Let $\alpha:\mathbb R^n\to (0,\infty)$, $b:\mathbb R^n\times\mathbb R\to\mathbb R^n$ and $c:\mathbb R^n\times\mathbb R\to\mathbb R$ be smooth functions. Then the differential equations $$ b(x,u(x))Du(x)+c(x,u(x))=0 \quad\mathrm{in}\ \mathbb R^n$$ and $$ b_\alpha (x,u(x))Du(x)+c_\alpha(x,u(x))=0\quad\mathrm{in}\ \mathbb R^n$$ (with $b_\alpha(x,z)=\alpha(x)b(x,z)$ and $c_\alpha$ accordingly) have the same solutions $u$.
Using the method of characteristics we obtain the characteristic equations \begin{align*}\tag{$*$} \frac{dz}{ds} & = -\alpha(x) c(x,z) \\ \frac{dx}{ds} & = \alpha(x)b(x,z) \end{align*}
for the scaled PDE and \begin{align*}\tag{$**$} \frac{dz}{ds} & = - c(x,z) \\ \frac{dx}{ds} & = b(x,z) \end{align*}
for the original PDE. It is known that solutions $(x,z) = (x(s),z(s))$ of these characteristic systems are solutions of the PDE, $u(x) = z(x(s))$.
Now I am wondering: How are the solutions of $(*)$ and $(**)$ connected? In other words: If I calculate a solution $(x_\alpha,z_\alpha)$ of the characteristic system for the scaled PDE, can I obtain a solution $(x,z)$ of $(**)$ from it?
I thought that maybe I can scale $s$ by introducing another map $t$, and considering $x(t(s))$. This got me to $$\frac{du}{ds} = \frac{du}{dx}\frac{dx}{dt}\frac{dt}{ds}$$ and thus by comparing this with the original PDE\begin{align*} \frac{dt}{ds} & = \alpha(x(t(s))) \\ \mathrm{s.t.} \quad t(0) & = 0 \end{align*}
But I do not know how to proceed. Can I solve this for $t$ and is this even the right way? I'd appreciate any advice on this. Thanks!