Solve the hypergeometric equation around all singularities:
x
(
1
−
x
)
y
″
+
{
γ
−
(
1
+
α
+
β
)
x
}
y
′
−
α
β
y
=
0
Let
P
0
(
x
)
=
−
α
β
,
P
1
(
x
)
=
γ
−
(
1
+
α
+
β
)
x
,
P
2
(
x
)
=
x
(
1
−
x
)
Then
P
2
(
0
)
=
P
2
(
1
)
=
0.
Hence, x = 0 and x = 1 are singular points. Let's start with x = 0. To see if it is regular, we study the following limits:
lim
x
→
a
(
x
−
a
)
P
1
(
x
)
P
2
(
x
)
=
lim
x
→
0
(
x
−
0
)
(
γ
−
(
1
+
α
+
β
)
x
)
x
(
1
−
x
)
=
lim
x
→
0
x
(
γ
−
(
1
+
α
+
β
)
x
)
x
(
1
−
x
)
=
γ
lim
x
→
a
(
x
−
a
)
2
P
0
(
x
)
P
2
(
x
)
=
lim
x
→
0
(
x
−
0
)
2
(
−
α
β
)
x
(
1
−
x
)
=
lim
x
→
0
x
2
(
−
α
β
)
x
(
1
−
x
)
=
0
Hence, both limits exist and x = 0 is a regular singular point. Therefore, we assume the solution takes the form
y
=
∑
r
=
0
∞
a
r
x
r
+
c
with a0 ≠ 0. Hence,
y
′
=
∑
r
=
0
∞
a
r
(
r
+
c
)
x
r
+
c
−
1
y
″
=
∑
r
=
0
∞
a
r
(
r
+
c
)
(
r
+
c
−
1
)
x
r
+
c
−
2
.
Substituting these into the hypergeometric equation, we get
x
∑
r
=
0
∞
a
r
(
r
+
c
)
(
r
+
c
−
1
)
x
r
+
c
−
2
−
x
2
∑
r
=
0
∞
a
r
(
r
+
c
)
(
r
+
c
−
1
)
x
r
+
c
−
2
+
γ
∑
r
=
0
∞
a
r
(
r
+
c
)
x
r
+
c
−
1
−
(
1
+
α
+
β
)
x
∑
r
=
0
∞
a
r
(
r
+
c
)
x
r
+
c
−
1
−
α
β
∑
r
=
0
∞
a
r
x
r
+
c
=
0
That is,
∑
r
=
0
∞
a
r
(
r
+
c
)
(
r
+
c
−
1
)
x
r
+
c
−
1
−
∑
r
=
0
∞
a
r
(
r
+
c
)
(
r
+
c
−
1
)
x
r
+
c
+
γ
∑
r
=
0
∞
a
r
(
r
+
c
)
x
r
+
c
−
1
−
(
1
+
α
+
β
)
∑
r
=
0
∞
a
r
(
r
+
c
)
x
r
+
c
−
α
β
∑
r
=
0
∞
a
r
x
r
+
c
=
0
In order to simplify this equation, we need all powers to be the same, equal to r + c − 1, the smallest power. Hence, we switch the indices as follows:
∑
r
=
0
∞
a
r
(
r
+
c
)
(
r
+
c
−
1
)
x
r
+
c
−
1
−
∑
r
=
1
∞
a
r
−
1
(
r
+
c
−
1
)
(
r
+
c
−
2
)
x
r
+
c
−
1
+
γ
∑
r
=
0
∞
a
r
(
r
+
c
)
x
r
+
c
−
1
−
(
1
+
α
+
β
)
∑
r
=
1
∞
a
r
−
1
(
r
+
c
−
1
)
x
r
+
c
−
1
−
α
β
∑
r
=
1
∞
a
r
−
1
x
r
+
c
−
1
=
0
Thus, isolating the first term of the sums starting from 0 we get
a
0
(
c
(
c
−
1
)
+
γ
c
)
x
c
−
1
+
∑
r
=
1
∞
a
r
(
r
+
c
)
(
r
+
c
−
1
)
x
r
+
c
−
1
−
∑
r
=
1
∞
a
r
−
1
(
r
+
c
−
1
)
(
r
+
c
−
2
)
x
r
+
c
−
1
+
γ
∑
r
=
1
∞
a
r
(
r
+
c
)
x
r
+
c
−
1
−
(
1
+
α
+
β
)
∑
r
=
1
∞
a
r
−
1
(
r
+
c
−
1
)
x
r
+
c
−
1
−
α
β
∑
r
=
1
∞
a
r
−
1
x
r
+
c
−
1
=
0
Now, from the linear independence of all powers of x, that is, of the functions 1, x, x2, etc., the coefficients of xk vanish for all k. Hence, from the first term, we have
a
0
(
c
(
c
−
1
)
+
γ
c
)
=
0
which is the indicial equation. Since a0 ≠ 0, we have
c
(
c
−
1
+
γ
)
=
0.
Hence,
c
1
=
0
,
c
2
=
1
−
γ
Also, from the rest of the terms, we have
(
(
r
+
c
)
(
r
+
c
−
1
)
+
γ
(
r
+
c
)
)
a
r
+
(
−
(
r
+
c
−
1
)
(
r
+
c
−
2
)
−
(
1
+
α
+
β
)
(
r
+
c
−
1
)
−
α
β
)
a
r
−
1
=
0
Hence,
a
r
=
(
r
+
c
−
1
)
(
r
+
c
−
2
)
+
(
1
+
α
+
β
)
(
r
+
c
−
1
)
+
α
β
(
r
+
c
)
(
r
+
c
−
1
)
+
γ
(
r
+
c
)
a
r
−
1
=
(
r
+
c
−
1
)
(
r
+
c
+
α
+
β
−
1
)
+
α
β
(
r
+
c
)
(
r
+
c
+
γ
−
1
)
a
r
−
1
But
(
r
+
c
−
1
)
(
r
+
c
+
α
+
β
−
1
)
+
α
β
=
(
r
+
c
−
1
)
(
r
+
c
+
α
−
1
)
+
(
r
+
c
−
1
)
β
+
α
β
=
(
r
+
c
−
1
)
(
r
+
c
+
α
−
1
)
+
β
(
r
+
c
+
α
−
1
)
Hence, we get the recurrence relation
a
r
=
(
r
+
c
+
α
−
1
)
(
r
+
c
+
β
−
1
)
(
r
+
c
)
(
r
+
c
+
γ
−
1
)
a
r
−
1
,
for
r
≥
1.
Let's now simplify this relation by giving ar in terms of a0 instead of ar−1. From the recurrence relation (note: below, expressions of the form (u)r refer to the Pochhammer symbol).
a
1
=
(
c
+
α
)
(
c
+
β
)
(
c
+
1
)
(
c
+
γ
)
a
0
a
2
=
(
c
+
α
+
1
)
(
c
+
β
+
1
)
(
c
+
2
)
(
c
+
γ
+
1
)
a
1
=
(
c
+
α
+
1
)
(
c
+
α
)
(
c
+
β
)
(
c
+
β
+
1
)
(
c
+
2
)
(
c
+
1
)
(
c
+
γ
)
(
c
+
γ
+
1
)
a
0
=
(
c
+
α
)
2
(
c
+
β
)
2
(
c
+
1
)
2
(
c
+
γ
)
2
a
0
a
3
=
(
c
+
α
+
2
)
(
c
+
β
+
2
)
(
c
+
3
)
(
c
+
γ
+
2
)
a
2
=
(
c
+
α
)
2
(
c
+
α
+
2
)
(
c
+
β
)
2
(
c
+
β
+
2
)
(
c
+
1
)
2
(
c
+
3
)
(
c
+
γ
)
2
(
c
+
γ
+
2
)
a
0
=
(
c
+
α
)
3
(
c
+
β
)
3
(
c
+
1
)
3
(
c
+
γ
)
3
a
0
As we can see,
a
r
=
(
c
+
α
)
r
(
c
+
β
)
r
(
c
+
1
)
r
(
c
+
γ
)
r
a
0
,
for
r
≥
0
Hence, our assumed solution takes the form
y
=
a
0
∑
r
=
0
∞
(
c
+
α
)
r
(
c
+
β
)
r
(
c
+
1
)
r
(
c
+
γ
)
r
x
r
+
c
.
