Question

(Y2+z2-x2)p-2xyq+2xz=0

Answer 1

Given equation is Lagrange's linear equation

Pp+Qq=R

where P=y²+z²-x², Q=-2xy, R=-2xz.

The auxiliary equation is

$\frac{dx}{P}=\frac{y}{Q}=\frac{dz}{R}$

$\frac{dx}{y^2+z^2-x^2}=\frac{y}{-2xy}=\frac{dz}{-2xz}$....................................................(1)

Taking last two ratios,

$\frac{dy}{-2xy}=\frac{dz}{-2xz}\newline \frac{dy}{y}=\frac{dz}{z}\newline \text{Integrating both side,}\newline logy=logz+logc,\space \text {where c is a constant.}\newline c=\frac{y}{z}$

Taking Lagrangian multipliers as, x,y,z, each ratios of (1),

$\frac{xdx+ydy+zdz}{-x(x^2+y^2+z^2)}$

Now take,

$\frac{dy}{-2xy}=\frac{xdx+ydy+zdz}{-x(x^2+y^2+z^2)}\newline \frac{dy}{y}=\frac{d(x^2+y^2+z^2)}{(x^2+y^2+z^2)}\newline \text{Integrating both side,}\newline logy=log(x^2+y^2+z^2)+logb\newline ∴\frac{y}{x^2+y^2+z^2}=b\newline ∴ \text{The general equation is}\newline ϕ(\frac{y}{z},\frac{y}{x^2+y^2+z^2})=0$