required solution is
x
(
t
2
5
cos
t
+
4
5
sin
t
−
2
5
cos
√
6
t
+
√
6
5
sin
√
6
t
−
sin 2
t,
y
(
t
4
5
cos
t
+
8
5
sin
t
+
1
5
cos
√
6
t
−
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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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