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I just want to make sure this is right because I'm doing the homework online and I'm on my last attempt and I'm pretty sure I got the other two right yet the computer program said no.

First at I have to find the time when the population becomes extinct when $p(0) = 710$. My answer was $2\ln\left(\frac{760}{50}\right)$ and its in months.

Then I have to find the initial population if they become extinct after $1$ year. Now does that mean I use $t = 12$ since the first answer is in months?

EDIT: When I edited it for the first time I deleted, unintentionally, some information of the title that I've already corrected, sorry.

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What is your solution to the ODE? – user7530 Jan 27 '13 at 22:49
Something isn't quite right. The solution to the ode you've given grows exponentially from that initial condition, so the population never becomes extinct. – icurays1 Jan 27 '13 at 22:51
p=760-50e^(t/2) – Gamecocks99 Jan 27 '13 at 22:51
I am very puzzled. If $\frac{dp}{dt}=5p-380$ and we start at $p=710$, then $p$ is increasing. Your solution above is not a solution, plug in and you will see it doesn't work. Perhaps your post does not correctly reflect the question you were given. – André Nicolas Jan 27 '13 at 22:51
I just checked the edit history. The first version was $(0.5)p -380$. It got changed while being LaTeX'ed. – André Nicolas Jan 27 '13 at 22:57

We have the DEQ:

$$\tag 1 \frac{dp}{dt} = 0.5p - 380$$

Solving $(1)$, yields (where c is an unknown constant):

$$\tag 2 \large p(t) = 760 + c e^{\frac{t}{2}}$$

From the initial condition, $p(0) = 710$, we solve for the unknown constant $c$, yielding:

$$p(0) = 760 + c e^{0} = 710 \rightarrow c = -50$$

Substituting that back into $(2)$ yields:

$$\tag 2 \large p(t) = 760 -50 e^{\frac{t}{2}}$$

Now, we want to know at what time, $t$, when $p(t) = 0$.

We get $\large t = 2~ \text{log} (\frac{76}{5}) = 5.44259$ months.

This checks out with your answer!


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Great work, and enlarged text for emphasis! – amWhy May 5 '13 at 2:06
Regards. ${}{}{}{}{}{}$ – Babak S. May 5 '13 at 8:11

The DE: $$\frac{dp}{dt} = \frac{1}{2}p - 380$$ $$\int \frac{dp}{\frac{1}{2}p - 380} = \int dt$$ $$2\int \frac{dp}{p - 760} = \int dt$$ $$2\ln |p-760| = t + C$$ $$\ln|p-760| = \frac{t}{2} + C$$ $$p-760 = Ce^{\frac{t}{2}}$$ Solving for $C$: $$C = 710-760 = -50$$ Solution to IVP: $$p = -50e^{\frac{t}{2}}+760$$

Ok, I've got the same thing you do here... Solving for $t$ when $p = 0$: $$0 = -50e^{\frac{t}{2}}+760$$ $$0 = -50e^{\frac{t}{2}}+760$$ $$e^{\frac{t}{2}} = \frac{76}{5}$$ $$t = 2\ln\frac{76}{5}\approx5.44$$

That's odd... I'm getting your same answer. This probably means its a problem with the entry into the online homework system.

Do you enter in the numeric result, or the symbolic result? If the symbolic result, go back to the "help" page and make sure you're entering in the natural log correctly. If the numeric result, enter in more sig figs. Check to make sure all the times are in months (for this part of the problem, at least).

I suggest emailing/calling/talking to your professor, and letting them know the problem you're encountering.

As to the second part of the question, yes, keep your units consistent across the problem.

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