Question

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

Answer 1

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

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

$x^2dx+y^2dy=0$

$x^3+y^3=c_1$

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

$ln(x-y)=lnz+lnc_2$

$\frac{x-y}{z}=c_2$

$F(c_1,c_2)=F(x^3+y^3,\frac{x-y}{z})=0$

for  the curve $xz=a^3,y=0$ :

$c_1=x^3$

$x/z=c_2$

$\implies z=x/c_2=\sqrt[3]{c_1}/c_2$

$x/c_2=a^3/x\implies x^2=c_2a^3$

$z=\frac{z\sqrt[3]{x^3+y^3}}{x-y}$

$\sqrt[3]{x^3+y^3}=x-y$

$z=a^3/x=a^3/\sqrt[3]{c_1}=\frac{a^3}{\sqrt[3]{x^3+y^3}}$