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jocelynevsq

  • 3 years ago

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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  1. zepdrix
    • 3 years ago
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    What part are you stuck on Ms Jocelyn? :) Were you able to find P(t)?

  2. jocelynevsq
    • 3 years ago
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    i don't know how to start it

  3. zepdrix
    • 3 years ago
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    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).

  4. zepdrix
    • 3 years ago
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    \[\large \frac{dP}{dt}=P\left(2-\frac{P}{5000}\right)\]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:

  5. zepdrix
    • 3 years ago
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    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! :)

  6. zepdrix
    • 3 years ago
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    We'll have to apply `partial fraction decomposition` to the fraction so we can integrate it.

  7. zepdrix
    • 3 years ago
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    \[\large \frac{5000}{P(10000-P)} \quad = \quad \frac{A}{P}+\frac{B}{10000-P}\]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.\[\large \frac{1}{2}\int\limits \frac{1}{P}+\frac{1}{10000-P}dP=dt\]

  8. zepdrix
    • 3 years ago
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    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,\[\large \ln\left(\frac{P}{10000-P}\right)=2t+c\]

  9. zepdrix
    • 3 years ago
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    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".\[\large \frac{P}{10000-P}=e^{2t+c}\]

  10. zepdrix
    • 3 years ago
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    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,\[\large \frac{P}{10000-P}=Ce^{2t}\]

  11. zepdrix
    • 3 years ago
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    Following any of this? :) Or this is way too confusing?

  12. zepdrix
    • 3 years ago
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    It's a ton of steps to get through this one... very easy to get lost :(

  13. jocelynevsq
    • 3 years ago
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    i get it..it's really helpful!

  14. zepdrix
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    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, \[\large P=\frac{10000Ce^{2t}}{1+Ce^{2t}}\]

  15. zepdrix
    • 3 years ago
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    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

  16. zepdrix
    • 3 years ago
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    They told us also that the `Initial population` is 3000. We can write it like this,\[P(0)=3000 \qquad \rightarrow \qquad \large 3000=\frac{10000Ce^{0}}{1+Ce^{0}}\]

  17. zepdrix
    • 3 years ago
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    And ummm hopefully that will get us the C value...

  18. zepdrix
    • 3 years ago
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    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:

  19. jocelynevsq
    • 3 years ago
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    i got 3/7 for C also

  20. zepdrix
    • 3 years ago
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    So if we take the limit, \[\huge \lim_{t \rightarrow \infty} \frac{10000\cdot\frac{3}{7}e^{2t}}{1+\frac{3}{7}e^{2t}}\]

  21. zepdrix
    • 3 years ago
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    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.

  22. zepdrix
    • 3 years ago
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    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

  23. jocelynevsq
    • 3 years ago
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    10,000 is the correct answer :D thank you so much this was very helpful!

  24. zepdrix
    • 3 years ago
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    err maybe the question was just looking for a number, so you don't have to say it all fancy like that :) lol

  25. zepdrix
    • 3 years ago
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    It was correct? yay team \c:/

  26. jocelynevsq
    • 3 years ago
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    yay for you actually lol xD

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