= Solution
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 $y\ne0$. Then $z'=1-z$, so $z=1+Ke^{-t}$. Therefore \b[all real solutions] are
$$
\boxed{y(t)\equiv0\quad\text{or}\quad y(t)=\frac{s}{\sqrt{1+Ke^{-t}}},\qquad s\in\{-1,1\},}
$$
on intervals where $1+Ke^{-t}>0$. This family includes $y\equiv\pm1$ when $K=0$. For a prescribed finite nonzero $y(0)=y_0$, $s=\operatorname{sgn}y_0$ and $K=y_0^{-2}-1>-1$, so the solution exists for all future time. The <cubic logistic differential equation> has
$$
\boxed{\lim_{t\to\infty}y(t)=\begin{cases}1&y_0>0,\\0&y_0=0,\\-1&y_0<0.\end{cases}}
$$
These 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 \b[$y(0)=1$]; \b[$y(0)=-1$] also works. The central <equilibrium point> is unstable and the two outer <equilibrium points> attract solutions from their respective half-lines.
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