Chapter 6
=
(
−
1)
n
+
k
W
[
y
1
, . . . , y
k
−
1
, y
k
+1
, . . . , y
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
) = cos 3
x
. All these functions, except
p
2
(
x
), are continuous on (
−∞
,
∞
), and
p
2
(
x
) is defined and continuous on (0
,
∞
). Thus, Theorem 1 guarantees the existence of
a unique solution on (0
,
∞
).
(b)
By dividing both sides of the given differential equation by
x
2
−
1, we rewrite the equation
in standard form, that is,
y
+
sin
x
x
2
−
1
y
+
√
x
+ 4
x
2
−
1
y
+
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
+ 4
x
2
−
1
,
p
3
(
x
) =
e
x
x
2
−
1
,
and
g
(
x
) =
x
2
+ 3
x
2
−
1
.
Functions
p
1
(
x
),
p
3
(
x
), and
g
(
x
) are defined and continuous on (
−∞
,
∞
) except
x
=
±
1;
p
2
(
x
) is defined and continuous on
{
x
≥ −
4
, x
=
±
1
}
. Thus, the common domain for
p
1
(
x
),
p
2
(
x
),
p
3
(
x
), and
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 Spring '10
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 Math, Differential Equations, Linear Algebra, Algebra, Equations, Sin, unique solution, AirTrain Newark

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