624 CHAPTER 32. SOLUTIONS NEAR A REGULAR SINGULAR POINT

Taking antiderivatives, another solution is

y2 ∈ y1

∫ 1y2

1e−Pdx

where P(x) ∈∫

x−1 p(x)dx. Thus

P(x) ∈∫ (b0

x+b1 +b2x+ · · ·

)dx = b0 lnx+b1x+b2x2/2+ · · ·

and so−P(x) = lnx−b0 + k (x)

for k (x) some analytic function, k (0) = 0. Therefore,

e−P(x) = eln(x−b0)+k(x) = x−b0g(x)

for g(x) some analytic function, g(0) = 1. Therefore,

y2 ∈ y1 (x)∫ 1

y21

(x−b0g(x)

)dx, g(0) = 1. (32.17)

Next it is good to understand y1 and r1 in terms of b0. Consider the zeros to the indicialequation,

r (r−1)+b0r+ c0 = r2− r+b0r+ c0 = 0.

It is given that r1 = r2+m where m is a non negative integer. Thus the left side of the aboveequals

(r− r2)(r− r2−m) = r2−2rr2− rm+ r22 + r2m

and so−2r2−m = b0−1

which implies

r2 =1−b0

2− m

2and hence

r1 = r2 +m =1−b0

2+

m2

y1 (x) = x1−b0+m

2

∞

∑n=0

anxn, a0 = 1 (32.18)

Now from Theorem 32.2.2 and looking at 32.18 y1 (x)−2 is of the form

1

x1−b0+m (∑∞n=0 anxn)2 = xb0−1−m (1+h(x))

where h(x) is analytic, h(0) = 0. Therefore, 32.17 is

y2 (x) ∈ y1 (x)∫

xb0−1−m (1+h(x))(

x−b0 g(x))

dx

y2 (x) ∈ y1 (x)∫

x−1−m (1+ l (x))dx, l (0) = 0 (32.19)