We are now ready to study the solutions corresponding to the different cases for c1 − c2 = γ − 1 (this reduces to studying the nature of the parameter γ: whether it is an integer or not).
Then y1 = y|c = 0 and y2 = y|c = 1 − γ. Since
y
=
a
0
∑
r
=
0
∞
(
c
+
α
)
r
(
c
+
β
)
r
(
c
+
1
)
r
(
c
+
γ
)
r
x
r
+
c
,
we have
y
1
=
a
0
∑
r
=
0
∞
(
α
)
r
(
β
)
r
(
1
)
r
(
γ
)
r
x
r
=
a
0
⋅
2
F
1
(
α
,
β
;
γ
;
x
)
y
2
=
a
0
∑
r
=
0
∞
(
α
+
1
−
γ
)
r
(
β
+
1
−
γ
)
r
(
1
−
γ
+
1
)
r
(
1
−
γ
+
γ
)
r
x
r
+
1
−
γ
=
a
0
x
1
−
γ
∑
r
=
0
∞
(
α
+
1
−
γ
)
r
(
β
+
1
−
γ
)
r
(
1
)
r
(
2
−
γ
)
r
x
r
=
a
0
x
1
−
γ
2
F
1
(
α
−
γ
+
1
,
β
−
γ
+
1
;
2
−
γ
;
x
)
Hence,
y
=
A
′
y
1
+
B
′
y
2
.
Let A′ a0 = a and B′ a0 = B. Then
y
=
A
2
F
1
(
α
,
β
;
γ
;
x
)
+
B
x
1
−
γ
2
F
1
(
α
−
γ
+
1
,
β
−
γ
+
1
;
2
−
γ
;
x
)
Then y1 = y|c = 0. Since γ = 1, we have
y
=
a
0
∑
r
=
0
∞
(
c
+
α
)
r
(
c
+
β
)
r
(
c
+
1
)
r
2
x
r
+
c
.
Hence,
y
1
=
a
0
∑
r
=
0
∞
(
α
)
r
(
β
)
r
(
1
)
r
(
1
)
r
x
r
=
a
0
2
F
1
(
α
,
β
;
1
;
x
)
y
2
=
∂
y
∂
c
|
c
=
0
.
To calculate this derivative, let
M
r
=
(
c
+
α
)
r
(
c
+
β
)
r
(
c
+
1
)
r
2
.
Then
ln
(
M
r
)
=
ln
(
(
c
+
α
)
r
(
c
+
β
)
r
(
c
+
1
)
r
2
)
=
ln
(
c
+
α
)
r
+
ln
(
c
+
β
)
r
−
2
ln
(
c
+
1
)
r
But
ln
(
c
+
α
)
r
=
ln
(
(
c
+
α
)
(
c
+
α
+
1
)
⋯
(
c
+
α
+
r
−
1
)
)
=
∑
k
=
0
r
−
1
ln
(
c
+
α
+
k
)
.
Hence,
ln
(
M
r
)
=
∑
k
=
0
r
−
1
ln
(
c
+
α
+
k
)
+
∑
k
=
0
r
−
1
ln
(
c
+
β
+
k
)
−
2
∑
k
=
0
r
−
1
ln
(
c
+
1
+
k
)
=
∑
k
=
0
r
−
1
(
ln
(
c
+
α
+
k
)
+
ln
(
c
+
β
+
k
)
−
2
ln
(
c
+
1
+
k
)
)
Differentiating both sides of the equation with respect to c, we get:
1
M
r
∂
M
r
∂
c
=
∑
k
=
0
r
−
1
(
1
c
+
α
+
k
+
1
c
+
β
+
k
−
2
c
+
1
+
k
)
.
Hence,
∂
M
r
∂
c
=
(
c
+
α
)
r
(
c
+
β
)
r
(
c
+
1
)
r
2
∑
k
=
0
r
−
1
(
1
c
+
α
+
k
+
1
c
+
β
+
k
−
2
c
+
1
+
k
)
.
Now,
y
=
a
0
x
c
∑
r
=
0
∞
(
c
+
α
)
r
(
c
+
β
)
r
(
c
+
1
)
r
2
x
r
=
a
0
x
c
∑
r
=
0
∞
M
r
x
r
.
Hence,
∂
y
∂
c
=
a
0
x
c
ln
(
x
)
∑
r
=
0
∞
(
c
+
α
)
r
(
c
+
β
)
r
(
c
+
1
)
r
2
x
r
+
a
0
x
c
∑
r
=
0
∞
(
(
c
+
α
)
r
(
c
+
β
)
r
(
c
+
1
)
r
2
{
∑
k
=
0
r
−
1
(
1
c
+
α
+
k
+
1
c
+
β
+
k
−
2
c
+
1
+
k
)
}
)
x
r
=
a
0
x
c
∑
r
=
0
∞
(
c
+
α
)
r
(
c
+
β
)
r
(
c
+
1
)
r
)
2
(
ln
x
+
∑
k
=
0
r
−
1
(
1
c
+
α
+
k
+
1
c
+
β
+
k
−
2
c
+
1
+
k
)
)
x
r
.
For c = 0, we get
y
2
=
a
0
∑
r
=
0
∞
(
α
)
r
(
β
)
r
(
1
)
r
2
(
ln
x
+
∑
k
=
0
r
−
1
(
1
α
+
k
+
1
β
+
k
−
2
1
+
k
)
)
x
r
.
Hence, y = C′y1 + D′y2. Let C′a0 = C and D′a0 = D. Then
y
=
C
2
F
1
(
α
,
β
;
1
;
x
)
+
D
∑
r
=
0
∞
(
α
)
r
(
β
)
r
(
1
)
r
2
(
ln
(
x
)
+
∑
k
=
0
r
−
1
(
1
α
+
k
+
1
β
+
k
−
2
1
+
k
)
)
x
r
The value of
γ
is
γ
=
0
,
−
1
,
−
2
,
⋯
. To begin with, we shall simplify matters by concentrating a particular value of
γ
and generalise the result at a later stage. We shall use the value
γ
=
−
2
. The indicial equation has a root at
c
=
0
, and we see from the recurrence relation
a
r
=
(
r
+
c
+
α
−
1
)
(
r
+
c
+
β
−
1
)
(
r
+
c
)
(
r
+
c
−
3
)
a
r
−
1
,
that when
r
=
3
that that denominator has a factor
c
which vanishes when
c
=
0
. In this case, a solution can be obtained by putting
a
0
=
b
0
c
where
b
0
is a constant.
With this substitution, the coefficients of
x
r
vanish when
c
=
0
and
r
<
3
. The factor of
c
in the denominator of the recurrence relation cancels with that of the numerator when
r
≥
3
. Hence, our solution takes the form
y
1
=
b
0
(
−
2
)
×
(
−
1
)
(
(
α
)
3
(
β
)
3
(
3
!
0
!
x
3
+
(
α
)
4
(
β
)
4
4
!
1
!
x
4
+
(
α
)
5
(
β
)
5
5
!
