Solve: $u_x^2+u_y^2+2(u_x-x)(u_y-y)-2u=0, u(x,0)=0$

Question

Solve :$u_x^2+u_y^2+2(u_x-x)(u_y-y)-2u=0, u(x,0)=0$

My attempt: $f(x,y,u,p,q)=p^2+q^2+2pq-2py-2xq+2xy-2u=0$

using charpits equation :

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

i cant go further can any one help

Answer 1

Hint: $$\frac{dx-dp}{-2p-2q+2y+2q-2y+2p}=\frac{dy-dq}{-2q-2p+2x+2p-2x+2q}=\frac{du}{-pdp-qdq} $$ which implies that : $$dx-dp=0 \Rightarrow p=x+a $$ and $$dy-dq=0 \Rightarrow q=y+b $$