Question

Find the differential equation of the family of surfaces ∅(x+y+z , x²+y²-z²) =0.
what is the order of this PDE ?

Answer 1

$u = x+y+z,v=x^2+y^2-z^2. \ F(u,v)=0. \newline 0 = F_x = \frac{\partial F}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial F}{\partial v}\frac{\partial v}{\partial x}. \newline 0 = F_y = \frac{\partial F}{\partial u}\frac{\partial u}{\partial y}+\frac{\partial F}{\partial v}\frac{\partial v}{\partial y}. \newline Let \ \frac{\partial u}{\partial x}=\frac{\partial u}{\partial x}\frac{\partial x}{\partial x} + \frac{\partial u}{\partial y}\frac{\partial y}{\partial x}+\frac{\partial u}{\partial z}\frac{\partial z}{\partial x}=1+p. \newline Let \ \frac{\partial u}{\partial y}=\frac{\partial u}{\partial x}\frac{\partial x}{\partial y} + \frac{\partial u}{\partial y}\frac{\partial y}{\partial y}+\frac{\partial u}{\partial z}\frac{\partial z}{\partial y}=1+q. \newline \frac{\partial v}{\partial x}=2x-2zp. \newline \frac{\partial v}{\partial y}=2y-2zq. \newline F_x=\frac{\partial F}{\partial u}(1+p)+\frac{\partial F}{\partial v}(2x-2pz)=0. \newline F_y=\frac{\partial F}{\partial u}(1+q)+\frac{\partial F}{\partial u}(2y-2qz)=0. \newline Eliminating \ \frac{\partial F}{\partial u} and \frac{\partial F}{\partial v} we \ get: \newline (1+p)(2y-2qz)-(1+q)(2x-2pz)=0. \newline y-qz+yp-qpz-x+pz-qx+qpz=0. \newline p(y+z)-q(x+z)=x-y \ - the \ resulting \ equation.$