2
!
x
5
+
⋯
)
=
b
0
(
−
2
)
2
∑
r
=
3
∞
(
α
)
r
(
β
)
r
r
!
(
r
−
3
)
!
x
r
=
b
0
(
−
2
)
2
(
α
)
3
(
β
)
3
3
!
∑
r
=
3
∞
(
α
+
3
)
r
−
3
(
β
+
3
)
r
−
3
(
1
+
3
)
r
−
3
(
r
−
3
)
!
x
r
.
If we start the summation at
r
=
0
rather than
r
=
3
we see that
y
1
=
b
0
(
α
)
3
(
β
)
3
(
−
2
)
2
×
3
!
x
3
2
F
1
(
α
+
3
,
β
+
3
;
(
1
+
3
)
;
x
)
.
The result (as we have written it) generalises easily. For
γ
=
1
+
m
, with
m
=
1
,
2
,
3
,
⋯
then
y
1
=
b
0
(
α
)
m
(
β
)
m
(
1
−
m
)
m
−
1
×
m
!
x
m
2
F
1
(
α
+
m
,
β
+
m
;
(
1
+
m
)
;
x
)
.
Obviously, if
γ
=
−
2
, then
m
=
3
. The expression for
y
1
(
x
)
we have just given looks a little inelegant since we have a multiplicative constant apart from the usual arbitrary multiplicative constant
b
0
. Later, we shall see that we can recast things in such a way that this extra constant never appears
The other root to the indicial equation is
c
=
1
−
γ
=
3
, but this gives us (apart from a multiplicative constant) the same result as found using
c
=
0
. This means we must take the partial derivative (w.r.t.
c
) of the usual trial solution in order to find a second independent solution. If we define the linear operator
L
as
L
=
x
(
1
−
x
)
d
2
d
x
2
−
(
α
+
β
+
1
)
x
d
d
x
+
γ
d
d
x
−
α
β
,
then since
γ
=
−
2
in our case,
L
c
∑
r
=
0
∞
b
r
(
c
)
x
r
=
b
0
c
2
(
c
−
3
)
.
(We insist that
b
0
≠
0
.) Taking the partial derivative w.r.t
c
,
L
∂
∂
c
c
∑
r
=
0
∞
b
r
(
c
)
x
r
+
c
=
b
0
(
3
c
2
−
6
c
)
.
Note that we must evaluate the partial derivative at
c
=
0
(and not at the other root
c
=
3
). Otherwise the right hand side is non-zero in the above, and we do not have a solution of
L
y
(
x
)
=
0
. The factor
c
is not cancelled for
r
=
0
,
1
and
r
=
2
. This part of the second independent solution is
[
∂
∂
c
b
0
(
c
+
c
(
c
+
α
)
(
c
+
β
)
(
c
+
1
)
(
c
−
2
)
x
+
c
(
c
+
α
)
(
c
+
α
+
1
)
(
c
+
β
)
(
c
+
β
+
1
)
(
c
+
1
)
(
c
+
2
)
(
c
−
2
)
(
c
−
1
)
x
2
)
]
|
c
=
0
.
=
b
0
(
1
+
α
β
1
!
×
(
−
2
)
x
+
α
(
α
+
1
)
β
(
β
+
1
)
2
!
×
(
−
2
)
×
(
−
1
)
x
2
)
=
b
0
∑
r
=
0
3
−
1
(
α
)
r
(
β
)
r
r
!
(
1
−
3
)
r
x
r
.
Now we can turn our attention to the terms where the factor
c
cancels. First
c
b
3
=
b
0
(
c
−
1
)
(
c
−
2
)
c
(
c
+
α
)
(
c
+
α
+
1
)
(
c
+
α
+
2
)
(
c
+
β
)
(
c
+
β
+
1
)
(
c
+
β
+
2
)
c
(
c
+
1
)
(
c
+
2
)
(
c
+
3
)
.
After this, the recurrence relations give us
c
b
4
=
c
b
3
(
c
)
(
c
+
α
+
3
)
(
c
+
β
+
3
)
(
c
+
1
)
(
c
+
4
)
)
.
c
b
5
=
c
b
3
(
c
)
(
c
+
α
+
3
)
(
c
+
α
+
4
)
(
c
+
β
+
3
)
(
c
+
β
+
4
)
)
(
c
+
2
)
(
c
+
1
)
(
c
+
5
)
(
c
+
4
)
.
So, if
r
≥
3
we have
c
b
r
=
b
0
(
c
−
1
)
(
c
−
2
)
(
c
+
α
)
r
(
c
+
β
)
r
(
c
+
1
)
r
−
3
(
c
+
1
)
r
.
We need the partial derivatives
∂
c
b
3
(
c
)
∂
c
|
c
=
0
=
b
0
(
1
−
3
)
3
−
1
(
α
)
3
(
β
)
3
0
!
3
!
[
1
1
+
1
2
+
1
α
+
1
α
+
1
+
1
α
+
2
+
1
β
+
1
β
+
1
+
1
β
+
2
−
1
1
−
1
2
−
1
3
]
.
Similarly, we can write
∂
c
b
4
(
c
)
∂
c
|
c
=
0
=
b
0
(
1
−
3
)
3
−
1
(
α
)
4
(
β
)
4
1
!
4
!
[
1
1
+
1
2
+
∑
k
=
0
k
=
3
1
α
+
k
+
∑
k
=
0
k
=
3
1
β
+
k
−
1
1
−
1
2
−
1
3
−
1
4
−
1
1
]
,
and
∂
c
b
5
(
c
)
∂
c
|
c
=
0
=
b
0
(
1
−
3
)
3
−
1
(
α
)
5
(
β
)
5
2
!
5
!
[
1
1
+
1
2
+
∑
k
=
0
k
=
4
1
α
+
k
+
∑
k
=
0
k
=
4
1
β
+
k
−
1
1
−
1
2
−
1
3
−
1
4
−
1
5
−
1
1
−
1
2
]
.
It becomes clear that for
r
≥
3
∂
c
b
r
(
c
)
∂
c
|
c
=
0
=
b
0
(
1
−
3
)
3
−
1
(
α
)
r
(
β
)
r
(
r
−
3
)
!
r
!
[
H
2
+
∑
k
=
0
k
=
r
−
1
1
α
+
k
+
∑
k
=
0
k
=
r
−
1
1
β
+
k
−
H
r
−
H
r
−
3
]
.
Here,
H
k
is the
k
th partial sum of the harmonic series, and by definition
H
0
=
0
and
H
1
=
1
.
Putting these together, for the case
γ
=
−
2
we have a second solution
y
2
(
x
)
=
log
x
×
b
0
(
−
2
)
2
∑
r
=
3
∞
(
α
)
r
(
β
)
r
r
!
(
r
−
3
)
!
x
r
+
b
0
∑
r
=
0
3
−
1
(
α
)
r
(
β
)
r
r
!
(
1
−
3
)
r
x
r
+
b
0
(
−
2
)
2
∑
r
=
3
∞
(
α
)
r
(
β
)
r
(
r
−
3
)
!
r
!
[
H
2
+
∑
k
=
0
k
=
r
−
1
1
α
+
k
+
∑
k
=
0
k
=
r
−
1
1
β
+
k
−
H
r
−
H
r
−
3
]
x
r
.
The two independent solutions for
γ
=
1
−
m
(where
m
is a positive integer) are then
y
1
(
x
)
=
1
(
1
−
m
)
m
−
1
∑
r
=
m
∞
(
α
)
r
(
β
)
r
r
!
(
r
−
m
)
!
x
r
and
y
2
(
x
)
=
log
x
×
y
1
(
x
)
+
∑
r
=
0
m
−
1
(
α
)
r
(
β
)
r
r
!
