So the equilibrium solutions are -1 and 1, but what are the non-equilibrium solutions?
I started with:
$\displaystyle dy/(1-y^2) = dt$
$\displaystyle -ln(1-y^2) = t + c$
Am I going in the right direction?
Unfortunately not. Remember, the integral of $\displaystyle dU/U$ is the natural log of the absolute value of U, but you don't have the differential of your denominator, $\displaystyle -(y^2-1)$ in the numerator. You'll have to use partial fraction decomposition.
A simple preliminary analysis of the DE...
$\displaystyle y^{'} = 1 - y^{2}$ (1)
... permits You to conclude that is...
a) $\displaystyle y^{'} >0 $ for $\displaystyle -1< y < 1$
b) $\displaystyle y^{'} = 0$ for $\displaystyle y=1$ or $\displaystyle y=-1$
c) $\displaystyle y^{'} < 0 $ for $\displaystyle y=1$ or $\displaystyle y=-1$
... so that is...
a) $\displaystyle \lim_{t \rightarrow \infty} y(t) = 1$ for $\displaystyle y(0) >-1$
b) $\displaystyle \lim_{t \rightarrow \infty} y(t) = - 1$ for $\displaystyle y(0) = -1$
c) $\displaystyle \lim_{t \rightarrow \infty} y(t) = -\infty$ for $\displaystyle y(0) <-1$
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$