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

Question

I am trying to solve the simultaneous differential equations $$\frac{dx}{y^2(x-y)}=\frac{dy}{x^2(x-y)}=\frac{dz}{z(x^2+y^2)}.$$ From the first two fractions, we can get the equation $x^3 - y^3 = c$, but I'm not able to get a second equation with $z$.

Answer 1

Let $y=mx$ and eliminate occurrences of $y$. Then the left and middle parts of the equality yield $$\frac{\mathrm{d}x}{x}=\frac{m^2 \,\mathrm{d}m}{1-m^3}$$

whereas the left and right parts give $$\begin{split}\frac{\mathrm{d}z}{z}&=\frac{1+m^2}{m^2(1-m)}\frac{\mathrm{d}x}{x}\\ &=\frac{(1+m^2)\mathrm{d}m}{(1-m^3)(1-m)}\text{.} \end{split}$$

These equations are solvable by the usual methods of integration by partial fractions.