Answer to Question #143101 in Differential Equations for Nikhil Singh

Question

Solve the following simultaneous differential equations
dx/y^2×(x-y)= dy/-x^2×(x-y)= dz/z×(x^2+y^2)

Answer 1

$\displaystyle\frac{\mathrm{d}x}{y^2(x - y)} = \frac{\mathrm{d}y}{-x^2(x - y)} = \frac{\mathrm{d}z}{z(x^2 + y^2)}\\ \textsf{Comparing the first two equations}\\ \frac{\mathrm{d}x}{y^2(x - y)} = \frac{\mathrm{d}y}{-x^2(x - y)}\\ \frac{\mathrm{d}x}{y^2} = \frac{\mathrm{d}y}{-x^2}\\ -x^2\, \mathrm{d}x = y^2\, \mathrm{d}y\\ \int -x^2\, \mathrm{d}x = \int y^2\, \mathrm{d}y\\ \frac{-x^3}{3} + C = \frac{y^3}{3}\\ \frac{x^3}{3} + \frac{y^3}{3} = C\\ x^3 + y^3 = C, y = \sqrt[3]{C - x^3}\\ \textsf{Comparing the first and last equation, we have;}\\ \frac{\mathrm{d}x}{x(C - x^3)^{\frac{2}{3}} - C + x^3} (x^2 + (C - x^3)^{\frac{2}{3}})= \frac{\mathrm{d}z}{z}\\ \textsf{Integrating both sides gives a non-standard}\\ \textsf{integral on the LHS.}\\ \textsf{We can choose}\,\left(\frac{1}{x - y}, \frac{1}{x - y}, \frac{1}{z}\right)\\ \textsf{as our multipliers,}\\ \frac{\mathrm{d}x}{x - y} - \frac{\mathrm{d}y}{x - y} - \frac{\mathrm{d}z}{z} = 0\\ \frac{\mathrm{d}x}{x - \sqrt[3]{C - x^3}} - \frac{\mathrm{d}y}{\sqrt[3]{C - y^3} - y} = \frac{\mathrm{d}z}{z}\\ \textsf{Integrating both sides, we have integrals}\\ \textsf{that cannot be expressed as standard}\\ \textsf{integrals except at}\, C = 0.\\ \textsf{Evaluating at}\, C = 0,\, \textsf{we have;}\\ \frac{\mathrm{d}x}{2x} - \frac{\mathrm{d}y}{-2y} = \frac{\mathrm{d}z}{z}\\ \int\, \frac{\mathrm{d}x}{2x} + \int\, \frac{\mathrm{d}y}{2y} = \int\,\frac{\mathrm{d}z}{z}\\ \frac{\ln{x}}{2} + \frac{\ln{y}}{2} + B = \ln{z}\\ \frac{\ln{Axy}}{2}= \ln{z} \\ \therefore z = F\sqrt{xy} \,\textsf{is a solution to}\\ \textsf{the partial differential equation.}\\ \textsf{Where}\, C, B, F \, \textsf{are arbitrary constants}\\ \textsf{Therefore, the solutions to the PDE are}\\ \phi\left(x^3 + y^3, \frac{z}{\sqrt{xy}}\right) = 0.$