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Chapter 6
W
2
(
x
)=(
−
1)
3
−
2
±
±
±
±
±
xx
−
1
1
−
x
−
2
±
±
±
±
±
=2
x
−
1
,
W
3
(
x
−
1)
3
−
3
±
±
±
±
±
5
15
x
4
±
±
±
±
±
=4
x
5
.
Now we divide both sides of the given equation by
x
3
to obtain an equation in standard form,
that is,
y
0
−
2
x
−
1
y
0
−
5
x
−
2
y
0
+5
x
−
3
y
=
x
−
5
.
Hence, the righthand side,
g
(
x
), in formula (1) on page 339 of the text equals to
x
−
5
. Applying
formula (11), page 340 of the text, yields
v
1
(
x
)=
Z
x
−
5
(
−
6
x
3
)
48
x
2
dx
=
−
1
8
Z
x
−
4
dx
=
1
24
x
−
3
,
v
2
(
x
Z
x
−
5
(2
x
−
1
)
48
x
2
dx
=
1
24
Z
x
−
8
dx
=
−
1
168
x
−
7
,
v
3
(
x
Z
x
−
5
(4
x
5
)
48
x
2
dx
=
1
12
Z
x
−
2
dx
=
−
1
12
x
−
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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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