I'm trying to show that the general solution of the hyperbolic PDE, $$u_{xx}-2u_{xy}-3u_{yy}=0,$$ is $u(x,t)=F(3x+y)+G(x-y)$.
I thought I could reduce the given PDE to form two ODEs. So far I have computed,
\begin{align} u_{xx}+u_{xy}-3u_{xy}-3u_{yy}=0 \\ (u_x+u_y)_x-3(u_x+u_y)_y=0 \end{align} Letting $v=u_x+u_y$ gives the first order PDE, $$v_x-3v_y=0.$$ I'm unsure if I am on the right track and how to continue further. Is there a more systematic way of solving this problem?
Thank you very much.