(
1
−
m
)
r
x
r
+
1
(
1
−
m
)
m
−
1
∑
r
=
m
∞
(
α
)
r
(
β
)
r
(
r
−
m
)
!
r
!
[
H
m
−
1
+
∑
k
=
0
k
=
r
−
1
1
α
+
k
+
∑
k
=
0
k
=
r
−
1
1
β
+
k
−
H
r
−
H
r
−
m
]
x
r
.
The general solution is as usual
y
(
x
)
=
A
y
1
(
x
)
+
B
y
2
(
x
)
where
A
and
B
are arbitrary constants. Now, if the reader consults a ``standard solution" for this case, such as given by Abramowitz and Stegun in §15.5.21 (which we shall write down at the end of the next section) it shall be found that the
y
2
solution we have found looks somewhat different from the standard solution. In our solution for
y
2
, the first term in the infinite series part of
y
2
is a term in
x
m
. The first term in the corresponding infinite series in the standard solution is a term in
x
m
+
1
. The
x
m
term is missing from the standard solution. Nonetheless, the two solutions are entirely equivalent.
The reason for the apparent discrepancy between the solution given above and the standard solution in Abramowitz and Stegun §15.5.21 is that there are an infinite number of ways in which to represent the two independent solutions of the hypergeometric ODE. In the last section, for instance, we replaced
a
0
with
b
0
c
. Suppose though, we are given some function
h
(
c
)
which is continuous and finite everywhere in an arbitrarily small interval about
c
=
0
. Suppose we are also given
h
(
c
)
|
c
=
0
≠
0
,
and
d
h
d
c
|
c
=
0
≠
0.
Then, if instead of replacing
a
0
with
b
0
c
we replace
a
0
with
b
0
h
(
c
)
c
, we still find we have a valid solution of the hypergeometric equation. Clearly, we have an infinity of possibilities for
h
(
c
)
. There is however a ``natural choice" for
h
(
c
)
. Suppose that
c
b
N
(
c
)
=
b
0
f
(
c
)
is the first non zero term in the first
y
1
(
x
)
solution with
c
=
0
. If we make
h
(
c
)
the reciprocal of
f
(
c
)
, then we won't have a multiplicative constant involved in
y
1
(
x
)
as we did in the previous section. From another point of view, we get the same result if we ``insist" that
a
N
is independent of
c
, and find
a
0
(
c
)
by using the recurrence relations backwards.
For the first
(
c
=
0
)
solution, the function
h
(
c
)
gives us (apart from multiplicative constant) the same
y
1
(
x
)
as we would have obtained using
h
(
c
)
=
1
. Suppose that using
h
(
c
)
=
1
gives rise to two independent solutions
y
1
(
x
)
and
y
2
(
x
)
. In the following we shall denote the solutions arrived at given some
h
(
c
)
≠
1
as
y
~
1
(
x
)
and
y
~
2
(
x
)
.
The second solution requires us to take the partial derivative w.r.t
c
, and substituting the usual trial solution gives us
L
∂
∂
c
∑
r
=
0
∞
c
h
(
c
)
b
r
x
r
+
c
=
b
0
(
d
h
d
c
c
2
(
c
−
1
)
+
2
c
h
(
c
)
(
c
−
1
)
+
h
(
c
)
c
2
)
.
The operator
L
is the same linear operator discussed in the previous section. That is to say, the hypergeometric ODE is represented as
L
y
(
x
)
=
0
.
Evaluating the left hand side at
c
=
0
will give us a second independent solution. Note that this second solution
y
~
2
is in fact a linear combination of
y
1
(
x
)
and
y
2
(
x
)
.
Any two independent linear combinations (
y
~
1
and
y
~
2
) of
y
1
and
y
2
are independent solutions of
L
y
=
0
.
The general solution can be written as a linear combination of
y
~
1
and
y
~
2
just as well as linear combinations of
y
1
and
y
2
.
We shall review the special case where
γ
=
1
−
3
=
−
2
that was considered in the last section. If we ``insist"
a
3
(
c
)
=
c
o
n
s
t
.
, then the recurrence relations yield
a
2
=
a
3
c
(
3
+
c
)
(
2
+
α
+
c
)
(
2
+
β
+
c
)
,
a
1
=
a
3
c
(
2
+
c
)
(
3
+
c
)
(
c
−
1
)
(
1
+
α
+
c
)
(
2
+
α
+
c
)
(
1
+
β
+
c
)
(
2
+
β
+
c
)
,
and
a
0
=
a
3
c
(
1
+
c
)
(
2
+
c
)
(
3
+
c
)
(
c
−
1
)
(
c
−
2
)
(
α
+
c
)
3
(
β
+
c
)
3
=
b
0
c
h
(
c
)
.
These three coefficients are all zero at
c
=
0
as expected. We have three terms involved in
y
2
(
x
)
by taking the partiial derivative w.r.t
c
, we denote the sum of the three terms involving these coefficients as
S
3
where
S
3
=
[
∂
∂
c
(
a
0
(
c
)
x
c
+
a
1
(
c
)
x
c
+
1
+
a
2
(
c
)
x
c
+
2
)
]
c
=
0
,
=
a
3
[
3
×
2
×
1
(
−
2
)
×
(
−
1
)
(
α
)
3
(
β
)
3
x
3
−
3
+
3
×
2
×
(
−
1
)
(
α
+
1
)
(
α
+
2
)
(
β
+
1
)
(
β
+
2
)
x
3
−
2
+
3
(
α
+
2
)
(
β
+
2
)
1
x
3
−
1
]
.
The reader may confirm that we can tidy this up and make it easy to generalise by putting
S
3
=
−
a
3
∑
r
=
1
3
(
−
3
)
r
(
r
−
1
)
!
(
1
−
α
−
3
)
r
(
1
−
β
−
3
)
r
x
3
−
r
.
Next we can turn to the other coefficients, the recurrence relations yield
a
4
=
a
3
(
3
+
c
+
α
)
(
3
+
c
+
β
)
(
4
+
c
)
(
1
+
c
)
a
5
=
a
3
(
4
+
c
+
α
)
(
3
+
c
+
α
)
(
4
+
c
+
β
)
(
3
+
c
+
α
(
5
+
c
)
(
4
+
c
)
(
1
+
c
)
(
2
+
c
)
Setting
c
=
0
gives us
y
~
1
(
x
)
=
a
3
x
3
∑
r
=
0
∞
(
α
+
3
)
r
(
β
+
3
)
r
(
3
+
1
)
r
r
!
x
r
=
a
3
x
3
2
F
1
(
α
+
3
,
β
+
3
;
(
1
+
3
)
;
z
)
.
This is (apart from the multiplicative constant
(
a
)
3
(
b
)
3
/
2
) the same as
y
1
(
x
)
. Now, to find
y
~
2
we need partial derivatives
∂
a
4
∂
c
|
c
=
0
=
a
3
[
(
3
+
c
+
α
)
(
3
+
c
+
β
)
(
4
+
c
)
(
1
+
c
)
(
1
α
+
3
+
c
+
1
β
+
3
+
c
−
1
4
+
c
−
1
1
+
c
)
]
c
=
0
=
a
3
(
3
+
α
)
1
(
3
+
β
)
1
(
1
+
3
)
1
×
1
(
1
α
+
3
+
1
β
+
3
−
1
4
−
1
1
)
.
Then
∂
a
5
∂
c
|
c
=
0
=
a
3
(
3
+
α
)
2
(
3
+
β
)
2
(
1
+
3
)
2
×
1
×
2
(
1
α
+
3
+
1
α
+
4
+
1
β
+
3
+
1
β
+
4
−
1
4
−
1
5
−
1
1
−
1
2
)
.
we can re-write this as
∂
a
5
∂
c
|
c
=
0
=
a
3
(
3
+
α
)
2
(
3
+
β
)
2
(
1
+
3
)
2
×
2
!
