Page 151 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Differential and Integral Equation

Notes Integrating
logu log(1 v ) log z = logc (c being constant)

uz
or = c
1 v
2
or u z = c (z zv )
or xz = c (y z ) ...(6)

is the solution of the equation (1).

Example 2: Solve
( z z y )dx z (z x )dy x (x y )dz = 0 ...(1)

Here P = z (z y ), Q ( z z x ), R ( x x y ) ...(2)

P Q
= , z z
y x

R D
= 2x y , 2z y ...(3)
x z
Q R
= 2z x , x
z y

The integrability condition

dQ R R P P Q
P Q R = 0 ...(4)
dz y x z y x
L.H.S. of equation (4) is

= z (z y ) 2z x x ( z z x )[2x y 2z y ] x (x y )[ z z ]

2
= 2 (z z y ) z (z x )(2x 2y 2 ) 2zx (x y )
z
2
2
2
= 2z 3 2z y 2z x 2zx 2 2yz 2 2xyz 2z 3 2z x 2zx 2 2xyz 0 = R.H.S.
So condition (4) is satisfied
Let
x = uz dx z du udz
,
...(5)
,
y = vz dy z dv vdz
Substituting in equation (1)
2
)
z 2 (1 v )[zdu udz ] z (1 u )[zdv vdz ] z u (u v dz = 0
)z
or (1 v du z (1 u )dv u (1 v ) v (1 u ) u (u v dz = 0
)

144 LOVELY PROFESSIONAL UNIVERSITY