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I encountered this problem in last week's GRE subject test, and I could not solve it. So I venture to post it in here. The question is as follows:


asking me $$\displaystyle \lim_{t\rightarrow \infty} \frac{y}{t}$$

I am not sure if my memory is totally correct. If any others take the same exam and found I remembered the problem wrong, please point out in comments.

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up vote 2 down vote accepted

To solve the ODE, you use the method of integrating factors. The proof is straight forward and can be found here. You write the left hand side of what SL2 rearranged as the derivative of a product $$ y' - \dfrac{1}{t}y = te^{-t} $$

So to do this, compute (using the notation in the wiki article) $P(x) = e^{\int\frac{-1}{t}dt} = \dfrac{1}{t}$. Now what we can say is that

$$\dfrac{y'}{t} - \dfrac{y}{t^2} = e^{-t} $$

By construction however, we now can write the left hand side as the derivative of the product of two functions:

$$\left(\dfrac{y}{t}\right)' = e^{-t}$$

We can readily integrate both sides now, and we see

$$\dfrac{y}{t} = -e^{-t} + C, C\in\mathbb{R}$$

so that

$$y = Ct-te^{-t}$$

Compute $\displaystyle\lim_{t\to\infty} \dfrac{y}{t} = C-e^{-t} = \boxed{C}$

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This is a first order linear ODE: $$y'-\frac1t y=te^{-t}$$

The solution to this has a nice closed form (see for example, this Wiki article), so we can actually compute $y$, and then take the desired limit.

In particular, using the method on the liked page, we can compute $y=Ct-te^{-t}$, so $\lim_{t\to\infty}\frac yt=\lim_{t\to\infty} C-e^{-t}=C$, so the limit will depend on the value of $C$ you take in the solution. If the problem were in fact an IVP, then we could determine $C$, and we would get an actual number for the solution.

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Thanks you a lot for this. I learnt this technique five years ago while I was in high school, and now it is completely forgotten. It is nice to see this again. – Kerry Oct 18 '11 at 2:44

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