Reparametrizing characteristic curves for PDE's

I'm looking for solutions for a PDE that looks like this $$\nabla u(\vec x) \cdot f(\vec x) = k.$$ For some clarification, $u$ is a log-probability. And this arises from a Fokker-Plank-like equation, and I'd like to find the marginals $u(x_1)$ for example. The method of characteristics tells me that if i take characteristic curves parametrized by $$\frac{d \vec x}{dt} = f(\vec x),$$ the variation of $u(\vec x)$ along these curves will be $$\frac{du}{dt}= k.$$ But in my particular case, $f$ determines a dynamic system $\dot{\vec x} = f$ which converges to a fixed point only in infinite time. Since I'd like to be able to have $u(x)$ instead of $u(t;x_0)$ I've come to the idea of parametrizing the characteristic curves with constant velocity, that is, take $$\frac{d \vec x}{ds} = c \frac{f(\vec x)}{\|f(\vec x)\|},$$ which will result in a new parametrization s which goes from 0 to the length of the characteristic curve from the boundary condition to the equilibrium point. This will lead to an equation for $u(s)$ $$\frac{du}{ds}= \frac{ck}{\|f(\vec x)\|}.$$ Clearly this only corresponds to the original equation for a correct choice of the constant $c$, yet it seems to me that the PDE is satisfied for any value of $c$. I'm a little confused. Am I doing something terribly stupid along the way, or am I just missing the place where I should constrain the constant $c$.

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Thanks for pointing out the typo! But why doesn't the second set of equations solve the pde? If i have $u(s)$ given by the solution to the last equation, and just take $u(t) = u(t(s))$ where $t(s)$ is the change of variables, then $u(t(s))$ will solve the first equation as far as I could think it through. – alexsuse Feb 8 '12 at 10:14