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The equilibrium points are $y=0,1,-1$. To obtain the other solutions of this separable differential equation without losing their signs, set $z=y^{-2}$ on an interval where $y\ne 0$. Then $z^{\prime }=1-z$, so $z=1+K{e}^{-t}$. Therefore all real solutions areon intervals where $1+K{e}^{-t}>0$. This family includes $y\equiv \pm 1$ when $K=0$. For a prescribed finite nonzero $y(0)=y_0$, $s=\mathrm{sgn}{y}_0$ and $K=y_0^{-2}-1>-1$, so the solution exists for all future time. The cubic logistic differential equation hasThese are the only possible limits. The sign cannot change by uniqueness, and the denominator tends to one. A nonzero initial condition producing a constant solution is $y(0)=1$; $y(0)=-1$ also works. The central equilibrium point is unstable and the two outer equilibrium points attract solutions from their respective half-lines.