Deriving general solution of a PDE $ \frac{ \partial u }{\partial x } + \frac{ \partial u }{\partial y } = \sqrt{ u } $. We have the PDE

$$ \frac{ \partial u }{\partial x } + \frac{ \partial u }{\partial y } = \sqrt{ u } $$

The characteristic curves are 
$$ dx = dy = \frac{ d z }{\sqrt{z} } $$
Solving this system, we obtain that 
$$ x = 2 \sqrt{z} + A $$
$$ y = 2 \sqrt{z} + B $$
Thus, we have the 
$$ g(x,y,u) = x - 2 \sqrt{u} $$
$$ h(x,y,u) = y - 2\sqrt{u} $$
Thus, solution $u = u(x,y)$ satisfies 
$$ F( x - 2\sqrt{u}, y - 2 \sqrt{u}) $$
where $F(g,h) $ is a smooth function satisfying $F_g^2 + F_h^2 \neq 0$.
Is this correct? How can I write a formula for $u $ explicitly? thanks
 A: You treat your equation as a quasilinear,
and so your solution is a way too complicated.
Better, think of the equation as a semilinear
one. It is known that if $(a,b) \ne (0,0),$ then the change of variables
$$
\begin{cases}
s = ax+by,\\
t= bx-ay
\end{cases}
$$
transforms the linear PDE 
$$
a u_x + bu_y=0
$$
to
$$
(a^2+b^2) w_s=0. 
$$
In effect, the change of variables
$$
\begin{cases}
s = x+y,\\
t= x-y
\end{cases}
$$
transforms your PDE to
\begin{equation*} \tag{$*$}
2 w_s =\sqrt{w}
\end{equation*} 
which is reducible to the separable ODE
$$
2z'(s)=\sqrt{z(s)} \iff 2\sqrt{z(s)}-\frac s2 = C
$$
where $C$ is an arbitrary constant. It follows that
any solution of $(*)$ satisfies
$$
2\sqrt{w(s,t)}-\frac s2=f(t)
$$
(why?) and hence any solution of the original PDE satisfes
$$
2\sqrt{u(x,y)}-\frac {x+y}2=f(x-y),
$$
where $f$ is a continuously differentiable function 
(you can consult my book
on PDEs to see what are the standard methods of solution
of first-order semilinear and quasilinear PDEs.)
