Question

Find the general solution of

(y-xu)p+(x+yu)q=x²+y²

Answer 1

Given, $(y-xu)p+(x+yu)q = x^{2}+y^{2}$.

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

Here, $P = y-xu, Q=(x+yu), R = x^{2}+y^{2}$.

The subsidiary equations are

$\dfrac{dx}{P}=\dfrac{dy}{Q}=\dfrac{du}{R}\\ \dfrac{dx}{y+xu}=\dfrac{dy}{-(x+yu)}=\dfrac{du}{x^{2}+y^{2}}~~~~~~~~~~~-(1)$.

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

Let us consider,

$\dfrac{xdx+ydy}{(x^{2}-y^{2})u} = \dfrac{du}{x^{2}-y^{2}}\\ xdx+ydy=udu\\ \text{Integrating, }\\ \int xdx + \int ydy = \int zdz\\ \dfrac{x^{2}}{2}+\dfrac{y^{2}}{2} =\dfrac{z^{2}}{2}+\dfrac{c_{1}}{2}\\ x^{2}+y^{2}-u^{2}=c_{1}$

Consider,

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

The general solution is,

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