[
∑
k
=
0
1
(
1
α
+
3
+
k
+
1
β
+
3
+
k
)
+
∑
k
=
1
3
1
k
−
∑
k
=
1
5
1
k
−
1
1
−
1
2
]
.
The pattern soon becomes clear, and for
r
=
1
,
2
,
3
,
⋯
∂
a
r
+
3
∂
c
|
c
=
0
=
a
3
(
3
+
α
)
r
(
3
+
β
)
r
(
1
+
3
)
r
×
r
!
[
∑
k
=
0
r
−
1
(
1
α
+
3
+
k
+
1
β
+
3
+
k
)
+
∑
k
=
1
3
1
k
−
∑
k
=
1
r
+
3
1
k
−
∑
k
=
1
r
1
k
]
.
Clearly, for
r
=
0
,
∂
a
3
∂
c
|
c
=
0
=
0.
The infinite series part of
y
~
2
is
S
∞
, where
S
∞
=
x
3
∑
r
=
1
∞
∂
a
r
+
3
∂
c
|
c
=
0
x
r
.
Now we can write (disregarding the arbitrary constant) for
γ
=
1
−
m
y
~
1
(
x
)
=
x
3
2
F
1
(
α
+
m
,
β
+
m
;
1
+
m
;
z
)
y
~
2
(
x
)
=
y
~
1
(
x
)
log
x
−
∑
r
=
1
m
(
−
m
)
r
(
r
−
1
)
!
(
1
−
α
−
m
)
r
(
1
−
β
−
m
)
r
x
m
−
r
.
+
x
3
∑
r
=
0
∞
(
α
+
m
)
r
(
β
+
m
)
r
(
1
+
m
)
r
×
r
!
[
∑
k
=
0
r
−
1
(
1
α
+
m
+
k
+
1
β
+
m
+
k
)
+
∑
k
=
1
3
1
k
−
∑
k
=
1
r
+
3
1
k
−
∑
k
=
1
r
1
k
]
x
r
.
Some authors prefer to express the finite sums in this last result using the digamma function
ψ
(
x
)
. In particular, the following results are used
H
n
=
ψ
(
n
+
1
)
+
γ
e
m
.
Here,
γ
e
m
=
0.5772156649
=
ψ
(
1
)
is the Euler-Mascheroni constant. Also
∑
k
=
0
n
−
1
1
z
+
k
=
ψ
(
z
+
n
)
−
ψ
(
z
)
.
With these results we obtain the form given in Abramamowitz and Stegun §15.5.21, namely
y
~
2
(
x
)
=
y
~
1
(
x
)
log
x
−
∑
r
=
1
m
(
−
m
)
r
(
r
−
1
)
!
(
1
−
α
−
m
)
r
(
1
−
β
−
m
)
r
x
m
−
r
.
+
x
3
∑
r
=
0
∞
(
α
+
m
)
r
(
β
+
m
)
r
(
1
+
m
)
r
×
r
!
[
ψ
(
α
+
r
+
m
)
−
ψ
(
α
+
m
)
+
ψ
(
β
+
r
+
m
)
−
ψ
(
β
+
m
)
−
ψ
(
r
+
1
+
m
)
−
ψ
(
r
+
1
)
+
ψ
(
1
+
m
)
+
ψ
(
1
)
]
x
r
.
In this section, we shall concentrate on the ``standard solution", and we shall not replace
a
0
with
b
0
(
c
−
1
+
γ
)
. We shall put
γ
=
1
+
m
where
m
=
1
,
2
,
3
,
⋯
. For the root
c
=
1
−
γ
of the indicial eqauation we had
A
r
=
[
A
r
−
1
(
r
+
α
−
1
+
c
)
(
r
+
β
−
1
+
c
)
(
r
+
c
)
(
r
+
c
+
γ
−
1
)
]
c
=
1
−
γ
=
A
r
−
1
(
r
+
α
−
γ
)
(
r
+
β
−
γ
)
(
r
+
1
−
γ
)
(
r
)
,
where
r
≥
1
in which case we are in trouble if
r
=
γ
−
1
=
m
. For instance, if
γ
=
4
, the denominator in the recurrence relations vanishes for
r
=
3
. We can use exactly the same methods that we have just used for the standard solution in the last section. We shall not (in the instance where
γ
=
4
) replace
a
0
with
b
0
(
c
+
3
)
as this will not give us the standard form of solution that we are after. Rather, we shall ``insist" that
A
3
=
c
o
n
s
t
.
as we did in the standard solution for
γ
=
−
2
in the last section. (Recall that this defined the function
h
(
c
)
and that
a
0
will now be replaced with
b
0
(
c
+
3
)
h
(
c
)
.) Then we may work out the coefficients of
x
0
to
x
2
as functions of
c
using the recurrence relations backwards. There is nothing new to add here, and the reader may use the same methods as used in the last section to find the results of §15.5.18 and §15.5.19, these are
y
1
=
2
F
1
(
α
,
β
;
1
+
m
;
x
)
,
and
y
2
=
2
F
1
(
α
,
β
;
1
+
m
;
x
)
log
x
+
z
m
∑
r
=
1
∞
(
α
)
r
(
β
)
r
r
!
(
1
+
m
)
r
[
ψ
(
α
+
r
)
−
ψ
(
α
)
+
ψ
(
β
+
k
)
−
ψ
(
β
)
−
ψ
(
m
+
1
+
r
)
+
ψ
(
m
+
1
)
−
ψ
(
r
+
1
)
+
ψ
(
1
)
]
z
r
−
∑
k
=
1
m
(
k
−
1
)
!
(
−
m
)
k
(
1
−
α
)
k
(
1
−
β
)
k
z
−
r
.
Note that the powers of
z
in the finite sum part of
y
2
(
x
)
are now negative so that this sum diverges as
z
→
0
$
Let us now study the singular point x = 1. To see if it is regular,
lim
x
→
a
(
x
−
a
)
P
1
(
x
)
P
2
(
x
)
=
lim
x
→
1
(
x
−
1
)
(
γ
−
(
1
+
α
+
β
)
x
)
x
(
1
−
x
)
=
lim
x
→
1
−
(
γ
−
(
1
+
α
+
β
)
x
)
x
=
1
+
α
+
β
−
γ
lim
x
→
a
(
x
−
a
)
2
P
0
(
x
)
P
2
(
x
)
=
lim
x
→
1
(
x
−
1
)
2
(
−
α
β
)
x
(
1
−
x
)
=
lim
x
→
1
(
x
−
1
)
α
β
x
=
0
Hence, both limits exist and x = 1 is a regular singular point. Now, instead of assuming a solution on the form
y
=
∑
r
=
0
∞
a
r
(
x
−
1
)
r
+
c
,
we will try to express the solutions of this case in terms of the solutions for the point x = 0. We proceed as follows: we had the hypergeometric equation
x
(
1
−
x
)
y
″
+
(
γ
−
(
1
+
α
+
β
)
x
)
y
′
−
α
β
y
=
0.
Let z = 1 − x. Then
d
y
d
x
=
d
y
d
z
×
d
z
d
x
=
−
d
y
d
z
=
−
y
′
d
2
y
d
x
2
=
d
d
x
(
d
y
d
x
)
=
d
d
x
(
−
d
y
d
z
)
=
d
d
z
(
−
d
y
d
z
)
×
d
z
d
x
=
d
2
y
d
z
2
=
y
″
Hence, the equation takes the form
z
(
1
−
z
)
y
″
+
(
α
+
β
−
γ
+
1
−
(
1
+
α
+
β
)
z
)
y
′
−
α
β
y
=
0.
