if $ \frac{dx}{dt}= x(a^2-x^2) $ find and classify the stability of the equilibrium points, and draw the bifurcation diagram.
I have found the stability of the function at $x =0,-a,a$ I believe $0$ is a unstable equilibrium while $-a,a$ are both stable equilibrium. I then drew the bifurication diagram, it looks like the absolute value graph but sideways, is this correct?
Thanks!