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The population P(t) of a species satisfies the logistic differential equation dP/dt= P(2-(P/5000)) where the initial population 3,000 and t is the time in years. What is the lim t>infinity P(t)?

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What part are you stuck on Ms Jocelyn? :) Were you able to find P(t)?
i don't know how to start it
There is some function P(t) which satisfies the given differential equation. So we'll start with our differential equation, move some stuff around, and integrate to solve for P(t).

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Other answers:

\[\large \frac{dP}{dt}=P\left(2-\frac{P}{5000}\right)\]This is a separable differential equation, so we'll move all of the P's to one side, and move the dt to the other. Hmm it's going to take some work though D:
We'll start by getting a common denominator in the brackets, giving us this, \[\large \frac{dP}{dt}=P\left(\frac{10000-P}{5000}\right)\]Let's go ahead and move that extra P on top as well.\[\large \frac{dP}{dt}=\frac{P(10000-P)}{5000}\]We'll multiply both sides by the `reciprocal` oh the giant term on the right.\[\large \frac{5000}{P(10000-P)}\frac{dP}{dt}=1\]Then we'll "multiply" the dt to the other side,\[\large \frac{5000}{P(10000-P)}dP=dt\]We've successfully separated the P's and t's! :)
We'll have to apply `partial fraction decomposition` to the fraction so we can integrate it.
\[\large \frac{5000}{P(10000-P)} \quad = \quad \frac{A}{P}+\frac{B}{10000-P}\]Multiplying through by the denominator gives us,\[\large 5000=A(10000-P)+BP\] Hopefully I did my math correctly, I came up with \(A=1/2 \quad B=1/2\). So we can write the left side like this,\[\large \frac{\frac{1}{2}}{P}+\frac{\frac{1}{2}}{10000-P}dP=dt\]We'll factor out the 1/2, and take the integral.\[\large \frac{1}{2}\int\limits \frac{1}{P}+\frac{1}{10000-P}dP=dt\]
Integrating will give you a couple of natural logs,\[\large \frac{1}{2}\left[\ln P-\ln\left(10000-P\right)\right]=t+c\]Multiplying both sides by 2 gives us,\[\large \left[\ln P-\ln\left(10000-P\right)\right]=2t+c\](The 2 was distributed but the other one got absorbed into the c). We'll combine our logs using a rule of logs,\[\large \ln\left(\frac{P}{10000-P}\right)=2t+c\]
We'll exponentiate both sides, rewrite them with a base of e,\[\huge e^{\ln\left(\frac{P}{10000-P}\right)}=e^{2t+c}\] On the left, the exponential e, and the natural log are inverse operations of one another, so they're "cancel out".\[\large \frac{P}{10000-P}=e^{2t+c}\]
Recalling a rule of exponents allows us to write the right side like this,\[\large \frac{P}{10000-P}=e^{2t}\cdot e^c\]e^c is just another constant, let's call it such,\[\large \frac{P}{10000-P}=Ce^{2t}\]
Following any of this? :) Or this is way too confusing?
It's a ton of steps to get through this one... very easy to get lost :(
i get's really helpful!
We'll multiply both sides by the denominator on the left.\[\large P=(10000-P)Ce^{2t}\]Multiply out the right side,\[\large P=10000Ce^{2t}-PCe^{2t}\]Move the clunky P term to the left side by adding it,\[\large P+PCe^{2t}=10000Ce^{2t}\]Factor out a P on the left,\[\large P\left(1+Ce^{2t}\right)=10000Ce^{2t}\]Then do some division to solve for P, \[\large P=\frac{10000Ce^{2t}}{1+Ce^{2t}}\]
I checked my work a few minutes ago on Wolfram, and they came up with something `very slightly` different. They don't have the constant C attached to the 10,000 on the top. So I hope I didn't make a silly mistake somewhere in there.. hmm
They told us also that the `Initial population` is 3000. We can write it like this,\[P(0)=3000 \qquad \rightarrow \qquad \large 3000=\frac{10000Ce^{0}}{1+Ce^{0}}\]
And ummm hopefully that will get us the C value...
I came up with C=3/7. I dunno this problem is sooooo long, I have a feeling I may have made an error somewhere along the way lol. I hope that at least gives you an idea of how to progress through this of problem D:
i got 3/7 for C also
So if we take the limit, \[\huge \lim_{t \rightarrow \infty} \frac{10000\cdot\frac{3}{7}e^{2t}}{1+\frac{3}{7}e^{2t}}\]
Ok so the exponentials are increasing at the same rate. The +1 on the bottom will become insignificant as t->infty. So this is approaching a value of \(1\) times the coefficients on the exponentials.\[\large \frac{10000\cdot\frac{3}{7}}{\frac{3}{7}}\] There are some little algebra tricks you can do to convince yourself of this, but I didn't feel like being that thorough XD heh.
As times progresses on and on and on, the population of the given species will approach but not exceed 10,000. I would word it something like that. And again, I would strongly suggest you go through this a bit more and make sure I didn't make any mistakes :D
10,000 is the correct answer :D thank you so much this was very helpful!
err maybe the question was just looking for a number, so you don't have to say it all fancy like that :) lol
It was correct? yay team \c:/
yay for you actually lol xD

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