Page 331 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Differential and Integral Equation

Notes
dx dy dz adx bdy c dz
= ...(6)
c c a b 0
where stands for (ax by cz ),
so that

ax by dz = K
1
dx dy
and =
c c (K 1 )
Integrating

x (K 1 ) = y + K .
2
y x (ax by cz ) = (ax by cz ). [as K = (K )]
2 1

Example 3: Solve by Monge’s method r (a b )s abt xy .
Solution: Putting

dp s dy dq s dx
r = , and r ,
dx dy
dp s dy dq s dx
(a b )s ab = xy
dx dy

or dp dy ab dq dx xy dxdy [ s dy 2 (a b )dx dy abdx 2 ]

The subsidiary equations are
dy 2 (a b )dx dy ab dx 2 = 0 ...(1)

and dp dy ab dq dx xydx dy = 0. ...(2)
From (1)

dy a dx = 0, ...(3)
dy b dx = 0, ...(4)
Whence y ax = c , and y bx = c .
1 2
Combining these with (2), we get
adp ab dq ax (c ax )dx = 0
1
and bdp abq bx (c 2 bx )dx = 0

x 2 ax 3
or p bq c 1 = A,
2 3

x 2 bx 3
p aq c = B
2
2 3

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