Since z = 1 − x, the solution of the hypergeometric equation at x = 1 is the same as the solution for this equation at z = 0. But the solution at z = 0 is identical to the solution we obtained for the point x = 0, if we replace each γ by α + β − γ + 1. Hence, to get the solutions, we just make this substitution in the previous results. Note also that for x = 0, c1 = 0 and c2 = 1 − γ. Hence, in our case, c1 = 0 while c2 = γ − α − β. Let us now write the solutions. In the following we replaced each z by 1 - x.
To simplify notation from now on denote γ − α − β by Δ, therefore γ = Δ + α + β.
y
=
A
{
2
F
1
(
α
,
β
;
−
Δ
+
1
;
1
−
x
)
}
+
B
{
(
1
−
x
)
Δ
2
F
1
(
Δ
+
β
,
Δ
+
α
;
Δ
+
1
;
1
−
x
)
}
y
=
C
{
2
F
1
(
α
,
β
;
1
;
1
−
x
)
}
+
D
{
∑
r
=
0
∞
(
α
)
r
(
β
)
r
(
1
)
r
2
(
ln
(
1
−
x
)
+
∑
k
=
0
r
−
1
(
1
α
+
k
+
1
β
+
k
−
2
1
+
k
)
)
(
1
−
x
)
r
}
y
=
E
{
1
(
−
Δ
+
1
)
Δ
−
1
∑
r
=
1
−
Δ
−
α
−
β
∞
(
α
)
r
(
β
)
r
(
1
)
r
(
1
)
r
−
Δ
(
1
−
x
)
r
}
+
+
F
{
(
1
−
x
)
Δ
∑
r
=
0
∞
(
Δ
)
(
Δ
+
α
)
r
(
Δ
+
β
)
r
(
Δ
+
1
)
r
(
1
)
r
(
ln
(
1
−
x
)
+
1
Δ
+
∑
k
=
0
r
−
1
(
1
Δ
+
α
+
k
+
1
Δ
+
β
+
k
−
1
Δ
+
1
+
k
−
1
1
+
k
)
)
(
1
−
x
)
r
}
y
=
G
{
(
1
−
x
)
Δ
(
Δ
+
1
)
−
Δ
−
1
∑
r
=
−
Δ
∞
(
Δ
+
α
)
r
(
Δ
+
β
)
r
(
1
)
r
(
1
)
r
+
Δ
(
1
−
x
)
r
}
+
+
H
{
∑
r
=
0
∞
(
Δ
)
(
Δ
+
α
)
r
(
Δ
+
β
)
r
(
Δ
+
1
)
r
(
1
)
r
(
ln
(
1
−
x
)
−
1
Δ
+
∑
k
=
0
r
−
1
(
1
α
+
k
+
1
β
+
k
−
1
−
Δ
+
1
+
k
−
1
1
+
k
)
)
(
1
−
x
)
r
}
Finally, we study the singularity as x → ∞. Since we can't study this directly, we let x = s−1. Then the solution of the equation as x → ∞ is identical to the solution of the modified equation when s = 0. We had
x
(
1
−
x
)
y
″
+
(
γ
−
(
1
+
α
+
β
)
x
)
y
′
−
α
β
y
=
0
d
y
d
x
=
d
y
d
s
×
d
s
d
x
=
−
s
2
×
d
y
d
s
=
−
s
2
y
′
d
2
y
d
x
2
=
d
d
x
(
d
y
d
x
)
=
d
d
x
(
−
s
2
×
d
y
d
s
)
=
d
d
s
(
−
s
2
×
d
y
d
s
)
×
d
s
d
x
=
(
(
−
2
s
)
×
d
y
d
s
+
(
−
s
2
)
d
2
y
d
s
2
)
×
(
−
s
2
)
=
2
s
3
y
′
+
s
4
y
″
Hence, the equation takes the new form
1
s
(
1
−
1
s
)
(
2
s
3
y
′
+
s
4
y
″
)
+
(
γ
−
(
1
+
α
+
β
)
1
s
)
(
−
s
2
y
′
)
−
α
β
y
=
0
which reduces to
(
s
3
−
s
2
)
y
″
+
(
(
2
−
γ
)
s
2
+
(
α
+
β
−
1
)
s
)
y
′
−
α
β
y
=
0.
Let
P
0
(
s
)
=
−
α
β
,
P
1
(
s
)
=
(
2
−
γ
)
s
2
+
(
α
+
β
−
1
)
s
,
P
2
(
s
)
=
s
3
−
s
2
.
As we said, we shall only study the solution when s = 0. As we can see, this is a singular point since P2(0) = 0. To see if it is regular,
lim
s
→
a
(
s
−
a
)
P
1
(
s
)
P
2
(
s
)
=
lim
s
→
0
(
s
−
0
)
(
(
2
−
γ
)
s
2
+
(
α
+
β
−
1
)
s
)
s
3
−
s
2
=
lim
s
→
0
(
2
−
γ
)
s
2
+
(
α
+
β
−
1
)
s
s
2
−
s
=
lim
s
→
0
(
2
−
γ
)
s
+
(
α
+
β
−
1
)
s
−
1
=
1
−
α
−
β
.
lim
s
→
a
(
s
−
a
)
2
P
0
(
s
)
P
2
(
s
)
=
lim
s
→
0
(
s
−
0
)
2
(
−
α
β
)
s
3
−
s
2
=
lim
s
→
0
(
−
α
β
)
s
−
1
=
α
β
.
Hence, both limits exist and s = 0 is a regular singular point. Therefore, we assume the solution takes the form
y
=
∑
r
=
0
∞
a
r
s
r
+
c
with a0 ≠ 0. Hence,
y
′
=
∑
r
=
0
∞
a
r
(
r
+
c
)
s
r
+
c
−
1
y
″
=
∑
r
=
0
∞
a
r
(
r
+
c
)
(
r
+
c
−
1
)
s
r
+
c
−
2
Substituting in the modified hypergeometric equation we get
(
s
3
−
s
2
)
y
″
+
(
(
2
−
γ
)
s
2
+
(
α
+
β
−
1
)
s
)
y
′
−
(
α
β
)
y
=
0
And therefore:
(
s
3
−
s
2
)
∑
r
=
0
∞
a
r
(
r
+
c
)
(
r
+
c
−
1
)
s
r
+
c
−
2
+
(
(
2
−
γ
)
s
2
+
(
α
+
β
−
1
)
s
)
∑
r
=
0
∞
a
r
(
r
+
c
)
s
r
+
c
−
1
−
(
α
β
)
∑
r
=
0
∞
a
r
s
r
+
c
=
0
i.e.,
∑
r
=
0
∞
a
r
(
r
+
c
)
(
r
+
c
−
1
)
s
r
+
c
+
1
−
∑
r
=
0
∞
a
r
(
r
+
c
)
(
r
+
c
−
1
)
x
r
+
c
+
(
2
−
γ
)
∑
r
=
0
∞
a
r
(
r
+
c
)
s
r
+
c
+
1
+
(
α
+
β
−
1
)
∑
r
=
0
∞
a
r
(
r
+
c
)
s
r
+
c
−
α
β
∑
r
=
0
∞
a
r
s
r
+
c
=
0.
