bifurcation in differential equations I have an equation
$$
y' = (2+y)(k-y^2)
$$
and I am asked to find the equilibrium solutions and bifurcation values for all values of $k$.
My approach was that... there exists $2$ equlibrium solutions for $k>0$, and $1$ equilibrium soltuion for $k = 0$ and $0$ equilibrium solution for $k < 0$.
After differentiating the function to
$$
f' = -4y - 3y^2
$$
I do not know how to continue. Do I plug in $\sqrt k$ and $\sqrt {-k}$, which is the equilibrium solution to the differentiated function...?
Thanks.
 A: Hints:
To find the critical points, we set up and solve $y' = 0$, hence:
$$y' = (2+y)(k-y^2) = 0 \implies y = -2, ~\pm ~\sqrt{k}$$
Now that we have these critical points, how do they behave for various values of $k$?
Can you determine how to find the bifurcation points from this?
Look at the direction field plot and what happens to  number of critical points for $k = 4, 1, 0, -1$:



Another point of interest is $k = 4$, as look at the direction field plot (collapses to two equilibrium points):

A: Find the equilibrium points as pointed out above by solving for $y'=0$ . As for bifurcation, we have to find the values of parameter $k$ at which there will be a qualitative change in dynamics.For $k>0$ , there will be three equilibrium points. 
Rewriting $y'$ as $y' =(2+y)(\sqrt(k)+y)(\sqrt(k)-y)$
Now if k >0 , there are three roots for this polynomial and two are negative. Also if k is not equal to 4, the roots will be distinct. When k=4 , two roots will be -2 and one will be 2. When roots are distinct , the smallest root will be a stable equilibrium point followed by an unstable and another stable equilibrium point in that order. But at K=4, the stable and unstable equilibrium points coalesce at y =-2 and the equilibrium will be stable in one direction and unstable in the other. Hence there is a qualitative change in dynamics at k=4.
Once again, as k becomes zero. the second and third equilibrium points will coalesce and produce a similar situation as discussed above but only with direction of stability reversed. Finally for k <0 , there will be only one equilibrium point and it will be stable . Hence there is a qualitative change in dynamics at $k =0$ too.
Hence my guess would be that the roots are -2, $\sqrt(k)$,$-\sqrt(k)$ 
and bifurcation points are $k=4 ,0$
