Solve the following differential equation:\(x^2\cdot y\,dx-(x^3+y^3)\cdot dy=0\)

Answer: \(\log|y|-\frac{x^3}{3y^3}=c\)

\(\frac{dy}{dx}=\frac{x^2y}{x^3+y^3}\) ... (1). This is a homogeneous equation. Put \(y=vx\), so \(\frac{dy}{dx}=v+x\frac{dv}{dx}\).

\(v+x\frac{dv}{dx}=\frac{x^3v}{x^3+v^3x^3}=\frac{v}{1+v^3}\)
\(\therefore x\frac{dv}{dx}=\frac{v}{1+v^3}-v=\frac{v-v-v^4}{1+v^3}=\frac{-v^4}{1+v^3}\)

Separating the variables: \(\frac{1+v^3}{v^4}dv=-\frac{dx}{x}\), i.e. \(\left(\frac{1}{v^4}+\frac{1}{v}\right)dv=-\frac{dx}{x}\).

Integrating: \(\frac{v^{-3}}{-3}+\log|v|=-\log|x|+c\)
\(\therefore -\frac{1}{3v^3}+\log|v|+\log|x|=c\)
\(\therefore -\frac{x^3}{3y^3}+\log\left|\frac{y}{x}\cdot x\right|=c\)

\(\therefore \log|y|-\frac{x^3}{3y^3}=c\) is the general solution.