Finding the general solution of $\frac{dx}{xy^3-2x^4}=\frac{dy}{2y^4-x^3y}=\frac{dz}{2z(x^3-y^3)}$

Question

Find the general solution of the equation: $$\frac{dx}{xy^3-2x^4}=\frac{dy}{2y^4-x^3y}=\frac{dz}{2z(x^3-y^3)}$$

Answer 1

$$\frac{dx}{xy^3-2x^4}=\frac{dy}{2y^4-x^3y}=\frac{dz}{2z(x^3-y^3)}$$ This is probably the Charpit-Lagrange system of ODEs in order to solve the PDE : $$(xy^3-2x^4)\frac{\partial z}{\partial x}+(2y^4-x^3y)\frac{\partial z}{\partial y}=2(x^3-y^3)z \tag 1$$ A first characteristic equation comes from solving $\frac{dx}{xy^3-2x^4}=\frac{dy}{2y^4-x^3y}$

$\frac{dy}{dx}=\frac{y}{x}(\frac{2(\frac{y}{x})^3-1}{(\frac{y}{x})^3-2})\quad$ is an homogeneous ODE. The usual change of function is $y(x)=xu(x)$

$u+x\frac{du}{dx}=u\frac{2u^3-1}{u^3-2}$

$\frac{dx}{x}=\frac{u^3-2}{u(u^3+1)}du$

$\ln|x|=\ln|u^2-u+1|+\ln|u+1|-2\ln|u|+$constant.

$\frac{1}{xu^2}(u^2-u+1)(u+1)=c_1$ $$\boxed{\frac{x}{y^2}\left((\frac{y}{x})^2-\frac{y}{x}+1\right)(\frac{y}{x}+1)=c_1}$$

A second characteristic comes from solving $\frac{\frac{1}{x}dx}{y^3-2x^3}=\frac{\frac{1}{y}dy}{2y^3-x^3}=\frac{dz}{2z(x^3-y^3)}$

Using wellknown properties of equal fractions : $\frac{\frac{1}{x}dx}{y^3-2x^3}=\frac{\frac{1}{y}dy}{2y^3-x^3}=\frac{\frac{1}{x}dx+\frac{1}{y}dy}{(y^3-2x^3)+(2y^3-x^3)}$

$\frac{\frac{1}{x}dx+\frac{1}{y}dy}{3(y^3-x^3)}=\frac{dz}{2z(x^3-y^3)}$

$2(\frac{1}{x}dx+\frac{1}{y}dy)=3\frac{dz}{z}$

$3\ln|z|-2\ln|xy|=$constant $$\boxed{\frac{z^3}{x^2y^2}=c_2}$$

General solution of the PDE, equation $(1)$, on the implicit form is $c_2=\Phi(c_1)$ where $\Phi$ is an arbitrary function (to be determined according to some boundary conditions not specified in the wording of the question): $$z^3=x^2y^2\:\Phi\left(\frac{x}{y^2}\left((\frac{y}{x})^2-\frac{y}{x}+1\right)(\frac{y}{x}+1) \right)$$