General solution of PDE $u_x + u_y = \sqrt{u} $ 
Consider the equation $u_x + u_y = \sqrt{u} $ 
Derive the general solution 
Observe that the trivial solution $u(x,y) ≡ 0 $ is not covered by the
  general solution.

This is my solution:
Characteristic equation;
${\frac {dx}{1}}$=${\frac {dy}{1}}$=${\frac {du}{\sqrt{u}}}$
solving the ﬁrst equality
$x-y=c_1$
$dx={\frac {du}{\sqrt{u}}}$
$c_2=2\sqrt{u}-x$
$F(c_1,c_2)=0$ then $c_2=f(c_1)$
$F(x-y,2\sqrt{u}-x)=0$
$2\sqrt{u}-x=f(x-y)$ 
then
$u={\frac {(x+f(x-y))^2}{4}}$
is my solution true? i am not sure and i don't understand how to observe that trivial condition is not covered by the general solution.  i would appreciate it if you help.
 A: The derivation and the proposed solution look correct. Indeed,
we have
$$
u_x+u_y=\frac{x+f(x-y)}{2}=\sqrt{u}
$$
for $x+f(x-y)\geq 0$. Note that $u\equiv 0$ would imply $x+f(x-y) = 0$ for all such $x$, $y$, which is impossible.
A: Your solution is correct. I did this problem in my PDE course and have posted my own solution.  
We first write the characteristic equations as
$$\frac{dx}{1}=\frac{dy}{1}=\frac{dz}{\sqrt{z}} $$
Then, solving the first equality,
$$dx = dy ~\Rightarrow~ x = y+c_1 ~\Rightarrow~ x-y = c_1$$
So, we have found a function $\phi(x,y,z)$ such that  $\phi(x,y,z)=x-y=$ constant and can verify that this satisfies $a\phi_x+b\phi_y+c\phi_z=0$.  We now need to find some $\psi(x,y,z)=$ constant which is independent of $\phi$. By the first and last equations,
$$dx = \frac{dz}{\sqrt{z}} ~\Rightarrow~ dx= z^{-\frac{1}{2}}dz ~\Rightarrow~ x+c_2=2\sqrt{z} ~\Rightarrow~ c_2=2\sqrt{z}-x$$
Therefore our second function is $\psi(x,y,z)=2\sqrt{z}-x=$ constant. We have that $a\psi_x+b\psi_y+c\psi_z=0$. Hence, we have satisfied $F(\phi,\psi)=0$ for an arbitrary $F\in C^1(\mathbb R^2)$. So, we can let $\psi=f(\phi)$ where $f\in C^1(\mathbb R)$ is an arbitrary function. Then,
\begin{align*}
\begin{split}
2\sqrt{z}-x=f(x-y) &~\Rightarrow~ 2\sqrt{z}=x+f(x-y) \\&~\Rightarrow~
\sqrt{z}=\frac{x+f(x-y)}{2} \\&~\Rightarrow~
z=u(x,y)=\frac{(x+f(x-y))^2}{4}
\end{split}    
\end{align*}
To see that $u(x,y)\equiv 0$ isn't covered by the general solution, we can observe that for some arbitrary $f$, where $u(x,y)$ is defined as
$$u(x,y)=\frac{(x+f(x-y))^2}{4}$$
isn't equal to $0$ as $f$ is arbitrary. We are not free to choose $f$ so that $u(x,y)\equiv 0$. Therefore, the trivial solution $u(x,y)\equiv 0$ isn't covered by the general solution.
