GRE math question

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

Answer 1

$y(0) = 0\\\ y'(0) = - e^{-1}$

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.

Answer 2

$$y'=(y^2-1)e^{2012 y-1}\implies e^{2012 y-1}=\frac{y'}{y^2-1}.$$ 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, $-1<y(t)<1$.

Answer 3

Note that $y(0)=0$ and $y'(0) = -e^{-1} < 0$. Thus, $y$ is negative in a neighborhood around $0$. This means that we can never have $y>0$, since in order for that to happen we need $y$ to be increasing somewhere, and in particular increasing somewhere where $-1<y<0$, which is impossible since $|y|<1\implies y'<0$. This eliminates options (a) and (b).

This leaves us with (c), (d), or (e). Now suppose the statement in (d) were true, i.e. $-1<y(t)<1$ for all $t$. Then $y'(t)<0$ for all $t$, i.e. $y$ is decreasing, and since $y$ is bounded from below, this means that $y$ decreases to some limit as $t\rightarrow\infty$. Furthermore, for $y$ to decrease to some finite limit, we need $\lim\limits_{t\rightarrow\infty}{y'(t)} = 0$, and if $\lim\limits_{t\rightarrow\infty}{y(t)} = l$, then $\lim\limits_{t\rightarrow\infty}{y'(t)} = (l^2-1)e^{2012l-1}$. This would imply $l = \pm 1$, and clearly $l = 1$ is impossible. So if the statement in (d) were true, the statement in (c) would also be true, and the answer would be (e). If the statement in (d) were false, then the statement in (e) would also be false, and so the answer would be (c). So we've narrowed it down to two choices, depending on whether the statement in (d) is true or not. If you don't have much time, taking a 50-50 guess wouldn't be a bad idea.

Now, if you do have time, you can play around with it to get $$ \frac{y'}{y+1} = (y-1)e^{2012y-1}\implies \ln{|y(t)+1|} = \int\limits_{0}^{t}{(y(s)-1)(e^{2012y(s)-1})\text{ d}s}. $$ Initially, we have $-1<y<1$. Now, suppose at some point we no longer have $-1<y<1$, and let $t_0 = \inf\{t>0: y(t)\not\in (-1,1)\}$. Since $y$ is continuous, we must have $y(t_0) = \pm 1$ (i.e. at the boundary of the set $(-1,1)$), and since $y(t_0) = 1$ is impossible, we have $y(t_0) = -1$. Furthermore, by definition we have $y(t)\in (-1,1)$ for $t<t_0$. As $t\rightarrow t_0^{-}$, we have $y(t)\rightarrow y(t_0) = -1$, and hence $\ln{|y(t)+1|}\rightarrow -\infty$. However, since $-1<y(s)<0$ for $s<t_0$, the integrand is uniformly bounded for $t<t_0$, and so the integral cannot diverge to $-\infty$. More specifically, we have $$\left|(y(s)-1)e^{2012y(s)-1}\right| = (1-y(s))e^{2012y(s)-1}\le 2e^{-1}$$ for $s<t_0$, and hence \begin{align}|\ln{|y(t)+1|}| = \left|\int\limits_{0}^{t}{(y(s)-1)(e^{2012y(s)-1})\text{ d}s}\right| &\le \int\limits_{0}^{t}{\left|(y(s)-1)(e^{2012y(s)-1})\right|\text{ d}s}\\ &\le \int\limits_{0}^{t}{2e^{-1}\text{ d}s} = 2e^{-1}t \end{align} and so $\lim\limits_{t\rightarrow t_0^{-}}{|\ln{|y(t)+1|}|}\le\lim\limits_{t\rightarrow t_0^{-}}{2e^{-1}t} = 2e^{-1}t_0<\infty$, contradicting the statement that $\ln{|y(t)+1|}\rightarrow -\infty$ as $t\rightarrow t_0^{-}$.

Thus, we can never have $y(t)\not\in(-1,1)$, i.e. $-1<y(t)<1$ for all $t$. The statement in (d) is thus true, so the correct option is $\boxed{\text{(e)}}$.

Answer 4

It's given that when $t=0$ then $y=0$, so $y' = (0^2-1)e^\text{something}$, and that is negative. Thus $y$ is decreasing as $t$ increases. And as long as $y$ remains $>-1$, you have $y^2-1$ negative and the exponential function is positive, so $y'$ is negative and so $y$ is decreasing. If $y$ should reach $-1$, then you'd have $y^2-1 = (-1)^2 - 1=0$, so $y'$ would be $0$ and $y$ would not be changing at all. The value of $y$ cannot go from $-1$ to anything less than $-1$, because when $y<-1$ then $y^2-1$ is positive and so $y'$ is positive and $y$ is increasing toward $-1$.

Thus $y\to -1$ as $t\to\infty$.