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

$$y'=(y^2-1)e^{2012 y-1}\implies e^{2012 y-1}=\frac{y'}{y^2-1}.$$
Since the LHS is continuous, the RHS is also continuous. By the way, $y'(0)<0$, and thus, $y^2-1$ must be negatif (because the LHS is positif). Therefore, $-1<y(t)<1$.
