Question

(x^2-y^2 -zy)p +(x^2-y^2-zx)q =z(x-y)

Answer 1

By lagrange's auxilary equation - $\dfrac{dx}{x^{2}-y^{2}-zy}=\dfrac{dy}{x^{2}-y^{2}-zx}=\dfrac{dz}{z(x-y)}$

now in first two ratio , deviding numerator and denominator by x and y respectively ,

$= \dfrac{xdx}{x^{3}-xy^{2}-xzy}=\dfrac {ydy}{x^{2}y-y^{3}-zxy}$ = $\dfrac {{dz}/{z}}{{x-y}}$

$=$ $\dfrac{xdx-ydy}{x^{3}+y^{3}-xy(x+y)}$ $=\dfrac{dz/z}{x-y}$

$=\dfrac{xdx-ydy}{(x+y)(x-y)^2}=\dfrac{dz/z}{x-y}$

$=\dfrac{xdx-ydy}{x^{2}-y^{2}}=\dfrac{dz}{z}$

$=\dfrac{d(x^{2}-y^{2})}{2(x^{2}-y^{2})}$ $=\dfrac{dz}{z}$ , now integerating both sides we get ,

$=z^{2}=x^{2}-y^{2}+c_1$ which is required solution.