Solution
= Solution
Separate variables and integrate on $|y|<1$:
$$
\int\frac{dy}{1-y^2}=x+C,\qquad
\operatorname{artanh}y=\frac12\log\frac{1+y}{1-y}=x+C.
$$
Hence the <general solution> in this region is
$$
\boxed{y(x)=\tanh(x+C),\qquad C\in\mathbb R.}
$$
Every such solution increases from values near $-1$ towards $1$, remains strictly between the two <equilibria>, and is defined for all real $x$.