Question

Find the integral surface of the partial differential equation
(X-y)y^2p+(y-x)x^2q=(x^2+y^2)z

Answer 1

dx/((x-y)y²)=dy/((y-x)x²)=dz/((x²+y²)z)

dx/((x-y)y²)=dy/((y-x)x²)

$\int$ y²dy = $\int$ -x²dx

y³ = -x³+c₁

dx/((x-y)y²)=dz/((x²+y²)z)

$\int$ y²/z dz = $\int$ (x²+y²)/(x-y) dx

$\intop$ (x²+y²)/(x-y) dx = $\int$ ((2y²/(x-y)+x+y)dx = 2y²$\intop$ 1/(x-y)dx +$\intop$ x dx + y $\intop$ 1 dx =

= 2 y² ln(x-y) + x²/2 + xy + c₂

y²ln(z) = 2 y² ln(x-y) + x²/2 + xy + c₂

integral surface:

y³ = -x³+c₁

y²ln(z) = 2 y² ln(x-y) + x²/2 + xy + c₂