Answer in Differential Equations for Smrutismita Lenka #138947
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}\ndfrac{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”
