Answer to Question #252293 in Differential Equations for pde

Question

Determine the general solution to the lagrange equation

y²zp-x²zq=x²y

Answer 1

$\displaystyle y^2zp - x^2zq=x^2y\\ \text{Divide through by z to obtain}\\ y^2p-x^2q=\frac{x^2y}{z}\\ \text{The subsidiary equation is }\\ \frac{dx}{y^2} =\frac{dy}{-x^2}=\frac{zdz}{x^2y}\\ \text{Taking the the first equation}\\ \frac{dx}{y^2} =\frac{dy}{-x^2}\\ \text{Integrating both sides,we have}\\ \frac{x}{y^2}= -\frac{y}{x^2}+c_1\\ \implies c_1 = \frac{x}{y^2}+ \frac{y}{x^2}\\ \text{Next, we take another pair of ratios}\\ \frac{dx}{y^2}=\frac{zdz}{x^2y}\\ \text{Integrating, we have}\\ \frac{x}{y^2}=\frac{z^2}{2x^2y}+c_2\\ \implies c_2 = \frac{x}{y^2}-\frac{z^2}{2x^2y}\\ \text{Hence the general solution is}\\ \phi (\frac{x}{y^2}+ \frac{y}{x^2}), \frac{x}{y^2}-\frac{z^2}{2x^2y}=0$