Answer to Question #179786 in Differential Equations for kavya

Question

find the integral surface of the equation (x-y)y^2*p + (y-x)x^2*q = (x^2+y^2)z through the curve xz = a^3 and y = 0

Answer 1

Auxiliary equations

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

Taking dx and dy

$\frac{dy}{dx}=-\frac{x^2}{y^2}$

$y^2dy=-x^2dx$

$C_1=x^3+y^3$

$\frac{dx}{(x-y)y^2}=\frac{dz}{(x^2+y^2)z}$

$\frac{(x^2+y^2)dx}{(x-y)y^2}=\frac{dz}{z}$

$lnz=\frac{1}{y^2}F(x,y)+C_2$

Here, y cannot be 0 in this equation that is why there is no intersection with curve

xz=a³, y=0