In order to simplify this equation, we need all powers to be the same, equal to r + c, the smallest power. Hence, we switch the indices as follows
∑
r
=
1
∞
a
r
−
1
(
r
+
c
−
1
)
(
r
+
c
−
2
)
s
r
+
c
−
∑
r
=
0
∞
a
r
(
r
+
c
)
(
r
+
c
−
1
)
s
r
+
c
+
(
2
−
γ
)
∑
r
=
1
∞
a
r
−
1
(
r
+
c
−
1
)
s
r
+
c
+
+
(
α
+
β
−
1
)
∑
r
=
0
∞
a
r
(
r
+
c
)
s
r
+
c
−
α
β
∑
r
=
0
∞
a
r
s
r
+
c
=
0
Thus, isolating the first term of the sums starting from 0 we get
a
0
(
−
(
c
)
(
c
−
1
)
+
(
α
+
β
−
1
)
(
c
)
−
α
β
)
s
c
+
∑
r
=
1
∞
a
r
−
1
(
r
+
c
−
1
)
(
r
+
c
−
2
)
s
r
+
c
−
∑
r
=
1
∞
a
r
(
r
+
c
)
(
r
+
c
−
1
)
x
r
+
c
+
+
(
2
−
γ
)
∑
r
=
1
∞
a
r
−
1
(
r
+
c
−
1
)
s
r
+
c
+
(
α
+
β
−
1
)
∑
r
=
1
∞
a
r
(
r
+
c
)
s
r
+
c
−
α
β
∑
r
=
1
∞
a
r
s
r
+
c
=
0
Now, from the linear independence of all powers of s (i.e., of the functions 1, s, s2, ...), the coefficients of sk vanish for all k. Hence, from the first term we have
a
0
(
−
(
c
)
(
c
−
1
)
+
(
α
+
β
−
1
)
(
c
)
−
α
β
)
=
0
which is the indicial equation. Since a0 ≠ 0, we have
(
c
)
(
−
c
+
1
+
α
+
β
−
1
)
−
α
β
)
=
0.
Hence, c1 = α and c2 = β.
Also, from the rest of the terms we have
(
(
r
+
c
−
1
)
(
r
+
c
−
2
)
+
(
2
−
γ
)
(
r
+
c
−
1
)
)
a
r
−
1
+
(
−
(
r
+
c
)
(
r
+
c
−
1
)
+
(
α
+
β
−
1
)
(
r
+
c
)
−
α
β
)
a
r
=
0
Hence,
a
r
=
−
(
(
r
+
c
−
1
)
(
r
+
c
−
2
)
+
(
2
−
γ
)
(
r
+
c
−
1
)
)
(
−
(
r
+
c
)
(
r
+
c
−
1
)
+
(
α
+
β
−
1
)
(
r
+
c
)
−
α
β
)
a
r
−
1
=
(
(
r
+
c
−
1
)
(
r
+
c
−
γ
)
)
(
(
r
+
c
)
(
r
+
c
−
α
−
β
)
+
α
β
)
a
r
−
1
But
(
r
+
c
)
(
r
+
c
−
α
−
β
)
+
α
β
=
(
r
+
c
−
α
)
(
r
+
c
)
−
β
(
r
+
c
)
+
α
β
=
(
r
+
c
−
α
)
(
r
+
c
)
−
β
(
r
+
c
−
α
)
.
Hence, we get the recurrence relation
a
r
=
(
r
+
c
−
1
)
(
r
+
c
−
γ
)
(
r
+
c
−
α
)
(
r
+
c
−
β
)
a
r
−
1
,
∀
r
≥
1
Let's now simplify this relation by giving ar in terms of a0 instead of ar−1. From the recurrence relation,
a
1
=
(
c
)
(
c
+
1
−
γ
)
(
c
+
1
−
α
)
(
c
+
1
−
β
)
a
0
a
2
=
(
c
+
1
)
(
c
+
2
−
γ
)
(
c
+
2
−
α
)
(
c
+
2
−
β
)
a
1
=
(
c
+
1
)
(
c
)
(
c
+
2
−
γ
)
(
c
+
1
−
γ
)
(
c
+
2
−
α
)
(
c
+
1
−
α
)
(
c
+
2
−
β
)
(
c
+
1
−
β
)
a
0
=
(
c
)
2
(
c
+
1
−
γ
)
2
(
c
+
1
−
α
)
2
(
c
+
1
−
β
)
2
a
0
As we can see,
a
r
=
(
c
)
r
(
c
+
1
−
γ
)
r
(
c
+
1
−
α
)
r
(
c
+
1
−
β
)
r
a
0
∀
r
≥
0
Hence, our assumed solution takes the form
y
=
a
0
∑
r
=
0
∞
(
c
)
r
(
c
+
1
−
γ
)
r
(
c
+
1
−
α
)
r
(
c
+
1
−
β
)
r
s
r
+
c
We are now ready to study the solutions corresponding to the different cases for c1 − c2 = α − β.
Then y1 = y|c = α and y2 = y|c = β. Since
y
=
a
0
∑
r
=
0
∞
(
c
)
r
(
c
+
1
−
γ
)
r
(
c
+
1
−
α
)
r
(
c
+
1
−
β
)
r
s
r
+
c
,
we have
y
1
=
a
0
∑
r
=
0
∞
(
α
)
r
(
α
+
1
−
γ
)
r
(
1
)
r
(
α
+
1
−
β
)
r
s
r
+
α
=
a
0
s
α
2
F
1
(
α
,
α
+
1
−
γ
;
α
+
1
−
β
;
s
)
y
2
=
a
0
∑
r
=
0
∞
(
β
)
r
(
β
+
1
−
γ
)
r
(
β
+
1
−
α
)
r
(
1
)
r
s
r
+
β
=
a
0
s
β
2
F
1
(
β
,
β
+
1
−
γ
;
β
+
1
−
α
;
s
)
Hence, y = A′y1 + B′y2. Let A′a0 = A and B′a0 = B. Then, noting that s = x−1,
y
=
A
{
x
−
α
2
F
1
(
α
,
α
+
1
−
γ
;
α
+
1
−
β
;
x
−
1
)
}
+
B
{
x
−
β
2
F
1
(
β
,
β
+
1
−
γ
;
β
+
1
−
α
;
x
−
1
)
}
Then y1 = y|c = α. Since α = β, we have
y
=
a
0
∑
r
=
0
∞
(
c
)
r
(
c
+
1
−
γ
)
r
(
(
c
+
1
−
α
)
r
)
2
s
r
+
c
Hence,
y
1
=
a
0
∑
r
=
0
∞
(
α
)
r
(
α
+
1
−
γ
)
r
(
1
)
r
(
1
)
r
s
r
+
α
=
a
0
s
α
2
F
1
(
α
,
α
+
1
−
γ
;
1
;
s
)
y
2
=
∂
y
∂
c
|
c
=
α
To calculate this derivative, let
M
r
=
(
c
)
r
(
c
+
1
−
γ
)
r
(
(
c
+
1
−
α
)
r
)
2
Then using the method in the case γ = 1 above, we get
∂
M
r
∂
c
=
(
c
)
r
(
c
+
1
−
γ
)
r
(
(
c
+
1
−
α
)
r
)
2
∑
k
=
0
r
−
1
(
1
c
+
k
+
1
c
+
1
−
γ
+
k
−
2
c
+
1
−
α
+
k
)
Now,
y
=
a
0
s
c
∑
r
=
0
∞
(
c
)
r
(
c
+
1
−
γ
)
r
(
(
c
+
1
−
α
)
r
)
2
s
r
=
a
0
s
c
∑
r
=
0
∞
M
r
s
