How to solve $ \frac{dy}{dx} = \frac{x^3 -y^3}{x-y}$ which is a seemingly homogenous equation.

Question

How to solve $$ \frac{dy}{dx} = \frac{x^3 -y^3}{x-y}?$$

I can show $\frac{dy}{dx} = x^2 +y^2 +xy,$ which may be converted into an exact differential or any other process might be helpful ?

Answer 1

This is a Riccati equation, usually they are hard to solve unless you can see or guess a solution.

One can transform this into a second order linear DE using $y=-\frac{u'}{u}$ leading to $$ \frac{-u''}{u}+\frac{u'^2}{u^2}=\left(-\frac{u'}{u}\right)^2+x\left(-\frac{u'}{u}\right)+x^2\\~\\ \implies u''-xu'+x^2u=0 $$ which by Mathematica/WolframAlpha has a normalized form with solutions by named functions (Hermite polynomials and Kummer hypergeometric functions) in the variable $z=\frac{\sqrt[4]3}2(1+i)x$.