# Explicit Runge-Kutta with Butcher Tableau not stricty positive

I have a hard time figuring out how explicit Runge-Kutta methods work, so this is a "what do I misunderstand?" kind of question.

Let $S$ be a system, $Y$ its variables vector and $f(t, Y)$ the function returning the derivative of each variables at a given time and a given state of the system.

All explicit Runge-Kutta methods imply to compute the value of $f$ for multiple $(t, Y)$ conditions, and then to perform a linear combination of all these derivatives in order to obtain a "better" derivative. The linear combination to perform is defined by a vector which can be found at the bottom of the Butcher tableau of the method. Lets call this vector $B$.

My question is :

as negative values can be found in some $B$ vectors (like those used by the Runge-Kutta-Fehlberg method) it is possible that the final derivative will be negative whereas all values produced by f are positive, and vice versa.

It seems that in some circumstances, the explicit Runge-Kutta methods including negative coefficients in their $B$ vector can produce irrelevant results by changing the sign of the final derivative. So how comes those methods can be used to produce meaningful results?

For "normal", "tame" (i.e., non-stiff) ODE one expects that the $k$ vectors $$k_i=f(t+h·c_i,y+h·\sum a_{ij}k_j)$$ differ by $O(h)$. Since $\sum b_i=1$, if the components of the $k$ vectors are well away from zero, there is no risk of sign errors. And anyway, in many ODE the relative size does not matter, it is simply a question of translation.