Page 148 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Unit 8: Total Differential Equations, Simultaneous Equations

i.e. not involving x, y. Now we take Notes
z
u = ( ) ...(3)
where ( )z is the function of z alone as the solution of the given equation. Now taking the
differential of both sides of equation (3), we must get the given equation.

d d
On equating the two, we may get the value of . Eliminating x, y from the value of , using
dz dz
(3), and then integrating we can obtain the value of ( ).z Substituting the value of  in (3), we get
required solution.

Example 1: Solve

2
2
3x dx 3y dy (x 3 y 3 e 2z )dz = 0 ...(1)
by regarding one variable as constant.
Solution:
Let z be constant so that
dz = 0 ...(2)

Then (1) gives
2
2
3x dx 3y dy = 0 ...(3)
This gives

x 3 y 3 = constant = (z) (say) ...(4)
Taking the differential of (4) we have

2
2
3x dx 3y dy = d ...(5)
Comparing (5) with (1) we have
3
3
dz (x  y  e 2 z ) = d ...(6)
or eliminating x, y from (6) we have

d
(  e 2z ) =
dz
d
or   = e 2z ...(7)
dz

This equation is linear in , whose . .I F e z . So


z
e   z =  e 2z .e dz + constant
z

=  e dz C (say)
2z
z
Thus  ( ) = e  Ce z

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