Question

(x²-y²-z²) p+2xyz=2xz

Answer 1

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,

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

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