GRE math question I have the following problem in a GRE practice exam, I was wondering if someone could help me figure it out.
Suppose $y(t) = y$ solves $y' = (y^2-1)e^{2012y-1}$ with initial condition $y(0)=0$, then which of the following are true?
a) $\lim_{t \to \infty} y(t) = \infty$
b) $\lim_{t \to \infty} y(t) = 1$
c) $\lim_{t \to \infty} y(t) = -1$
d) $-1 < y(t) < 1$ for all $t$
e) Both C and D

$y(0) = 0\\\
y'(0) = - e^{-1}$
As long as $y^2(t) < 1, y'(t) < 0$
As $y'(t)$ approaches $-1, y'(t)$ approaches $0.$
Then it gets stuck.  Once $y'(t) = 0,$ then any increase in $t$ is not causing any changes to $y.$
As $t\to \infty, y(t) \to -1$ and for all $t, -1<y\le 0$
While d) is a more general statement than that, it is not wrong.
