Question

Solve: $x^{2}ydx-(x^3+y^3)dy=0$

Answer 1

$x^{2}ydx-(x^3+y^3)dy=0\frac{dx}{dy}=\frac{x^3+y^3}{x^{2}y}$

Putting $x=vy$

$\frac{dx}{dy}=v+y\frac{dv}{dy}\frac{dx}{dy}=\frac{x^3+y^3}{x^{2}y}\Rightarrow v+y\frac{dv}{dy}=\frac{v^3{y}^3+y^3}{v^2{y}^3}=\frac{v^3+1}{v^2}y\frac{dv}{dy}=\frac{v^3+1}{v^2}-v=\frac{v^3+1-v^3}{v^2}=\frac{1}{v^2}$

$\Rightarrow v^{2}dv=\frac{dy}{y}$ (variable separable method)

Integrating both sides

$\int v^{2}dv=\int \frac{dy}{y}\Rightarrow \frac{v^3}{3}=\ln y+C$

Putting $v=\frac{x}{y}$

$\frac{x^3}{3y^3}=\ln y+C$