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Exercises 5.2
Substituting these expressions into the nonhomogeneous equation given in (5.3) yields
2
x
0
p
−
2
x
p
=2
(
−
A
sin
t
+
B
cos
t
)
−
2(
A
cos
t
+
B
sin
t
)
=(
2
B
−
2
A
)cos
t
+(
−
2
A
−
2
B
)sin
t
=
−
cos
t.
By equating coeﬃcients we obtain
2
B
−
2
A
=
−
1a
n
d
−
2
A
−
2
B
=0
.
By solving these two equations simultaneously for
A
and
B
,weseethat
A
=
1
4
and
B
=
−
1
4
.
Thus, a particular solution to the nonhomogeneous equation given in (5.3) will be
x
p
(
t
)=
1
4
cos
t
−
1
4
sin
t
and a general solution to the nonhomogeneous equation (5.3) will be
x
(
t
x
h
(
t
)+
x
p
(
t
C
1
e
t
+
1
4
cos
t
−
1
4
sin
t.
We now must Fnd a function
y
(
t
). To do this, we subtract the second of the two di±erential
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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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