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anonymous

  • 5 years ago

differential equation, i dont get book's answer. question: dP(t)/dt = kP(t) ( 1 - P(t)/M) , where P is population at time t, M is carrying capacity.books solution: P(t) = MCe^(kt)/ ( M + Ce^kt). ok here is what i got. first 1 - P/M = (M- P ) / M, so dP/dt = kP( M-P)/M , which is seperable autonomous. we have integral M / [ P (M-P)] dP = integral k dt. using partial fractions i get integral 1 / P + 1 / ( M-P) = kt + C ln P - ln M-P = kt + C ln | P / (M-P) | = kt + C P /( M-P) = e^(kt + c) P = (M-P) Ce^kt , where C = e^c P = M Ce^kt -P*Ce^kt P + P Ce^kt = M Ce^kt P ( 1 + Ce^kt) = MCe^kt P

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  1. anonymous
    • 5 years ago
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    this problem is posted here http://www.sosmath.com/diffeq/first/application/population/population.html

  2. anonymous
    • 5 years ago
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    so i got P = MCe^kt / ( 1 + Ce^kt) , but the website has P = MCe^(kt)/ ( M + Ce^kt) theres a glaring discrepancy, i am missing an M

  3. dumbcow
    • 5 years ago
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    ok ill try to work it out and see if i get anything different

  4. anonymous
    • 5 years ago
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    the only difference in the website is that they kept 1 - P/M in the partial fractions

  5. anonymous
    • 5 years ago
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    whatcha got?

  6. anonymous
    • 5 years ago
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    the suspense is killing me

  7. dumbcow
    • 5 years ago
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    ok yeah im getting the same result you are. i tried both factoring out the 1/m and leaving it in when integrating fractions

  8. anonymous
    • 5 years ago
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    this is so freaking frustrating!!!

  9. anonymous
    • 5 years ago
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    last step P = MCe^(kt)/ ( 1 + Ce^kt), website has an extra M in there http://www.sosmath.com/diffeq/first/application/population/population.html both ways seem sound . im stumped

  10. anonymous
    • 5 years ago
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    Unless I'm missing something, surely it doesn't matter which you got? The constant comes out different anyway, so it (probably) leads to the same final results.

  11. anonymous
    • 5 years ago
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    it should matter

  12. dumbcow
    • 5 years ago
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    ahh that could be it, by us splitting up the fraction we may have changed the constant

  13. anonymous
    • 5 years ago
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    we get two different answers

  14. anonymous
    • 5 years ago
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    After you have solved for the constant? (You may well do, I haven't tried after that)

  15. anonymous
    • 5 years ago
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    oh i didnt try that

  16. anonymous
    • 5 years ago
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    you have a point though, suppose i multiplied top and bottom by M, the C could absorb one of the M's

  17. anonymous
    • 5 years ago
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    for my solution

  18. anonymous
    • 5 years ago
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    no you cant do that, then the C's would be different

  19. anonymous
    • 5 years ago
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    there is something seriously wrong here and i cant find it

  20. anonymous
    • 5 years ago
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    OK, I have carried on from your answer (got a different constant than in the website's solution), and it leads to the same final p(t) - unless I made a mistake. Try it and see what happens. In general, something can integrate to two different things (with a constant in both). The fact they both are general solutions / differentiate to the same thing does not imply they are the same. e.g. \[\int 2 \sec^2x\ \tan x = \tan^2x + c = \sec^2x + k \] but \[\tan^2x\ \not= \sec^2x \]

  21. anonymous
    • 5 years ago
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    yes but i did precise partial fraction integration, this isnt like a trig antiderivative which is differing by a constant

  22. anonymous
    • 5 years ago
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    what did you get ?

  23. anonymous
    • 5 years ago
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    can i see what you got

  24. anonymous
    • 5 years ago
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    thanks for your help btw

  25. anonymous
    • 5 years ago
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    I carried on from your value: P = MCe^(kt)/(1+Ce^(kt)). When t = 0, solve for C in terms of P_O and get C = P_0/(M-P_0) Put this back in and it results in the same value for P(t) in the final answer.

  26. anonymous
    • 5 years ago
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    \[P = \frac{MCe^{kt}}{1-Ce^{kt}} \frac{}{} \implies C = \frac{P_0}{M-P_0} \] My LaTeX is rendering horribly today, but maybe that shows up OK for you?

  27. anonymous
    • 5 years ago
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    Sorry, first is 1+ Ce ... in the fraction :/

  28. anonymous
    • 5 years ago
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    so ok

  29. anonymous
    • 5 years ago
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    i see the latex

  30. anonymous
    • 5 years ago
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    here is an idea, but it seems silly, define a new constant

  31. anonymous
    • 5 years ago
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    multiply top and bottom by M, then define a new constant C* = M*C

  32. anonymous
    • 5 years ago
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    P = M* M Ce^kt / [M( 1 + Ce^kt)] = M (MC)e^kt / [ M + MC e^kt ] , let C* = MC, so you have P = MC* e^kt / (M + C*e^kt)

  33. anonymous
    • 5 years ago
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    oh , thats bad symbols, one sec

  34. anonymous
    • 5 years ago
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    P = M M Ce^kt / [M( 1 + Ce^kt)] = M (MC)e^kt / [ M + MC e^kt ] , let C` = MC, so you have P = MC` e^kt / (M + C`e^kt)

  35. anonymous
    • 5 years ago
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    Yes, I don't see why you can't do that (but it seems unnecessary).

  36. anonymous
    • 5 years ago
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    If you just solve for C as it is, you get the value they did divided by M, so all you're doing is getting the same constant times M. Like I said, unnecessary, but if it helps you think about why they're the same then fair enough).

  37. anonymous
    • 5 years ago
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    well it doesnt explain why i get the same answer using partial fractions

  38. anonymous
    • 5 years ago
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    also you should have a plus there, not a minus

  39. anonymous
    • 5 years ago
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    you wrote P = M Ce^kt / ( 1 - Ce^kt), in latex

  40. anonymous
    • 5 years ago
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    Oh, I corrected it the post below

  41. anonymous
    • 5 years ago
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    ok thanks :) so the final answer comes out the same ,

  42. anonymous
    • 5 years ago
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    Indeed :D

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