Answer to Question #194366 in Differential Equations for Gaurav dofe

Question

yz(y+z)dx+xz(x+z)dy+xy(x+y)dz=0

Answer 1

This is total differentiation equation and also homogeneous.

yz(y+z)dx+xz(x+z)dy+xy(x+y)dz=0-----(1)

Let x=uz, y=vz

Thus dx=udz+zdu and dy=vdz+zdv

Put in equation 1

[v(v+1)du+u(u+1)dv]z⁴+2uv(u+v+1)z³dz=0

Divided by uv(u+v+1)z⁴=0 we get

{(v+1)/u(u+v+1)}du+{(u+1)/v(u+v+1)}dv+2dz/z=0

{(1/u)-(1/(u+v+1)}du+(1/v)-(1/(u+v+1)}dv+2dz/z=0

On integrating

logu+logv+2logz-(1/(u+v+1)d(u+v+1)=logc

uvz²/(u+v+1)=c

(x/z)(y/z)z²/((x/z)+(y/z)+1)=c

xyz=c(x+y+z)