General solution to a first-order partial differential equation $$
\begin{cases}
\displaystyle u(x+u)\frac {\partial }{\partial } - y(y+u)\frac {\partial }{\partial }  = 0 \\
u=\sqrt y ,x =1
\end{cases}
$$
my idea: Can we solve by the method : 
$$\frac {x}{u(+u )} =  \frac{y}{-y( +u)}$$
 A: This problem is solvable by using the method of characteristics. Letting $t$ be the characteristic coordinate, we obtain this system of differential equations
$$\frac{dx}{dt} = u(x+u), \quad\frac{dy}{dt} = -y(y+u), \quad\frac{du}{dt} = 0$$
We can easily solve for $u$ provided the initial condition that $u(x(0),y(0))=\sqrt{y_0}$,
$$u = \sqrt{y_0}$$
Plugging this result into the differential equation for $x$ and integrating gives
$$x(t)=ce^{\sqrt{y_0}t}-\sqrt{y_0}$$
Using the initial condition that $x(0)=1$ gives
$$x(t) = \left(1+\sqrt{y_0}\right)e^{\sqrt{y_0}t}-\sqrt{y_0}$$
Upon solving the differential equation for $y$, we see after applying the initial condition $y(0)=y_0$ (and after some algebra) that
$$y(t)=\frac{y_0}{\left(1+\sqrt{y_0}\right)e^{\sqrt{y_0}t}-\sqrt{y_0}}$$
We can rewrite the equation for $y$ simply in terms of $u$ and $x$ thereby inverting the equation and eliminating the characteristic coordinate
$$y=\frac{u^2}{x}$$
Rearrangement gives the simple solution to the complicated-appearing PDE
$$\boxed{u(x,y) = \sqrt{xy}}$$
Plugging this back into the original PDE verifies the result. May the Fourth be with you.
A: Comparing
$$
\left\{
\begin{array}{rcl}
u(x+u)u_x-y(y+u)u_y & = & 0\\
u_x dx + u_y dy & = & du
\end{array}
\right.
$$
we conclude $u(x,y) = C_0$ along the characteristic curves and
$$
\frac{dx}{x+C_0}= - \frac{C_0dy}{y(y+C_0)}
$$
Integrating both sides
$$
\ln(C_0+x)+\ln(y)-\ln(y+C_0) = C_1
$$
From now on please follow the LutzL comments below. 

(LutzL copy from comment)
$$
\frac{(x+C_0)y}{y+C_0}=\pm e^{C_1}=C_2=\phi(C_0),
$$ 
as the solution is a one-parameter family of characteristic curves, that is there is only one degree of freedom between the two integration constants. Then insert the initial condition $C_0=u(1,y)=\sqrt{y}=t$, $y=t^2$, to get 
$$
\phi(t)=\frac{(1+t)t^2}{t^2+t}=t
$$ 
so that (using again that $C_0=u(x,y)$)
$$
(x+C_0)y=C_0(y+C_0)\iff xy=C_0^2=u(x,y)^2.
$$
