Answer to Question #152792 in Differential Equations for Abbasali

Question

dx/y^2 z=dy/xz=dz/yz solution ?

Answer 1

$\displaystyle \frac{\mathrm{d}x}{y^2 z} = \frac{\mathrm{d}y}{xz} = \frac{\mathrm{d}z}{yz}\\ \textsf{Comparing the first and the second equation,}\\ \frac{\mathrm{d}x}{y^2 z} = \frac{\mathrm{d}y}{xz}\\ \frac{\mathrm{d}x}{y^2} = \frac{\mathrm{d}y}{x}\\ x\mathrm{d}x - y^2\mathrm{d}y = 0\\ \int\,x\mathrm{d}x - \int\,y^2\mathrm{d}y = 0\\ \frac{x^2}{2} - \frac{y^3}{3} = C\\ \textsf{At}\,\, C = 0\\ \frac{x^2}{2} - \frac{y^3}{3} = 0,\,\, \frac{3x^2}{2} = y^3,\,\, y = \sqrt[3]{\frac{3x^2}{2}}\\ \textsf{Comparing the first and the third equation,}\\ \frac{\mathrm{d}x}{y^2 z} = \frac{\mathrm{d}z}{yz}\\ \frac{\mathrm{d}x}{y} = \mathrm{d}z\\ \frac{\mathrm{d}x}{\sqrt[3]{\frac{3x^2}{2}}} = \mathrm{d}z\\ \int\frac{\mathrm{d}x}{\sqrt[3]{\frac{3x^2}{2}}} = \int\mathrm{d}z\\ \int\,\, \sqrt[3]{\frac{2}{3}} x^{-\frac{2}{3}}\mathrm{d}x = \int\mathrm{d}z\\ \sqrt[3]{\frac{2}{3}} \cdot 3x^{\frac{1}{3}} + C = z\\ z = 2^{\frac{1}{3}} 3^{\frac{2}{3}} + C = 18^{\frac{1}{3}} \cdot x^{\frac{1}{3}} + C\\ z = \sqrt[3]{18x} + C\\ \therefore\phi\left(z - \sqrt[3]{18x},\,\, \frac{x^2}{2} - \frac{y^3}{3}\right) = 0\,\,\textsf{is a solution to the PDE}$