General solution of $(x^2-y^2)dx + (3xy)dy = 0$ Find the general solution to the homogeneous differential equation $$(x^2-y^2)dx + (3xy)dy = 0$$ The differential equation does not seem to be separable, and I'm having a tough time to put it in the general form of $$\frac{dy}{dx} + p(x)y = f(x)$$ Would someone mind offering me some assistance? Thank you

As @user84413 suggested. This approach should work. \begin{align} (x^2-y^2)\mathrm{dx} + (3xy) \mathrm{dy}&=0 &&\;\left|\cdot \frac{1}{x^2} \right.\\\ \left(1-\left(\frac{y}{x}\right)^2\right)\mathrm{dx} + 3\frac{y}{x}\mathrm{dy}&=0 &&\;\left|\cdot \frac{1}{\mathrm{dx}}\right.\\\ &\vdots\\\ \frac{\mathrm{dy}}{dx} &= \left(\left(\frac{y}{x}\right)^2-1\right)\frac{1}{3}\frac{x}{y} &&\;\left|u = \frac{y}{x},\frac{\mathrm{dy}}{\mathrm{dx}} =\frac{\mathrm{du}}{\mathrm{dx}}x + u\right.\\\ \frac{\mathrm{du}}{\mathrm{dx}}x + u &= \frac{u^2-1}{3u} \end{align} Hope you can work out the rest of the work.