Understanding Bifurcation Points in the Function dy/dt = y^3 + ay^2

[sammycaps]

Can anyone help me find the bifurcation value of dy/dt = y^3 + ay^2 where a is the parameter. I found that the bifurcation value is 0 but at that equilibrium point the phase graph shows a source, not a node, so I'm not totally sure. Someone help please!

 

[jackmell]

Sure looks like a bifurcation point to me. You have $y^2(y+a)$. Just look at it in terms of a polynomial: when a=0, it has one equilibrium point at y=0. Anything other than a=0, then it has two equilibrium points at y=0 and y=-a. Not sure why you feel it has to be a node. This is a nice way in Mathematica to look at it although Manipulate is kinda' slow to up date it in real time:

Manipulate[
 StreamPlot[{1, y^3 - a y^2}, {x, -5, 5}, {y, -5, 5}, 
  PlotRange -> {{-5, 5}, {-5, 5}}], {a, -1, 1}]