Equilibrium points of the ODE $y'=\sin y−\frac{y}{2}$. 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.
 A: Hints:


*

*We find the equilibrium points by finding the point(s) where $y' = 0 \rightarrow \sin y - y/2 = 0$. You get the three roots mentioned.

*Here is a phase portrait for the three points.



Do you see the direction fields and how they behave for the three critical points? Conclusions? You just need to add direction to your result and maybe more initial points. Your conclusions are correct, but without the direction fields, I am not sure how you arrived at them.
A: 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.
