Page 41 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Differential and Integral Equation

Notes Equation (F) being identity, we can equate to zero the coefficients of various powers of x.
s
Equating to zero the coefficients of highest powers of x i.e. of x , we have
C (n s )(n s 1) = 0 ...(G)
0
Since C 0 0, so the indicial equation is

(n s )(n s 1) = 0 ...(H)
The solutions of equation (H) are
s = n and s = n 1 ...(I)

Equating to zero the coefficient of the next lower power of x i.e. of x s 1 , we have

a 1 (n s 1)(n s ) = 0 ...(J)

So a 1 0, as its coefficient is not zero for both s n and s n 1.

Again equating to zero the coefficient of the general term i.e. of x k r , we have

C s 2 (s r 2)(s r 1) (n s r )(n s r 1) C = 0
r
or

(s r 2)(s r 1)
C = C r 2 ...(K)
r (n s r )(n s r 1)
Putting r 2

( )(s 1)
s
C 2 = (n s 2)(n s 1) C 0

Putting r = 3
(s 1)(s 2)C
C 3 = 1 = 0, as C 1 0
(n s 3)(n s 2)

Thus
C = C 3 C 5 ... 0 (L)
1
Now there are two values for s i.e.
s = n and s = n 1 (I)
We first take s = n, then the general recurrence relation (K) becomes

(n r 2)(n r 1)
C = C r 2 (M)
r r (2n r 1)

,
,
Putting r = 2, 4, 6, ... we obtain the coefficients C C C 6 ... in terms of C i.e.
4
2
o
( n n 1)
C = C 0
2 2(2n 1)

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