Answer to Question #138947 in Differential Equations for Smrutismita Lenka

Question

Find the general solution of the pde
(y-zx)p+(x+yz)=x^2+y^2

Answer 1

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

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

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

The subsidiary equations are

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

Using the method of multipliers in each fraction,

$\dfrac{ydx}{y(y-xz)}=\dfrac{xdy}{x(x+yz)}=\dfrac{-dz}{-1(x^{2}+y^{2})}$

=$\dfrac{ydx+xdy-dz}{y(y-xz)+x(x+yz)-(x^2+y^2)}$

=$\dfrac{ydx+xdy-dz}{y^2-yxz+x^2+xyz-x^2-y^2}$

=$\dfrac{ydx+xdy-dz}{0}$

$\to ydx+xdy-dz=0$

$\to d(xy)=dz$

Integrating Both the sides

$\int d(xy)=\int dz$

$xy=z+c$ $_1$

c$_1$ =xy-z

Again using method of multiplier

=$\dfrac{xdx}{x(y-xz)}=\dfrac{-ydy}{-y(x+yz)}=\dfrac{zdz}{z(x^{2}+y^{2})}$

=$\dfrac{xdx-ydy+zdz}{xy-x^2z-yx-y^2z+zx^2+zy^2}$

$= \dfrac{xdx-ydy+zdz}{0}$

$\to xdx-ydy+zdz=0$

Integrating above equation we get,

$\to \dfrac{x^2}{2}-\dfrac{y^2}{2}+\dfrac{z^2}{2}+c=0$

$\to c_2=y^2-x^2-z^2$

The solution of the given equation are

$xy-z=c_1$ and $x^2-y^2+z^2+c_2=0$