Method of characteristics - eliminating variables I am trying to follow a guide for the method of characteristics; quoting the first example:

We use the method of characteristics to solve the problem
$ 2u_x - u_y = 0, \;\; u(x, 0) = f(x) $
(...) we can easily solve [the characteristic equations] to get:
$x = 2s + x_0 \;\;\; y= -s + y_0 \;\;\; u = u_0$

Up to here I have no problem. However, the guide then substitutes in the initial condition and says this:

we now eliminate $x_0$ and $s$ to find that:
$ y = -s \;\;\; \Rightarrow \;\;\; u = f(x_0) = f(x - 2s) = f(x + 2y)$

My question is: where did $y_0$ and $u_0$ go?
 A: You have characteristics given as functions $x(s)$ and $y(s)$ which depend on parameter $s$:
\begin{align}
x = x(s) &= x_0 + 2s && \implies& x_0 &= x - 2s   \label{1}\tag{1}
\\ 
y = y(s) &= y_0 - s  && \implies& y_0 &= y + s   \label{2}\tag{2}
\end{align} 
The initial condition $u\left(x_0, 0\right) = f\left(x_0\right)$ is given for $y_0=0$, so in equation $\eqref{2}$ we set $y_0$ to be equal  $0$:
$$0 = y_0 = y + s \qquad\;\implies \quad\bbox[5pt, border: 1.5pt solid #DD0712]{y = -s^{\,\!}}  \label{3}\tag{3}$$
But then we can write general solutions as 
$$
u\left(x_0, y_0\right) = u\left(x_0, 0\right) = f\left(x_0\right) = f\left(x-2s\right) \overset{\eqref{3}}{=} f\left(x+2s\right)
$$

Now, one can ask why do we say that $u\left(x_0, y_0\right)$ is the general solution?
Well, if you recall the definition of characteristics, which are the 

$\;\big(\cdots\big)\;$ parametric curves along which solution $u\left(x, y\right)$ remains constant,

you will see that $u\left(x, y\right) = u\left(x_0, y_0\right)$ as long as $(x,y)$ and $\left(x_0,y_0\right)$ lie on the same curve $\eqref{1}$–$\eqref{2}$.
Then we choose $y_0$ to be equal $0$ in order to incorporate initial condition $u(x,0) = f(x)$.
A: The characteristic equations are :
$$\frac{dx}{2}=\frac{dy}{-1}=\frac{du}{0}$$
which leads to the equations of chararacteristics :
$$\begin{cases}u=c_1\\x+2y=c_2 \end{cases}$$
The general solution on implicit form with is :
$$\Phi\left(u\:,\:x+2y\right)=0$$
where $\Phi$ is any derivable function of two variables.
This is equivalent to :
$$u=F(x+2y)$$
where $F$ is any derivable function.
The bounding condition leads to :
$$u(x,0)=f(x)=F(x+0)=F(x)$$
Hense $F=f$ is now determined and the solution is :
$$u(x,y)=F(x+2y)=f(x+2y)$$ 
