Question

Solve the partial differential equation :
z (z^2-xy )(px-qy) = x^4

Answer 1

Given Equation is $z(z^2-xy) (px-qy) = x^4$

then $px-qy = \frac{x^4}{z(z^2-xy)}$

comparing with general equation $Pp+Qq=R$

Then $\frac{dx}{x} = \frac{dy}{-y} = \frac{dz}{ \frac{x^4}{z(z^2-xy)}}$

Taking first two terms,

$\frac{dx}{x} = \frac{dy}{-y}$

Integrating both sides,

$\int \frac{dx}{x} = \int\frac{dy}{-y}$

$log (x) = -log(y) + logC$

$xy = C$ (1)

Taking first and last term,

$\frac{dx}{x} = \frac{dz}{ \frac{x^4}{z(z^2-C)}}$

Integrating both sides, $\int x^3{dx} =\int z(z^2-C){dz}$

$x^4 = z^4-2z^2C + C' = z^4-2xyz^2 + C'$

$x^4 -z^4+2xyz^2=C'$ (2)

So required solution is

$f(xy,x^4-z^4+2xyz^2) = 0$