Differential Equations: Stable, Semi-Stable, and Unstable I am trying to identify the stable, unstable, and semistable critical points for the following differential equation: $\dfrac{dy}{dt} = 4y^2 (4 - y^2)$.
If I understand the definition of stable and unstable critical points, then it seems to me that the semi-stable point is at $y = 0$ since in a neighborhood around $y=0$ we have the slope as positive. It also seems that $y = -2,2$ is a stable critical point since points around its neighborhood are negative for decreasing values and positive for increasing values. So it appears there are no unstable critical points. Am I on the right track here?
I also need to find $k$ for $y(t) \rightarrow k$ given $y(0) = 1.4$ and $k$ for $y(t) \rightarrow k$ given $y(0) = -3.2$.
Any help anyone can provide is appreciated. 
 A: Let me first say that which fixed points are stable/unstable and then the reasons.
From the equation $y'=4y^2(4-y^2)$, the fixed points are $0$, $-2$, and $2$. The first one is inconclusive, it could be stable or unstable depending on where you start your trajectory. $-2$ is unstable and $2$ is stable.
Now, there are two ways to investigate the stability. Since we have a one-dimensional system, the better way would be to draw the graph of 
$$f(y)=4y^2(4-y^2)$$
1) On each fixed point, you have two regions (left and right) with your fixed point as the middle point. For example, on the fixed point $y=-2$, the left hand region corresponds to $y<-2$ while the right hand region corresponds to $y>-2$.
2) Let's look at the fixed point $y=-2$. If $f(y)<0$ when $y<-2$, we draw a right-hand-arrow on the left hand region; otherwise draw a left-hand-arrow instead. Repeat the same method on the right hand region. 
3) Repeat step (2) for all fixed points.
For our case, you should get a left-hand arrow on $y<-2$ and right-hand arrow on $y>-2$. So, this implies that the fixed point $y=-2$ is unstable. The same method reveals then that $y=2$ is stable. For $y=0$, you would get right-hand arrow on both regions, so this indicates that we have the so called structural stability. For our case, if we start on $y<0$, we will be attracted towards $y=0$; if we start on $y>0$, we will be repelled from $y=0$. 
Another method would be to calculate the derivative of $f(y)$ and evaluate it at all the fixed points.
1) If $f'(y)<0$, the fixed point is stable.
2) If $f'(y)>0$, the fixed point is unstable.
3) if $f'(y)=0$, it is inconclusive. 