r
=
a
0
s
c
(
ln
(
s
)
∑
r
=
0
∞
(
c
)
r
(
c
+
1
−
γ
)
r
(
(
c
+
1
−
α
)
r
)
2
s
r
+
∑
r
=
0
∞
(
c
)
r
(
c
+
1
−
γ
)
r
(
(
c
+
1
−
α
)
r
)
2
{
∑
k
=
0
r
−
1
(
1
c
+
k
+
1
c
+
1
−
γ
+
k
−
2
c
+
1
−
α
+
k
)
}
s
r
)
Hence,
∂
y
∂
c
=
a
0
s
c
∑
r
=
0
∞
(
c
)
r
(
c
+
1
−
γ
)
r
(
(
c
+
1
−
α
)
r
)
2
(
ln
(
s
)
+
∑
k
=
0
r
−
1
(
1
c
+
k
+
1
c
+
1
−
γ
+
k
−
2
c
+
1
−
α
+
k
)
)
s
r
Therefore:
y
2
=
∂
y
∂
c
|
c
=
α
=
a
0
s
α
∑
r
=
0
∞
(
α
)
r
(
α
+
1
−
γ
)
r
(
1
)
r
(
1
)
r
(
ln
(
s
)
+
∑
k
=
0
r
−
1
(
1
α
+
k
+
1
α
+
1
−
γ
+
k
−
2
1
+
k
)
)
s
r
Hence, y = C′y1 + D′y2. Let C′a0 = C and D′a0 = D. Noting that s = x−1,
y
=
C
{
x
−
α
2
F
1
(
α
,
α
+
1
−
γ
;
1
;
x
−
1
)
}
+
D
{
x
−
α
∑
r
=
0
∞
(
α
)
r
(
α
+
1
−
γ
)
r
(
1
)
r
(
1
)
r
(
ln
(
x
−
1
)
+
∑
k
=
0
r
−
1
(
1
α
+
k
+
1
α
+
1
−
γ
+
k
−
2
1
+
k
)
)
x
−
r
}
From the recurrence relation
a
r
=
(
r
+
c
−
1
)
(
r
+
c
−
γ
)
(
r
+
c
−
α
)
(
r
+
c
−
β
)
a
r
−
1
we see that when c = β (the smaller root), aα−β → ∞. Hence, we must make the substitution a0 = b0(c − ci), where ci is the root for which our solution is infinite. Hence, we take a0 = b0(c − β) and our assumed solution takes the new form
y
b
=
b
0
∑
r
=
0
∞
(
c
−
β
)
(
c
)
r
(
c
+
1
−
γ
)
r
(
c
+
1
−
α
)
r
(
c
+
1
−
β
)
r
s
r
+
c
Then y1 = yb|c = β. As we can see, all terms before
(
c
−
β
)
(
c
)
α
−
β
(
c
+
1
−
γ
)
α
−
β
(
c
+
1
−
α
)
α
−
β
(
c
+
1
−
β
)
α
−
β
s
α
−
β
vanish because of the c − β in the numerator.
But starting from this term, the c − β in the numerator vanishes. To see this, note that
(
c
+
1
−
α
)
α
−
β
=
(
c
+
1
−
α
)
(
c
+
2
−
α
)
⋯
(
c
−
β
)
.
Hence, our solution takes the form
y
1
=
b
0
(
(
β
)
α
−
β
(
β
+
1
−
γ
)
α
−
β
(
β
+
1
−
α
)
α
−
β
−
1
(
1
)
α
−
β
s
α
−
β
+
(
β
)
α
−
β
+
1
(
β
+
1
−
γ
)
α
−
β
+
1
(
β
+
1
−
α
)
α
−
β
−
1
(
1
)
(
1
)
α
−
β
+
1
s
α
−
β
+
1
+
⋯
)
=
b
0
(
β
+
1
−
α
)
α
−
β
−
1
∑
r
=
α
−
β
∞
(
β
)
r
(
β
+
1
−
γ
)
r
(
1
)
r
(
1
)
r
+
β
−
α
s
r
Now,
y
2
=
∂
y
b
∂
c
|
c
=
α
.
To calculate this derivative, let
M
r
=
(
c
−
β
)
(
c
)
r
(
c
+
1
−
γ
)
r
(
c
+
1
−
α
)
r
(
c
+
1
−
β
)
r
.
Then using the method in the case γ = 1 above we get
∂
M
r
∂
c
=
(
c
−
β
)
(
c
)
r
(
c
+
1
−
γ
)
r
(
c
+
1
−
α
)
r
(
c
+
1
−
β
)
r
(
1
c
−
β
+
∑
k
=
0
r
−
1
(
1
c
+
k
+
1
c
+
1
−
γ
+
k
−
1
c
+
1
−
α
+
k
−
1
c
+
1
−
β
+
k
)
)
Now,
y
b
=
b
0
∑
r
=
0
∞
(
(
c
−
β
)
(
c
)
r
(
c
+
1
−
γ
)
r
(
c
+
1
−
α
)
r
(
c
+
1
−
β
)
r
s
r
+
c
)
=
b
0
s
c
∑
r
=
0
∞
M
r
s
r
Hence,
∂
y
∂
c
=
b
0
s
c
ln
(
s
)
∑
r
=
0
∞
(
c
−
β
)
(
c
)
r
(
c
+
1
−
γ
)
r
(
c
+
1
−
α
)
r
(
c
+
1
−
β
)
r
s
r
+
b
0
s
c
∑
r
=
0
∞
(
c
−
β
)
(
c
)
r
(
c
+
1
−
γ
)
r
(
c
+
1
−
α
)
r
(
c
+
1
−
β
)
r
(
1
c
−
β
+
∑
k
=
0
r
−
1
(
1
c
+
k
+
1
c
+
1
−
γ
+
k
−
1
c
+
1
−
α
+
k
−
1
c
+
1
−
β
+
k
)
)
s
r
Hence,
∂
y
∂
c
=
b
0
s
c
∑
r
=
0
∞
(
c
−
β
)
(
c
)
r
(
c
+
1
−
γ
)
r
(
c
+
1
−
α
)
r
(
c
+
1
−
β
)
r
(
ln
(
s
)
+
1
c
−
β
+
∑
k
=
0
r
−
1
(
1
c
+
k
+
1
c
+
1
−
γ
+
k
−
1
c
+
1
−
α
+
k
−
1
c
+
1
−
β
+
k
)
)
s
r
At c = α we get y2. Hence, y = E′y1 + F′y2. Let E′b0 = E and F′b0 = F. Noting that s = x−1 we get
y
=
E
{
1
(
β
+
1
−
α
)
α
−
β
−
1
∑
r
=
α
−
β
∞
(
β
)
r
(
β
+
1
−
γ
)
r
(
1
)
r
(
1
)
r
+
β
−
α
x
−
r
}
+
+
F
{
x
−
α
∑
r
=
0
∞
(
α
−
β
)
(
α
)
r
(
α
+
1
−
γ
)
r
(
1
)
r
(
α
+
1
−
β
)
r
(
ln
(
x
−
1
)
+
1
α
−
β
+
∑
k
=
0
r
−
1
(
1
α
+
k
+
1
α
+
1
+
k
−
γ
−
1
1
+
k
−
1
α
+
1
+
k
−
β
)
)
x
−
r
}
From the symmetry of the situation here, we see that
y
=
G
{
1
(
α
+
1
−
β
)
β
−
α
−
1
∑
r
=
β
−
α
∞
(
α
)
r
(
α
+
1
−
γ
)
r
(
1
)
r
(
1
)
r
+
α
−
β
x
−
r
}
+
+
H
{
x
−
β
∑
r
=
0
∞
(
β
−
α
)
(
β
)
r
(
β
+
1
−
γ
)
r
(
1
)
r
(
β
+
1
−
α
)
r
(
ln
(
x
−
1
)
+
1
β
−
α
+
∑
k
=
0
r
−
1
(
1
β
+
k
+
1
β
+
1
+
k
−
γ
−
1
1
+
k
−
1
β
+
1
+
k
−
α
)
)
x
−
r
}