Bifurcation diagrams. No idea what i'm doing...

The differential equation depends on a parameter $a\in\mathbb{R}$. Find the equilibrium points and determine whether they are source, sink or neither and sketch the bifurcation diagram. $$x'(t)=x^3-2ax^2+4x$$ where $a\geq 0$

Question

Bifurcation diagrams. No idea what i'm doing...

The differential equation depends on a parameter $a\in\mathbb{R}$. Find the equilibrium points and determine whether they are source, sink or neither and sketch the bifurcation diagram. $$x'(t)=x^3-2ax^2+4x$$ where $a\geq 0$

abb0t · Answer

If I remember correctly, to find the equilibrium points, find the zero's. So factor out an x...
$$x(x^2-2ax+4)=0$$

abb0t · Answer

plug in the equilibrium points for f'. and If f is less than the equilibrium points then it's a source, and sinks if it's greater than equilibrium point.

richyw · Answer

ok what I am struggling with is how to split it up. like for some values of a there are a different amount of equlibrium points. are there any resources that just explain what is going on?
i'm at my wits end with Hirsch, Differential equations. There isn't even a solution manual

swissgirl · Answer

I found something that may be helpful

swissgirl · Answer

http://www.google.ca/url?sa=t&rct=j&q=&esrc=s&source=web&cd=3&ved=0CEIQFjAC&url=http%3A%2F%2Fwww.math.utah.edu%2F~gustafso%2F2250phaseline.pdf&ei=HDb-UPrbO8Xd0QGDyICADA&usg=AFQjCNGFuGaShSKXuMS06cs1UZwS7v94Qw&sig2=lRJCHNClHKGY4Pa6Aa-yLQ&bvm=bv.41248874,d.dmQ

swissgirl · Answer

Go to page 53 Example 31