Page 341 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Differential and Integral Equation

Notes 2 2
Solution: Here R = (1 q ), S 2pq ,T (1 p ),
U = (1 p 2 q 2 ) 1/2 ,V (1 p 2 q 2 3/2 .
)
The equation in is (RT UV ) 2 US U 2 = 0

( 2pq ) 1
2
2
2
2
2
or [(1 p )(1 q ) (1 p q )] 2 2 = 0
1 p 2 q 2 1 p q
p q
or 2 2 2 (1 p 2 q 2 ) 2pq 1 p 2 q 2 1 = 0
1
or = (roots being equal).
pq (1 p 2 q 2 )
We get only one system which will give only one intermediate integral.
The system is U dy T dx U dp 0,

U dx Rdy U dq = 0,

1 (1 p 2 ) dp
dy dx 2 2 = 0
(1 p 2 q 2 ) pq (1 p 2 q 2 ) dq (1 p q )

1 (1 q 2 ) dq
dx dy 2 2 = 0
(1 p 2 q 2 ) pq (1 p 2 q 2 ) pq (1 p q )

dp
2
or pq dy (1 p )dx = 0,
(1 p 2 q 2 )
dq
2
pqdx (1 q )dy = 0.
(1 p 2 q 2 )
Eliminating
2 2
dy ,[(1 p 2 )(1 q 2 ) p q ]dx [(1 q 2 )dp pqdq ]/ (1 p 2 q 2 )

(1 q 2 )dp pq dq
or dx 2 2 3/2 = 0
(1 p q )

2
(1 p 2 q 2 )dp (p dp pq dq )
or dx 2 2 3/2 2 2 3/2 = 0
(1 p q ) (1 p q )
1
p (2p dp 2q dq )
or dx (1 p 2 q 2 ) 1/2 dp 2 = 0
)
(1 p 2 q 2 3/2
or x p (1 p 2 q 2 ) 1/2 = . ...(1)

,
Similarly eliminating dx y q (1 p 2 q 2 ) 1/2 ...(2)

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