Question

(𝑥 − 𝑦)𝑦2𝑝 + (𝑦 − 𝑥)𝑥2𝑞 = (𝑥2 + 𝑦2)𝑧,

Find the Integra surface when 𝑥𝑧 = 𝑎3 (𝑎 > 0), 𝑦 = 0.

Answer 1

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

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

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

$x^3+y^3=C_1$

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

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

$\frac{d(x-y)}{x-y}=\frac{dz}{z}$

$ln(x-y)=ln(z)+ln(C_2)$

$C_2=\frac{x-y}{z}$

$F(x^3+y^3,\frac{x-y}{z})=0$

For xz=a³, y=0

$x^3=C_1$

$x^2=a^3C_2$

The integral surface 

x=C