Equilibrium solutions of differential equations

We are looking for where the derivative $\dfrac{dy}{dt} = 0$.

This is satisfied when $y^2-2 = 0$ and you do not want $y^2 - 4 = 0$ (division by zero).

This leads to the two equilibrium points, $y = \pm \sqrt{2}$.

We can find a closed-form solution (ugly) for this DEQ:

$$y(t)-\dfrac{\ln(\sqrt{2}-y(t))}{\sqrt{2}}+\dfrac{\ln(y(t)+\sqrt{2})}{\sqrt{2}} = c+\dfrac{t^3}{3}-t$$

A direction field plot shows (look at $y = \pm~ \sqrt{2}$):

Additionally, if we look at $t = \pm 1$, on the following direction field plot, we can see the horizontal tangents as: