Answer to Question #154252 in Differential Equations for Prakash

Question

solve (y-xz)p+(yz-x)q=(x+y)(x-y)

Answer 1

Given, $(y-xz)p+(yz-x)q = (x+y)(x-y)$.

$\Rightarrow(y-xz)p+(yz-x)q = x^2-y^2$

This equation is of the form $Pp+Qq=R$ (Lagrange's linear partial differential equation).

Here, $P = y-xz, Q=yz-x, R = x^{2}-y^{2}$.

The subsidiary equations are

$\dfrac{dx}{P}=\dfrac{dy}{Q}=\dfrac{dz}{R}\\ \dfrac{dx}{y-xz}=\dfrac{dy}{yz-x}=\dfrac{dz}{x^{2}-y^{2}}~~~~~~~~~~~-(1)$.

Each ratio of (1) is equal to $\dfrac{xdx+ydy}{(y^{2}-x^{2})z} = \dfrac{dx+dy}{(1+z)(y-x)}$.

Let us consider,

$\dfrac{xdx+ydy}{(y^{2}-x^{2})z} = \dfrac{dz}{x^{2}-y^{2}}\\\Rightarrow xdx+ydy+zdz=0\\ \text{Integrating, }\\ \int xdx + \int ydy + \int zdz=0\\\Rightarrow \dfrac{x^{2}}{2}+\dfrac{y^{2}}{2} +\dfrac{z^{2}}{2}=\dfrac{c_{1}}{2}\\\;\\\Rightarrow x^{2}+y^{2}+z^{2}=c_{1}$

Consider,

$\dfrac{dx+dy}{(1+z)(y-x)}=\dfrac{dz}{(x+y)(x-y)}\\\Rightarrow \dfrac{dx+dy}{1+z}=\dfrac{dz}{-(x+y)}\\\Rightarrow -(x+y)d(x+y)=(1+z)dz\\ \text{Integrating,}\\ -\int (x+y)d(x+y)=\int (1+z)dz\\\Rightarrow -\dfrac{(x+y)^2}{2}=\dfrac{(1+z)^{2}}{2}-\dfrac{c_{2}}{2}\\\;\\ \Rightarrow (1+z)^{2}+(x+y)^{2}=c_{2}$

The general solution is,

$\phi(c_{1},c_{2})=0\\ \phi\left(x^{2}+y^{2}+z^{2},\left((1+z)^{2}+(x+y)^{2}\right)\right)=0.$