Find the equilibrium solutions of the following differential equation:
$$\dfrac{dy}{dt} = \dfrac{(t^2 - 1)(y^2 - 2)}{(y^2 -4)}$$
I'm not sure how to go about doing this since t appears explicitly on the right hand side. Would $y = \sqrt{2}$ or $-\sqrt{2}$ be solutions?
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$\begingroup$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:
I think your equilibrium solutions are $y=\pm \sqrt2$ if by equilibrium it is meant $\frac{dy}{dt}=0$, i.e. steady-state.
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