Page 149 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Differential and Integral Equation

Notes Now from (4) we have
3 3 2z z
x  y = e  Ce ...(8)
which is the required solution

Example 2: Solve
2 2
(2x  2xy  2xz  1)dx dy dz .2z = 0 ...(1)


by regarding one variable as constant.
Solution: Let x be constant, so that
dx = 0 ...(2)

Then dy  2z dz = 0

or ( d y z 2 ) = 0

so y z 2 = constant

x
=  ( ) (say) ...(3)
Taking differential of (3) we have

dy  2z dz = d ( ) ...(4)
x
Comparing (4) with (1) we have

2
2
x
 (2x  2xy  2x z  1)dx = d ( )
d
2
 = 2x  1 2 (y z 2 )

x

dx
d
2
or  = 2x  1 2x

dx
d 2 1

So  2x =  2x ...(5)
dx
  2x dx x 2
The equation (5) is linear in , so I.F. is e  e .
2 2
x
2
Thus  e x =   (2x  1) e dx C

2
2
x

=   x    2xe x    dx   e dx C
2 2
x
x
=  x e x   e dx   e dx C

2
=  x e x  C
So x C e x 2 . Thus y z 2   x C e  x 2 Q.E.D.

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