Question

x²p+y²q+z²=0

Answer 1

$\frac{dx}{-f_p}=\frac{dy}{-f_q}=\frac{dz}{-pf_p-qf_q}=\frac{dp}{f_x+pf_z}=\frac{dq}{f_y+qf_z}$

$\frac{dx}{-x^2}=\frac{dy}{-y^2}=\frac{dz}{px^2+qy^2}=\frac{dp}{2px+2pz}=\frac{dq}{2qy+2qz}$

$-1/y=-1/x+c_1$

$c_1=1/x-1/y$

$\frac{-2dx/x+2dy/y}{2x-2y}=\frac{dp/p-dq/q}{2x-2y}$

$-2lnx+2lny=lnp-lnq$

$y^2/x^2=p/q$

$x^2p+x^2p+z^2=2x^2p+z^2=0$

$\frac{dx}{2x^2}=-\frac{dz}{z^2}$

$1/z=-1/(2x)+c_2$

$c_2=1/z+1/(2x)$

$F(c_1,c_2)=F(1/x-1/y,1/z+1/(2x))=0$