Page 13 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Differential and Integral Equation

Notes As C 0, we have
0
2
K n 2 = 0 ...(14)
The equation (14) is called indicial equation.
So k = n or k = n

We first consider the case k = n, next equate the co-efficient of x k+1 to zero i.e.
2
2
C [(k + 1) n ] = 0
1
2
2
For k = n, (k + 1) n 0
So we have C = 0 ...(15)
1
Putting the co-efficient of x k+2 to zero, we get
2
C {(k + 2) (k + 1) + k + 2 n } + C = 0
2 0
2
2
or C [(k + 2) n ] + C = 0
2 0
C
or C = 0
2 (k 2) 2 n 2
C
= 0 2 2 for k = n
(n 2) n
C C
or = 0 0 ...(16)
(2n 2)(2) (n 1)2 2

Putting the co-efficient of x k+3 to zero, we get
2
2
C [(k + 3) n ] + C = 0
3 1
C
or C = 1 = 0, as C = 0
3 2 2 1
(n 3) n
Putting the co-efficient of x k+4 to zero, we get
2
2
C [(k + 4) n ] + C = 0
4 2
C
or C = 2
4 2 2
(n 4) n
C
= 2
(2n 4)(4)

C ( 1) 2 C
= 2 2
(n 2)2, 2 2 (n 1)(n 2)1.2(2) 4
Proceeding in the same way we get

C = 0 = C = C = C = ... ...(17)
1 3 5 7
( 1) k C
And C = 0 for 1, 2, 3 ...(18)
2k (n 1)(n 2)...1.2...(2 ) 2 2k
k

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