Answer to Question #138947 in Differential Equations for Smrutismita Lenka

Question #138947
Find the general solution of the pde
(y-zx)p+(x+yz)=x^2+y^2
1
Expert's answer
2020-10-22T18:00:36-0400

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}\\\\\n\\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"





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