Page 365 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Differential and Integral Equation

Notes 2
2 d R dR
r 2r ( n n 1)R = 0 ...(6)
dr 2 dr
1 d d 2
and sin ( n n 1) = 0 ...(7)
sin d d sin 2

To solve (6), let
p
r = e ,
dr p
so that = e r
dp

dR dR dp 1 dR
Therefore = .
dr dr dr r dp
d d
or r =
dr dp

d
Let us denote the operator by D, then
dp

2
d dR 2 d R dR
r r = r r
dr dr dr 2 dr
2
2 d R d dR dR
So r 2 = r r r
dr dr dr dr
d d
= r r 1 R
dr dr
= D (D 1)R

Using these values in (6), we get
D (D 1) 2D n (n 1) R = 0

or (D n )(D n 1)R = 0 ...(6a)

The solution of (6a) is

R = A e np B e ( n 1) p

or R = A r n B r (n 1) ...(5)
To solve (7) put cos µ

d d dµ d
so that = sin
d dµ d dµ
Substituting these values in (7) we have

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