Answer to Question #148142 in Differential Equations for Nikhil Singh

Question

Solve the partial differential equation: x^2p+y^2q=z^2

Answer 1

$x^2p + y^2q = z^2\\ \textsf{The lagrange auxiliary equation is}\\ \frac{\mathrm{d}x}{x^2} = \frac{\mathrm{d}y}{y^2} = \frac{\mathrm{d}z}{z^2} \\ \frac{\mathrm{d}x}{x^2} + \frac{\mathrm{d}y}{y^2} - 2\frac{\mathrm{d}z}{z^2} = 0\\ \textsf{Choosing}\,\,\left(\frac{1}{x^2}, \frac{1}{y^2}, -2\frac{1}{z^2}\right) \,\,\textsf{as multpliers.}\\ \int\frac{\mathrm{d}x}{x^2} + \int\frac{\mathrm{d}y}{y^2} - 2\int\frac{\mathrm{d}z}{z^2} = 0\\ -\frac{1}{x} - \frac{1}{y} + 2\frac{1}{z} = C\\ \frac{1}{x} + \frac{1}{y} - 2\frac{1}{z} = C\\ yz + xz - 2xy = Cxyz\\ \textsf{Let}\,z\,\textsf{be constant}.\\ \textsf{It implies}\,\, \frac{\mathrm{d}x}{x^2} - \frac{\mathrm{d}y}{y^2} = 0.\\ \int\frac{\mathrm{d}x}{x^2} - \int\frac{\mathrm{d}y}{y^2} = 0\\ -\frac{1}{x} = -\frac{1}{y} + \phi(z), \frac{1}{x} - \frac{1}{y} = -\phi(z)\\ y - x = \phi(z)xy\\ \frac{1}{x} - \frac{1}{y} = -\phi(z)\\ -\frac{1}{x^2} + \frac{1}{y^2}\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\mathrm{d}(\phi(z))}{\mathrm{d}z}\cdot\frac{\mathrm{d}z}{\mathrm{d}x}\\ \frac{\mathrm{d}x}{x^2} - \frac{\mathrm{d}y}{y^2} = -\frac{\mathrm{d}(\phi(z))}{\mathrm{d}z}\cdot\mathrm{d}z \\ \frac{\mathrm{d}x}{x^2} - \frac{\mathrm{d}y}{y^2} = 0\cdot\mathrm{d}z \\ \frac{\mathrm{d}(\phi(z))}{\mathrm{d}z} = 0\\ \mathrm{d}(\phi(z)) = 0\cdot\mathrm{d}z \\ \int\mathrm{d}(\phi(z)) = \int 0 \cdot\mathrm{d}z\\ \displaystyle\therefore \phi(z) = C\\ y - x = Cxy\\ \phi\left(\frac{1}{x} - \frac{1}{y}, \,\frac{1}{x} + \frac{1}{y} - \frac{2}{z}\right) = 0,\,\,\textsf{is the solution to the PDE}$