Solve partial differential equation using method of characteristics. Given:
$zp + yq = x$      
$x_{0} = s$ (x at time t = 0)
$y_{0} = 1$ (y at time t = 0)
$z_{0} = 2s$ (z at time t = 0)
Approach :
$ f=zp+yq-x$
Using Characteristic Equations :
$x'(t) = f_{p}$
$y'(t) = f_{q}$
$z'(t) = pf_{p} + qf_{q}$
$p'(t) = -f_{x}-pf_{z}$
$q'(t) = -f_{y}-qf_{z}$
I got : 
$y=e^{t}$
$p = (1+3e^{2t})/(3e^{2t}-1) $
But I am not able to proceed further.
 A: From the comment, the function to be found isn't on the form $f(x,y,z)$ , but $z(x,y)$.
So, the PDE is :
$$z\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y}=x \tag 1$$
Characteristic system : $\quad\frac{dx}{z}=\frac{dy}{y}=\frac{dz}{x}\tag 2$
One could parametrize in letting Eq.$(2)$ $=ds$ or $=dt$ depending the common symbolism used. But this case is so simple that it is even not necessary to parametrize.
A first characteristic equation comes from $\quad \frac{dx}{z}=\frac{dz}{x} \quad\to\quad z^2-x^2=c_1$
A second characteristic equation comes from $\frac{dx}{z}=\frac{dy}{y}=\frac{dx}{\sqrt{c_1+x^2}} \quad\to\quad \frac{\sqrt{c_1+x^2}+x}{y}=c_2=\frac{z+x}{y}$ 
The general solution of the PDE $(1)$ is expressed on the form of implicit equation :
$$z^2-x^2=F\left(\frac{z+x}{y}\right) \tag 3$$
where $F$ is any differentiable function.
Initial condition : $z(x_0,1)=2x_0\quad\to\quad (2x_0)^2-x_0^2=F\left(\frac{2x_0+x_0}{1}\right)=3x_0^2=F(3x_0)$
The function $F$ is determined : $\quad F(X)=\frac{1}{3}X^2$
Putting this function into Eq.$(3)$, with $X=\frac{z+x}{y}$ leads to : 
$$z^2-x^2=\frac{1}{3}\left(\frac{z+x}{y}\right)^2$$
Solving for $z$ :
$$z(x,y)=x\frac{3\:y^2+1}{3\:y^2-1}$$
