Page 146 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Unit 8: Total Differential Equations, Simultaneous Equations

Where  is only a function of z. Taking the differential of both the sides, we get Notes
y
(y z ) (dx dz ) (x z )dy ( y x z )(dy dz )
(y z ) 2 = d

or

(y 2 yz )dx dy (z 2 zx ) dz (y 2 zy yx yz )
= d ...(8)
(y z ) 2
Now from (1) and (8) we have,

d = 0 or k (constant)
Thus from (7)

( y x z )
= k
y z
or the solution is
( y x z ) = k (y z ) Q.E.D.

Example 2: Solve
2
2
(x y y 3 y 2 ) z dx (xy 2 x z x 3 )dy (xy 2 x 2 ) y dz = 0 ...(1)
2
2
2
Let P = x y y 3 y 2 , z Q xy 2 x z x 3 , R xy 2 x y
The condition of integrability is
Q R R P P Q
P Q R = 0 ...(2)
z y x z y x
So

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

R 2 P 2
= y 2xy , y
x z
P 2 2 Q 2 2
= x 3y 2yz , y 2xz 3x
y x
Substituting in (2) we have

2
2
= (x y y 3 y 2 ) z x 2 2xy x 2 xy 2 x z x 3 (y 2 2xy y 2 )
(xy 2 x 2 ) y x 2 3y 2 2yz y 2 2xz 3x 2

2
y
= y (x y )(x y ) yz 2 (x y ) ( x y x )(y x ) x z (2 )(x y )
x
2xy (x y )[2x 2 2y yz xz ]

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