TITLE: Equilibrium points of the ODE $y'=\sin y−\frac{y}{2}$.
QUESTION [2 upvotes]: Find the equilibrium points of the ODE, and investigate their stability:
$$y'=\sin y−\frac{y}{2}.$$
I know the equilibrium points are about $1.9$, $-1.9$, and $0$.  Not sure how to get to that point though.
I went ahead and tried to create a graph to find the stability of the problem.  I got this (forgive my poor artwork): http://i.imgur.com/5kxSb1g.png
From this I'm inferring that at $+1.9$ and $-1.9$ the solution is stable, and at $0$ it's unstable because of the split.  Would be great if someone confirmed this.
Thanks for the help.

REPLY [1 votes]: To find the equilibrium, set $y'=\sin y-\frac y2=0$ and you will obtain three points $0,y_+,y_-$.
To determine the stability, first we investigate $y^*=0$ and consider first-order approximation
$$y'\approx y-\frac y2=\frac12y$$
And solution is like $y=Ce^{x/2}$, we know that it's unstable at zero.
Then around $y_+$, we can write $y=\pi+\epsilon$ and obtain approximation equation
$$y'=\sin(\pi+\epsilon)-\frac y2=-\sin\epsilon-\frac y2\approx-\epsilon-\frac y2=\pi-\frac32y$$
With the same method, we know it's a stable point. Likewise you can determine the stability of $y_-$ and I leave it as an exercise.
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