I'm wondering how you know which branches are stable. For example taking $\alpha$ as the bifurcation parameter $$y'=\frac{y(1500-y)}{3200}-\alpha.$$
So I plot $\alpha=\frac{y(1500-y)}{3200}$ and flip the axes to get
From here I'm not sure how to work out the stability of the branches. I know the bottom branch is the unstable one, just not sure of the method why. Many thanks.
We are given:
$$y'=\dfrac{y(1500-y)}{3200}-\alpha$$
We first want to find the critical points of $y' = 0$, yielding:
$$y' = \dfrac{y(1500-y)}{3200}-\alpha = 0 \rightarrow y_1 = 10 \left(75-\sqrt{5625-32 \alpha}~\right), y_2 = 10 \left(75 + \sqrt{5625-32 \alpha}~\right)$$
We then want to evaluate the sign of the derivative of $f(y) = \dfrac{y(1500-y)}{3200}-\alpha$ at each critical point. We have:
$$f'(y) = \dfrac{750-y}{1600}$$
At the first critical point, we have:
$$f'(y_1) = \dfrac{750-10 \left(75-\sqrt{5625-32 \alpha}~\right)}{1600}$$
A plot of this shows:
Notice how this is always positive (you should be able to prove this), hence is a Source (unstable).
At the second critical point, we have:
$$f'(y_2) = \dfrac{750-10 \left(\sqrt{5625-32 \alpha}+75~\right)}{1600} $$
A plot of this shows:
Notice how this is always negative (you should be able to prove this), hence is a Sink (stable).
We also have a common point, Node (neither stable nor unstable) at:
$$\alpha = \dfrac{5625}{32}, y = 750$$
If we draw the bifurcation diagram (red = unstable, green = stable, and the node where these two intersect as described above), we have: