Page 39 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Differential and Integral Equation

Notes r s 2
Equating coefficients of x we get
C r (r s )(r s 1) (n r s 2)(n r s 1)C 2 r = 0

for r 0, 1, ... ...(D)
Since the leading term is C so that
o
C 1 0, C 2 0. Thus C satisfies
o
s
C 0 ( )(s 1) = 0 ...(E)
Since C 0 0, so the indicial equation is

( s s 1) = 0 ...(F)
giving the value s 0 and s 1.

Next putting r 1, we have

(s 1)s C 1 = 0 ...(G)

So for s 0, C and C are both arbitrary. Thus for s = 0, equation (D) becomes
0 1
(n r 2)(n r 1)
C = C r 2 ...(H)
r ( r r 1)
From equation (H),

( n n 1)
C 2 = C 0
1.2
(n 1)(n 2)
C 3 = C 1
3.2
(n 2)(n 3) ( n n 2)(n 1)(n 3)
C 4 = C 2 C 0
4.3 1.2.3.4
(n 3(n 4) (n 3)(n 1)(n 2)(n 4)
C 5 = C 3 C 1
5.4 1.2.3.4.5
...............................................................................................
...............................................................................................
...............................................................................................
and so
Substituting the above values of C, in equation (B) and using s = 0 value we have
0
( n n 1) 2 ( n n 2)(n 1)(n 3)x 4
F n ( ) = C 0 1 2 x 1.2.3.4 ...
x

(n 1)(n 2) (n 3)(n 1)(n 2)(n 4)
C 1 x x 3 x 4 ... ...(I)
1.2.3 1.2.3.4.5
By applying ratio test it may be shown that above two series converge in the interval (-1, 1)

32 LOVELY PROFESSIONAL UNIVERSITY