Answer to Question #137047 in Differential Equations for Nikhil

Question

Solve
dx/x+y-xy^2=dy/x^2-x-y=dz/z(y^2-x^2)

Answer 1

$\textsf{This is a simultaneous Total}\\\textsf{Differential equation.}\\ \textsf{We solve it by Lagrange's multiplier.}\\ \displaystyle\frac{(l\mathrm{d}x+m\mathrm{d}y +n\mathrm{d}z)}{(l\mathrm{P}+m\mathrm{Q}+n\mathrm{R})} =0.\\ l\mathrm{P}+m\mathrm{Q}+n\mathrm{R} = 0.\hspace{0.1cm}\textsf{So} \hspace{0.1cm}l\mathrm{d}x+m\mathrm{d}y +n\mathrm{d}z =0.\\ \textsf{Here}\hspace{0.1cm} l,m,n \hspace{0.1cm}\textsf{are multipliers.}\\ \textsf{The Lagrange’s auxiliary equations}\\\textsf{for given PDE is}\\ \displaystyle\frac{\mathrm{d}x}{x+y-xy^2}=\frac{\mathrm{d}y}{x^2y-x-y} = \frac{\mathrm{d}z}{z(y^2-x^2)}\\ \displaystyle\textsf{Choosing}\hspace{0.1cm} \left(x,y,\frac{1}{z}\right) \hspace{0.1cm}\\ \textsf{as multipliers, we have}\\ x\mathrm{d}x + y\mathrm{d}y + \frac{1}{z} \mathrm{d}z = 0\\ \textsf{Integrating both sides, we have thus,}\\ \int x\mathrm{d}x + \int y\mathrm{d}y + \int \frac{1}{z} \mathrm{d}z = 0\\ x² + y² + 2\ln(z) = C_1 \\ \textsf{From the first two equations, we have}\\ (x^2y-x-y) \mathrm{d}x - (x+y-xy^2)\mathrm{d}y= 0\\ \textsf{Solving the above DE, we have}\\ (x^2 y - x) \mathrm{d}x - (y - xy^2)\mathrm{d}y - y\mathrm{d}x - x\mathrm{d}y = 0\\ (x^2 y - x) \mathrm{d}x - (y - xy^2)\mathrm{d}y = y\mathrm{d}x + x\mathrm{d}y \\ x(xy - 1) \mathrm{d}x - y(1 - xy)\mathrm{d}y = y\mathrm{d}x + x\mathrm{d}y \\ (xy - 1)(x\mathrm{d}x + y\mathrm{d}y) = y\mathrm{d}x + x\mathrm{d}y \\ \displaystyle x\mathrm{d}x + y\mathrm{d}y = \frac{y\mathrm{d}x + x\mathrm{d}y}{xy - 1}\\ \displaystyle \int x\mathrm{d}x + \int y\mathrm{d}y = \int \frac{\mathrm{d}(xy)}{xy - 1}\\ \displaystyle\frac{x^2}{2} + \frac{y^2}{2} = \log{(xy - 1)} + C_2\\ \displaystyle\frac{x^2}{2} + \frac{y^2}{2} - \log{(xy - 1)} = C_2\\ \textsf{Thus, the solution of the given PDE is}\\ \phi\left(x² + y² + 2\ln(z),\right.\\\left.\frac{x^2}{2} + \frac{y^2}{2} - \log{(xy - 1)}\right) = 0 \hspace{0.2cm}\forall C_1, C_2\in \mathbb{R}$