Page 145 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Differential and Integral Equation

Notes Solution:
Let
P = y 2 yz , Q z 2 zx ,R y 2 xy ...(2)

The condition for integrability of equation (1) is

Q R R P P Q
P Q R = 0 ...(3)
z y x z y x
Now

Q R
= 2z x , 2y x
z y

R P
= , y y
x z
P Q
= 2y z , z
y x

Substituting in equation (3) we get
y
(y 2 yz )(2z 2x 2 ) (z 2 zx )( y y ) (y 2 xy )(2y z z )
or y 2 (2z 2x 2y 2 ) yz (2z 2x 2 ) 2 (z 2 zx ) 2xy 2
y
y
y
2
2
2
= 2y z 2xy 2 2yz 2 2xyz 2y z 2yz 2 2xyz 2xy = 0 = R.H.S.
So condition of integrability is verified.
Let z be constant, so that dz = 0. So from (1) we get
(y 2 yz )dx (z 2 zx )dy = 0 ...(4)

dx zdy
So = 0
x z y 2 yz

dx 1 1
or dy = 0 ...(5)
x z y z y
Integrating we get

y
log(x z ) log = Constant
y z

(x z )y
or log = constant ...(6)
y z
= log  (say)

( y x z )
so =  ...(7)
y z

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