Kenneth Kuttler

32.5. FINDING THE SOLUTION 625

Now suppose that m > 0. Then,

y2 (x)y1 (x)

=−x−m

m−1

∑n=1

Anxn−m

n−m+Am ln(x)+

∞

∑n=m+1

Anxn−m

n−m.

It follows

y2 = Am ln(x)y1 + x−m

(−1m

+∞

∑n=1

Bnxn

) y1︷︸︸︷xr1

∞

∑n=0

anxn.

Where Bn =An

n−m for n ̸= m. Therefore, y2 has the following form.

y2 = Am ln(x)y1 + xr2∞

∑n=0

Cnxn.

If m = 0 so there is a repeated zero to the indicial equation then 32.19 implies

y1= lnx+

∞

∑n=1

nxn +A0

where A0 is a constant of integration. Thus, the second solution is of the form

y2 = ln(x)y1 + xr2∞

∑n=0

Cnxn.

The following theorem summarizes the above discussion.

PROCEDURE 32.5.2 Let 32.4 be an equation with a regular singular point andlet r1 and r2 be real solutions of the indicial equation, 32.14 with r1 ≥ r2. Then if r1− r2 isnot equal to an integer, the general solution 32.4 may be written in the form :

∞

∑n=0

anxn+r1 +C2

∞

∑n=0

bnxn+r2

where we can have a0 = 1 and b0 = 1. If r1 = r2 = r then the general solution of 32.4 maybe obtained in the form

y1︷︸︸︷∞

∑n=0

anxn+r +C2

ln(x)

y1︷︸︸︷∞

∑n=0

anxn+r +∞

∑n=0

Cnxn+r

where we may take a0 = 1. If r1− r2 = m, a positive integer, then the general solution to32.4 may be written as

y1︷︸︸︷(∞

∑n=0

anxn+r1

)+C2

k ln(x)

y1︷︸︸︷(∞

∑n=0

anxn+r1

)+ xr2

∞

∑n=0

Cnxn

 ,

where k may or may not equal zero and we may take a0 = 1.

This procedure indicates what one should look for in the various cases. There is morediscussion in [25].

32.5. FINDING THE SOLUTION 625Now suppose that m > 0. Then,m—1+ Yi Ann=1—mxmn—-myo(x) —xyi (x) m+Amln(x)+ ) Anxn-mn=m+1 —mnIt followsYI——_—o1 2 oo= A, In (x x | — B, x" ) x" AnX".y2 m ( yyi+ (< +h n py nNWhere B, = a for n £m. Therefore, y2 has the following form.—my2 =Am In (x) y) +x? y C,x".n=0If m = 0 so there is a repeated zero to the indicial equation then 32.19 impliesA2 Inx+ yi x" +-AoYI n=1 nwhere Ag is a constant of integration. Thus, the second solution is of the formy2 =In(x)y) +x” y Cx".n=0The following theorem summarizes the above discussion.PROCEDURE 32.5.2 Ler 32.4 be an equation with a regular singular point andlet r; and rz be real solutions of the indicial equation, 32.14 with r, > rz. Then if r, — rz isnot equal to an integer, the general solution 32.4 may be written in the form :C\ y? Ax" +p y? b,x”n=0 n=0where we can have ag = | and by = 1. If r) = r2 =r then the general solution of 32.4 maybe obtained in the formy yCi y? anx"*" +C> | In(x) y? yx + y? Cxn=0 n=0 n=0where we may take ag = 1. If r) —r2 =m, a positive integer, then the general solution to32.4 may be written asv1 YIa aCi (x ent) +C> | kin(x) 2 en) +x” y Cx" |,n=0 n=0 n=0where k may or may not equal zero and we may take ay = 1.This procedure indicates what one should look for in the various cases. There is morediscussion in [25].