Answer to Question #157237 in Differential Equations for MOHIT RAIYANI

Question #157237

(x2-y2-z2) p+2xyz=2xz


1
Expert's answer
2021-01-26T03:16:33-0500

Given equation is Lagrange's linear equation "Pp+Qq=R"

The auxiliary equation is: "\\frac{dx}{x^2-y^2-z^2}=\\frac{dy}{2xy}=\\frac{dz}{2xz}"

Taking two last ratios:

"\\frac{dy}{2xy}=\\frac{dz}{2xz} \\implies \\frac{dy}{y}=\\frac{dz}{z}"

Integrating "\\ln{y}=\\ln{z}+\\ln{a} \\implies y=az"

Taking Lagrangian multipliers as, x,y,z, each ratios of "\\frac{dx}{x^2-y^2-z^2}=\\frac{dy}{2xy}=\\frac{dz}{2xz}"

"=\\frac{xdx+yd{y}+zdz}{x(x^2-y^2-z^2)+2xy^2+2xz^2}=\\frac{xdx+yd{y}+zdz}{x(x^2+y^2+z^2)}"

Now take,

"\\frac{dy}{2xy}=\\frac{xdx+ydy+zdz}{x(x^2+y^2+z^2)} \\implies \\frac{dy}{y}=\\frac{d(x^2+y^2+z^2)}{(x^2+y^2+z^2)}" "\\implies \\log{y}=\\log{x^2+y^2+z^2}" "\\implies y=x^2+y^2+z^2"

"f(\\frac{y}{z},\\frac{x^2+y^2+z^2}{y})=0"


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