problem solving PDE: $u*u_x + u_y = -u$ $$u*u_x + u_y = -u$$
$$u(0,t) = e^{-2t}$$
I tried solve with Lagrange and got 2 surfaces
$\phi(x,y,u) = x+u$ and $\psi(x,y,u) = y+ln(u)$ . when I used $u(0,t) = e^{-2t}$, I got a solution $$ u(x,y) = \dfrac{(1+\sqrt{1+4xe^{2y}})*e^{-2y}}{2} $$
but that seem to be wrong solution. I don't know what went wrong. your help is appreciated. 

Solving it with the method of charateristics, I obtained the general solution on implicit form :
$$\Phi\left(x+u\:,\:y+\ln(u)\right)=0$$
which is consistent with your result: $\phi(x,y,u) = x+u$ and $\psi(x,y,u) = y+ln(u)$
$\Phi(\phi,\psi)$ is any derivable function of two variables. An equivalent form is :
$$x+u=F\left(y+\ln(u)\right)$$
where $F$ is any derivable function.
So, the key point is the boundary condition $u(0,y)=e^{-2y}$
In $x=0$ we have : $$0+e^{-2y}=F\left(y+\ln(e^{-2y})\right) = F(-y)$$
This determines the function $\quad F(z)=e^{2z}\quad$ or any other symbol for the variable.
Puting it into the general solution gives :
$$x+u=e^{2\left(y+\ln(u)\right)} = u^2e^{2y}$$
Then, solving $e^{2y}u^2-u-x=0$ for $u$ leads to :
$$u=\frac{1\pm\sqrt{1+4e^{2y}x}}{2e^{2y}}$$
This is the correct result because bringing it back into $\quad u\:u_x+u_y+u =0\quad$ verifies the equality.
