Page 269 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Differential and Integral Equation

Notes 2
dz z l
.
d a

l
z ( /2 1)
, c
l
1 ( /2) a
1 1 ( /2) aY X x m 1 y n 1
l
z c a . c
2 l a ( a m 1) n 1

2
2
2
Example 5: Solve: z (p + q + 1) = c 2
Solution:
z 2
Put z dz = dZ i.e. Z =
2
Z dZ Z
. = zp = P (say)
x dz x

z dZ z
= zq = Q (say)
Y z Y
The given equation becomes
2
2
2Z + P + Q = c 2
now let Z = f (x + ay) + f (X)
Z dZ X dP
P = .
x X x dx
Z dZ X dZ
Q = . a
Y X y dX

dZ 2
2
2
(1 + a ) = c 2Z
dx
dZ (1 a 2 )
or 2 2 dx
(c a ) z

2
2
or [(1 + a )] [(c 2Z)] = X + c
2
2
2
or (1 + a ) (c z ) = (x + ay) + c
2
2
2
2
or (1 + a ) (c z ) = (x + ay + c) .
Self Assessment
Solve
2
14. p (1 + q ) = q(z a)
2
2
15. p = z (1 pq)

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