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Chapter 6
=(
−
1)
n
+
k
W
[
y
1
,...,y
k
−
1
,y
k
+1
n
](
x
)
.
Finally,
(
−
1)
n
+
k
−
1)
(
n
−
k
)+(2
k
)
−
1)
n
−
k
.
REVIEW PROBLEMS:
page 344
1. (a)
In notation of Theorem 1, we have
p
1
(
x
)
≡
0,
p
2
(
x
)=
−
ln
x
,
p
3
(
x
x
,
p
4
(
x
)
≡
2,
and
g
(
x
)=cos3
x
. All these functions, except
p
2
(
x
), are continuous on (
−∞
,
∞
), and
p
2
(
x
) is de±ned and continuous on (0
,
∞
). Thus, Theorem 1 guarantees the existence of
a unique solution on (0
,
∞
).
(b)
By dividing both sides of the given di²erential equation by
x
2
−
1, we rewrite the equation
in standard form, that is,
y
0
+
sin
x
x
2
−
1
y
0
+
√
x
+4
x
2
−
1
y
0
+
e
x
x
2
−
1
y
=
x
2
+3
x
2
−
1
.
Thus we see that
p
1
(
x
sin
x
x
2
−
1
,p
2
(
x
√
x
x
2
−
1
3
(
x
e
x
x
2
−
1
,
and
g
(
x
x
2
x
2
−
1
.
Functions
p
1
(
x
),
p
3
(
x
), and
g
(
x
) are de±ned and continuous on (
−∞
,
∞
) except
x
=
±
1;
p
2
(
x
) is de±ned and continuous on
{
x
≥−
4
,x
6
=
±
1
}
. Thus, the common domain for
p
1
(
x
),
p
2
(
x
),
p
3
(
x
), and
g
(
x
)is
{
x
4
6
=
±
1
}
, and, in addition, these functions are
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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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