For part (d) of the following question should I look at the integrated version of the equation in part (a) in order to look at how $u$ evolves i.e. take limits of $s$ or does it make sense to just look at the equation in part (a) i.e. the change in $u$ with respect to $\tau$.
Also, is taking the limit of $s$ the right path to follow? If not what is the right way to solve this question?
Any help would be really appreciated!
For part (d), when the activating signal is $0$ (e.g. $s= 0$), then the ode takes the form: \begin{equation} \frac{du}{d\tau} = - ru + \frac{u^2}{1+u^2} \end{equation} Note that since $u(0) = 0$, this means $\frac{du}{d\tau} = 0$, so $u(\tau) = 0$ for all $\tau$ given $s = 0$.
Now, the question wants to slowly increase $s$, so $s>0$. At some earlier time, then we can approximate $u \approx 0$ giving: \begin{equation} \frac{du}{d\tau} \approx s > 0 \end{equation} This implies $u(\tau) > 0$ at least for some time after $s>0$. Next, the question wants to know if $u(\tau)$ reverts back to $0$ once the activating signal is gone, e.g. $s \rightarrow 0$. Note that this gives us back: \begin{equation} \frac{du}{d\tau} = - ru + \frac{u^2}{1+u^2} \end{equation} However, this time $u(\text{initial}) = u(\text{immediately after signal is gone}) > 0$. This times, you will need to find the stability of the equilibrium $u^* = 0$. To do this, we differentiate $f(u) = -ru + \frac{u^2}{1+ u^2}$ to obtain: \begin{equation} f'(u) = -r + \frac{2u}{1+u^2} \end{equation} At equilibrium then, $f'(u^*) = -r$, hence $u^*$ is locally asymptotically stable. This means $u$ would still go back to $0$ once the activating signal is gone.