Question

Find the integral surface of the partial differential equation (x-y)y^2p+(y-x)x2q=(x^2+y^2) which passes through xz= a^3,y=0

Answer 1

$$(x-y)y^2p+(y-x)x^2q=x^2+y^2$$

Auxiliary equations

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

$${dx \over (x-y)y^2}={dy \over (y-x)x^2}$$ $$x^2dx=-y^2dy$$ $$x^3+y^3=C_1$$

$${dx-dy \over xy^2-y^3-yx^2+x^3}={dz \over x^2+y^2}$$ $${d(x-y) \over (x-y)(x^2+y^2)}={dz \over x^2+y^2}$$

$${d(x-y) \over x-y}={dz \over 1}$$ $$\ln |x-y|=z+\ln C_2$$ $$|x-y|=C_2e^z$$ $$F(x^3+y^2, |x-y|e^{-z})=0$$

For $xz=a^3, y=0$

$$x^3=C_1$$ $$|x|=C_2e^{a^3/x}$$

The integral surface 

$$x=C$$