Solution to Problem 4.15
March 14, 2011
Here,
R
(
x
) = 1,
P
(
x
) =
x
−
1,
Q
(
x
) =
−
αx
.
We therefore have
f
(
s
+
k
) = (
s
+
k
)(
s
+
k
−
1)
−
(
s
+
k
) = (
s
+
k
)(
s
+
k
−
2)
(1)
and
g
n
(
s
+
k
) = (
s
+
k
−
1)
−
α
if
n
= 1 and 0 if
n >
1.
(2)
The indicial equation is
f
(
s
) = (
s
)(
s
−
2) = 0,
(3)
so that
s
= 0 or 2.
For
k
≥
1, we therefore have
(
s
+
k
)(
s
+
k
−
2)
A
k
=
−
((
s
+
k
−
1)
−
α
)
A
k

1
.
(4)
Suppose
s
= 2. Then
k
(
k
+ 2)
A
k
=
−
(
k
+ 1
−
α
)
A
k

1
.
(5)
Or, since
k
(
k
+ 2) is positive for all
k
,
A
k
=
α
−
k
−
1
k
(
k
+ 2)
A
k

1
.
(6)
This uniquely determines all
A
k
(with only
A
0
arbitrary).
If
α
is an integer
≥
2, the series only has ±nitely many terms, and the
solution is a polynomial of degree
α
1
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View Full DocumentSuppose
s
= 0. Then
k
(
k
−
2)
A
k
=
−
(
k
−
1
−
α
)
A
k

1
.
(7)
For
k
= 1,
−
A
1
=
αA
0
.
(8)
For
k
= 2,
0 = (1
−
α
)
A
1
.
(9)
For
k
≥
3,
k
(
k
−
2) is positive, and
A
k
=
α
−
k
+ 1
k
(
k
−
2)
A
k

1
.
(10)
If
α
n
= 0 or 1, we have that
A
1
= 0,
A
0
= 0,
A
2
is arbitrary, and all
subsequent
A
k
. But this contradicts our assumption that
A
0
n
= 0. Therefore,
there is no regular solution with
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 Spring '11
 GLAGOLA

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