Page 313 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Differential and Integral Equation

Notes
1 1
= n 1 , (ax by )
(bD aD ) (bD aD )
1 x
= . (ax by )
(bD aD ) n 1 b
1 1 x
= (ax by )
(bD aD ) n 2 (bD aD b
)
1 x
= . (ax by )
(6D aD ) n 2 2b 2
1 1 1 2
= x (ax b )
2b 2 (bD aD ) n 3 (bD aD )

1 1 3
= . .x (ax b )
3b 3 (bD aD ) n 3
.......................................................................
.......................................................................
.......................................................................

1 1 x n
= n 1 n x . (ax b )
b (n 1)! (bD aD ) nb

x n
= (ax by )
n
b n
1 x n
Thus (ax by ) = n (ax b ) ...(18)
(bD aD ) n b n
When F ( , ) = 0
b
a
Example 1: Solve

2
(D 2 2aDD a D 2 )z = ( f y ax )
Solution: The auxiliary equation is
m 2 2am a 2 = 0

(m a ) 2 = 0
m = a, a
The complimentary function is

C.F. = F (y ax ) x F (y ax )
1 2
1
P.I. = 2 2 2 ( f y ax )
D 2aDD a D

1 x 2
= 2 ( f y ax ) ( f y ax )
(D aD ) 2

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