# 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

• Have you narrowed down the possible answers yet? Where are you stuck? Sep 7, 2016 at 21:13
• That's the thing, I don't really know how to approach this problem. I thought of trying to solve the differential equation, but didn't get very far. Sep 7, 2016 at 21:15
• The trick to multiple choice math problems is that there is usually and obvious and slow, and a subtle but fast method to each problem. Every time you take the slow approach there are two problems you don't get to before you are out of time. Sep 7, 2016 at 21:19

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 $\boxed{\text{(e)}}$.

• $y'<0$ in some neighborhood $(-a,a) \subset (-1,1)$ of $0$, so $y(t) > y(0) = 0$ for all $t\in (-a,0)$. So how can we never have $y>0$? Sep 7, 2016 at 23:00
• @BanachManifold presumably we are only concerned with $t>0$, so by "neighborhood" I meant "right neighborhood". However, the question didn't specify, and the answer doesn't change if we allow $t<0$ as well, so I might edit my answer to account for that. Sep 7, 2016 at 23:06
• @BanachManifold specifically, the first three options deal with the behavior as $t\rightarrow+\infty$, so only the behavior for $t>0$ matters. Only for option (d) does the issue of $t<0$ matter. Sep 7, 2016 at 23:10
• sorry, I completely forgot about the context of the question. Sep 7, 2016 at 23:16

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

• @DougM: It doesn't ! Why do you think it does ?
– Surb
Sep 7, 2016 at 21:23
• For all $t, -1<y(t)\le 0$ which is more specific than the answer provided. Not that the the answer provided isn't true. Sep 7, 2016 at 21:31
• @DougM: If $-1<y(t)\leq 0$, then automatically $-1<y(t)<1$. I don't understand your intervention... what's the problem ?
– Surb
Sep 7, 2016 at 21:32

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

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.

• "Then it gets stuck." That's great for intuition, but not exactly flawless. For example, the function defined by $f(x) = (x-1)^2$ for $x\ge 1$ and $f(x) = 0$ for $x<1$ satisfies $f' = 2\sqrt{f}$, but for $x<1$ the function is stuck at $0$, while at $x=1$ it magically gets "unstuck". Sep 7, 2016 at 21:50
• It is a multiple choice test. Intuition beats rigor. Sep 7, 2016 at 21:53
• Agreed. Let's hope the equation $f' = 2\sqrt{f}$ never makes it on the test then. :) Sep 7, 2016 at 21:54

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 $t\to\infty$.

• Sorry but I don't understand your proof of $y\to -1$ when $t\to \infty$. Of course $y'<0$, and thus $y$ is decreasing. The fact that $\lim_{t\to \infty }y'(t)=0$ and $y$ bounded just tell us that $\lim_{t\to \infty }y(t)$ exist, but not that it's $-1$, is it ?
– Surb
Sep 8, 2016 at 7:05
• @Surb : As long as $0>y>-1$, you have $y'<0$, but if $y<-1$ you have $y'>0$ so $y$ increases toward $-1$. Thus $y$ can never get less than $-1$. I don't know how you could have concluded that $y$ is bounded without going through something like that reasoning. $\qquad$ Sep 8, 2016 at 17:16