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Exercises 6.4
The auxiliary equation,
r
2
(
r
2
−
2) = 0, has roots
r
=
±
√
2 and a double root
r
= 0. Hence,
y
h
(
t
)=
c
1
+
c
2
t
+
c
3
e
√
2
t
+
c
4
e
−
√
2
t
is a general solution to the homogeneous equation coresponding to (6.22). A particular solution
to (6.22) has the form
y
p
(
t
Ae
3
t
. Substitution yields
D
2
(
D
2
−
2
)±
Ae
3
x
²
=
(
D
4
−
2
D
2
Ae
3
x
²
=81
Ae
3
x
−
(2)9
Ae
3
x
=63
Ae
3
x
=8
e
3
x
⇒
y
p
(
t
Ae
3
x
=
8
e
3
x
63
,
and so
y
(
t
y
p
(
t
)+
y
h
(
t
8
e
3
x
63
+
c
1
+
c
2
t
+
c
3
e
√
2
t
+
c
4
e
−
√
2
t
is a general solution to (6.22). We Fnd
x
(
t
) from the second equation in the original system.
x
(
t
e
3
t
+
y
(
t
)
−
y
0
(
t
)
=
e
3
t
+
³
8
e
3
x
63
+
c
1
+
c
2
t
+
c
3
e
√
2
t
+
c
4
e
−
√
2
t
´
−
³
72
e
3
x
63
+2
c
3
e
√
2
t
c
4
e
−
√
2
t
´
=
−
e
3
x
63
+
c
1
+
c
2
t
−
c
3
e
√
2
t
−
c
4
e
−
√
2
t
.
EXERCISES 6.4:
Method of Variation of Parameters, page 341
1.
To apply the method of variation of parameters, Frst we have to Fnd a fundamental solution
set for the corresponding homogeneous equation, which is
y
0
−
3
y
0
+4
y
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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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