Page 150 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Unit 8: Total Differential Equations, Simultaneous Equations

Self Assessment Notes

5. Solve the differential equation
2

yz dx zxdy  3xy dz  0
6. Solve

2(y z )dx  (x z )dy  (2y x z )dz  0




Method III: For Homogeneous Equations
Consider the equation


P dx Q dy  R dy = 0 ...(1)
If the functions P, Q and R are homogeneous functions of x, y, z then one variable say z, can be
separated from the other variables by substituting x = z u and y = zv, so that

dx = z du udz ,

dy = z dv v dz , ...(2)

in the given equation. Then transformed equation can be integrated as

v
u
v
u
du f 1 ( , ) f 2 ( , )dv  dz
u
v
F ( , ) z = 0 ...(3)
Now to integrate the first term, we find [ ( , )]d F u v and add and subtract it to numerator. After
doing so, the first term will also be integrable.
Example 1: Solve
(yz z 2 )dx xz dy xy dz = 0 ...(1)




Here yz z 2 , xz and xy are homogeneous in x, y, z. Let us put x = uz, and y = vz, so that
dx = z du u dz 

 ...(2)
dy = z dv v dz  
Substituting (2) in (1) we have

2
(vz 2 z 2 )(zdu udz ) uz 2 (zdv vdz ) uvz dz = 0
uv
z (v 1)du udv ( u v 1) dz = 0 ...(3)
uv

(v 1)du u dv dz
or ( u v 1) z = 0 ...(4)

Simplifying we have

du dv dz
= 0 ...(5)
u 1 v z

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