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Chapter 6
17.
Writing the given diferential equation,
x
3
y
0
−
3
x
2
y
0
+6
xy
0
−
6
y
=0
,
in standard Form (17), we see that its coeﬃcients,
−
3
/x
,6
/x
2
,and
−
6
/x
3
are continuous on
the speci±ed interval, which is
x>
0.
Next, substituting
x
,
x
2
x
3
into the diferential equation, we veriFy that these Functions
are indeed solutions.
x
3
(
x
)
0
−
3
x
2
(
x
)
0
x
(
x
)
0
−
6(
x
)=0
−
0+6
x
−
6
x
,
x
3
(
x
2
)
0
−
3
x
2
(
x
2
)
0
x
(
x
2
)
0
−
6(
x
2
−
6
x
2
+12
x
2
−
6
x
2
,
x
3
(
x
3
)
0
−
3
x
2
(
x
3
)
0
x
(
x
3
)
0
−
6(
x
3
)=6
x
3
−
18
x
3
+18
x
3
−
6
x
3
.
Evaluating the Wronskian yields
W
±
x, x
2
,x
3
²
(
x
)=
³
³
³
³
³
³
³
³
xx
2
x
3
12
x
3
x
2
026
x
³
³
³
³
³
³
³
³
=
x
³
³
³
³
³
2
x
3
x
2
26
x
³
³
³
³
³
−
³
³
³
³
³
x
2
x
3
x
³
³
³
³
³
=
x
(
6
x
2
)
−
(
4
x
3
)
=2
x
3
.
Thus
W
[
x, x
2
3
](
x
)
6
=0on(0
,
∞
)andso
{
x, x
2
3
}
is a Fundamental solution set For the
given diferential equation. We involve Theorem 2 to conclude that
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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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