I have the following relative simple differential equation:
${{dR}\over{dt}} = a-bR$
(with a steady state of $0=a-bR \Leftrightarrow R = {{a}\over{b}}$).
The integrated form of the equation can be calculated like this:
$\int {{dR}\over{dt}} dt = \int (a-bR) dt$ $\Leftrightarrow R = at-bRt + C$ $\Leftrightarrow R+bRt = at + C$ $\Leftrightarrow R(1+bt) = at + C$ $\Leftrightarrow R = {{at + C}\over{1+bt}}$
I now let t in the integrated function "go towards" $\infty$; like this I can check, whether the steady state is stable or unstable; (when the function goes towards $\pm \infty$ the function/steady state is unstable, else-wise the function should diverge towards the steady state for sufficient large t's - which basically means, that the steady state is stable, since the system has only a single steady state...)
$lim_{t\to\infty} {{at}\over{1+bt}} = lim_{t\to\infty} a/b = a/b$
So indeed, the limes indicates that the steady state should be stable.
However the function is only stable if $b > 0$, since:
$d/dx (a-bx)=-b$ $\Rightarrow$ system is stable for $b > 0$ $\Rightarrow$ system is unstable for $b < 0$
Where in the above calculation am I loosing the conclusion that the steady state is only stable for $b > 0$?
I think you made a mistake in calculating the integral \begin{align} \int (a-bR) dt \end{align} Because you forget to integrate $R$ itself. So suppose $R=t^2$. Than you have the following \begin{align} \int (a-bR) dt=\int (a-bt^2) dt=at-\frac{1}{3}bt^3\neq at-bt^3=at-bRt \end{align}