Solving $ u_x(x,y) u_y(x,y) = u(x,y)$ with characteristic method I want to solve the following nonlinear first order PDE:
$$ u_x(x,y) u_y(x,y) = u(x,y) $$
$$ u(0,y)=y^2 $$
I applied the characteristic method and found that:
$$ x(t)=e^t +c_2t$$
$$ y(t)=e^t+c_1t+y_0$$
$$ z(t)=u(x(t),y(t))=e^{2t} +(c_1+c_2)t+c_1c_2t+ y_0^2 $$
But I can not continue this to find the explicit solution.
 A: assume $u(x,y) = f(x + \lambda y) = f(u) \; \text{for some } \lambda > 0$
then the PDE is equivalent to the separable ode $\lambda (f') ^2 = f$
$$ \begin{align}
\lambda(f') ^2 = f \implies {df \over du} = { \sqrt f \over  \sqrt \lambda} \implies {df \over \sqrt f} = { du \over  \sqrt \lambda} \implies 2 \sqrt f = { u \over  \sqrt \lambda} + C \;\;\;\;\;\;\;\;\;\;\;\;\;\\
\text{we conclude that : }u(x,y) = f(x + \lambda y)= ( {x + \lambda y \over  2 \sqrt \lambda} + C' )^2 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \\
u(0,y) = y^2 \implies ( {\lambda y \over  2 \sqrt \lambda} + C' )^2 = y^2 \implies C' = 0, \; {\lambda \over  2 \sqrt \lambda} = ±1 \implies {\lambda ^2 \over  4 \lambda} = 1 \implies \lambda = 4 
\end{align}$$
so $$u(x,y) = ({x  \over 4} + y)^ 2$$
EDIT : in general if you have a non-linear PDE of the form $\Phi(u, u_x , u_y) = 0$
assume that $u(x,y) = f(x+ \lambda y) = f(v) \; \text{for some } \lambda > 0$ 
then $u_x = u_v $ and $u_y = \lambda u_v$
then when these are substituted in the original PDE it gives rise to an equivalent 1st order ODE which is easier to solve $\Phi_1 (u,u_v, \lambda u_v) = 0$ whose solution is the complete solution of the original PDE.
