Finding a general solution of pde $(x^2-y^2-u^2)\cdot u_x+(2xy)\cdot u_y=2xu$ how can I solve this partial dif. equation. I try to use Langarange methods
This is my solution;
$$\frac {dx}{x^2-y^2-u^2}=\frac {dy}{2xy}=\frac {du}{2xu}= λ$$
$$\frac {dy}{2xy}=\frac {du}{2xu}$$
$$\frac {du}{u}-\frac {dy}{y}=0$$
$$\ln u-\ln y=\ln c_1$$
$$\frac {u}{y}=c_1=w$$
then i know that we now need to ﬁnd another function. but i am stuck.
$$F(w,v)=0$$
$$w=f(v)$$
can u help me to find $v$ with steps, please?  i would appreciate it if you help.

There is a another way and this
$$\frac{du}{2xu}=\frac{xdx+ydy+udu}{x.(x^2-y^2-u^2)+y.2xy+u.2xu}$$
$$\frac{du}{2xu}=\frac{xdx+ydy+udu}{x^3-x.y^2-x.u^2+2xy^2+2xu^2}$$
$$\frac{du}{2xu}=\frac{xdx+ydy+udu}{x^3+x.y^2+x.u^2}$$ and if we simplify the equation
$$\frac{du}{2x.(u)}=\frac{xdx+ydy+udu}{x(x^2+y^2+u^2)}$$
$$\frac{du}{2x.(u)}=\frac{2xdx+2ydy+2udu}{2x(x^2+y^2+u^2)}$$ and cancled the $2x$ and use $d(x^2+y^2+u^2)=2dx+2dy+2du$ then we get
$$\frac{du}{u}=\frac{d(x^2+y^2+z^2)}{x^2+y^2+u^2}$$  and
$$ln|u|=ln|x^2+y^2+u^2|+lnd$$
$$ln|\frac{u}{x^2+y^2+u^2}|=lnd$$ then $$d=\frac{u}{x^2+y^2+u^2}=v$$ so $F(w,v)=F(\frac{u}{y},\frac{u}{x^2+y^2+u^2})=0$
