Method of characteristics - how to determine which one is the unique solution? I need to solve the following problem

$$xu_x + u_y = \sqrt{u}, \, u>0 $$
$$ u(x,0)=2+\sin(x).$$

Solving the characteristic equations implies

$x(t,s)=c_1(s)e^t, \, y(t,s)=t+c_2(s), \, u(t,s)=(\frac{t}{2} + c_3 (s) ) ^2 .$

Using $(x(0,s),y(0,s),u(0,s)) = (s,0,2+\sin(s))$ implies $c_1(s)=s, c_2(s)=0, c_3^2(s)=2+\sin(x)$, so that $u(x,y)=[\frac{y}{2}\pm \sqrt{2+\sin(xe^{-y})}]^2$.
My question is -

Which of the two options $c_3(s)=\sqrt{2+\sin(s)}$ or  $c_3(s)=-\sqrt{2+\sin(s)}$ is the unique solution?
(Transversality condition in this case implies that we have a unique solution )

Is it true that if we add the constraint $y>0$, then from $\sqrt{u}=\frac{t}{2}+c_3(s)$ we can deduce that $t>-2c_3(s)$, so that we must have $c_3(s)=\sqrt{2+\sin(s)}$ for a continuous solution?
Thank you!

When you solve for $z(t)=u(x(t),y(t))$ along the characteristic curve through $(s,0)$ for some fixed $s$, you have
$$
dz/dt = \sqrt{z}
,\qquad
z(0) = 2+\sin(s)
.
$$
This gives
$$
\frac{d}{dt} \sqrt{z} = \frac{dz/dt}{2 \sqrt{z}} = \frac12
,
$$
which after integration from $0$ to $t$ becomes
$$
\sqrt{z(t)} - \sqrt{2+\sin(s)} = \frac{t}{2}
,
$$
and the spurious solution with the wrong sign never even appears; you get
$$
z(t) = \left( \frac{t}{2} + \sqrt{2+\sin(s)} \right)^2
$$
right away.
