I have been looking at this past exam question for a while and I'm unsure how to answer: $$x ̇= r^2x−x^5,r ∈ {\rm I\!R}$$
a) Find the fixed points of the equation
b) Determine the stability of the fixed points
I know you have to set the equation equal to zero, and I feel as though I'm being quite dense in not being able to answer this question. Any help would be greatly appreciated. Thanks in advance!
Imposing $\dot x=0$ you get five solutions, $x=0$, $x=\pm \sqrt{r}$ and $x=\pm\sqrt{-r}$. These are the 5 fixed points.
To know their stability, substitute $x=x_0+\epsilon$ into the equation, where $x_0$ is a fixed point and $\epsilon$ is very small. Then linearize the equation in $\epsilon$. The resulting approximate equation will tell you the stability. If it tends to make $\epsilon$ grow, the point is unstable. If if tends to make $\epsilon$ decrease, it is stable.