Question

Eliminate the arbitrary function and hence obtain the partial differential equation:

φ (
z
x
3
,
y
x
) = 0

Answer 1

$\phi=(zx^3,yx)=0$

$zx^3=c_1$

$yx=c_2$

$dz=-\frac{3c_1dx}{x^4}, dx=-\frac{c_2dy}{y^2}\implies dz=\frac{3c_1c_2dy}{x^4y^2}$

$dz=-\frac{3c_1dx}{x^4}=\frac{3c_1c_2dy}{x^4y^2}$

$\frac{dz}{3c_1c_2}=-\frac{dx}{c_2x^4}=\frac{dy}{x^4y^2}$

$x^4y^2q-c_2x^4p=3c_1c_2$