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Logistic equation problem

  1. May 20, 2005 #1
    I just want to check my answer for this question:

    The population P(t) of a city suburb is governed by the initial value problem

    P' = P((10^-2) - (10^-6)P) P(0) = 2000

    Time measured in months.
    a) What is the limiting value of the population at t approaches infinity?
    b) How long before the population is one-half of this limiting value?

    a) I used the formula for the general solution
    birth-rate/death-rate = 10^-2/10^-6 = 10^4

    But I'm not sure if that is what I was supposed to do...

    b) Solving the DE
    P' = P((10^-2) - (10^-6)P)

    [tex]\int \frac{dP}{P(10^{-2}-10^{-6})} = \int dt[/tex]

    Using partial fractions, the left can be seperated

    [tex]\int \frac{10^2}{P} + \frac{10^{-4}}{10^{-2}-10^{-6}P} = \int dt[/tex]

    [tex]10^2 * ln(P-ln(10^{-2}-10^{-6}P) = t + c[/tex]

    Using the initial value P(0) = 2000, C = 250,000

    Half of the limiting value = (10^4)/2 = 5000

    Substituting into the equation, t = 139 months (3sf)
  2. jcsd
  3. May 20, 2005 #2


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    Check your work. Looks like you have lost some terms. You are missing a P in the equation before partial fractions, but your partial fraction expansion looks OK. Your parentheses need some work in

    [tex]10^2 * ln(P-ln(10^{-2}-10^{-6}P) = t + c[/tex]

    Consequently your C does not look right.

    Part a) looks OK. P = [tex]10^4 [/tex] is where the derivative goes to zero, so that is the limiting value of P.
  4. May 20, 2005 #3


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    So, what up Shan? What about when you correct your formula as Dan said, how about solving it "explicitly" for P(t). You can do that? Don't want to? I don't know. However if you did, it would clearly show what the limiting value is as t goes to infinity. Why not try and report it here. :smile:

    Edit: Plot too but suppose I shouldn't push it.
    Last edited: May 20, 2005
  5. May 20, 2005 #4
    lol sorry, sorry, it was a typo. It's supposed to be

    [tex]\int \frac{dP}{P(10^{-2}-10^{-6}P)} = \int dt[/tex]

    Another typo... V_V

    [tex]10^2 * (lnP-ln(10^{-2}-10^{-6}P)) = t + c[/tex]

    so when I rearrange...

    [tex]lnP-ln(10^{-2}-10^{-6}P)) = \frac{t + c}{10^2}[/tex]

    [tex]ln\frac{P}{10^{-2}-10^{-6}P} = \frac{t + c}{10^2}[/tex]

    [tex]\frac{P}{10^{-2}-10^{-6}P} = e^{\frac{t + c}{10^2}} = Ce^{\frac{t}{100}}[/tex]

    One thing I'm not sure about, can I solve for c using the second equation or the third one? Or does it not matter? Because I used the third equation so that I end up with

    [tex]\frac{2000}{10^{-2}-10^{-6}*2000} = Ce^0[/tex]


    And c ends up being 250,000

    lol, it's the second case. When I rearrange the equation up the top (presuming that it's right), I got

    [tex]P = Ce^{\frac{t}{100}} * (10^{-2}-10^{-6}P)[/tex]

    And I don't know how to get rid of the P on the right... (yes, my algebra is bad lol)
  6. May 20, 2005 #5


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    Dude, you gotta' get those parenthesis right:

    [tex]ln(P)-ln(10^{-2}-10^{-6}P) = \frac{t + c}{10^2}[/tex]

    We're good here. Actually leaving as:

    [tex]10^2\{ln(P)-ln(10^{-2}-10^{-6}P)\} = t + c[/tex]

    makes it easy to calcualate c but leave it as 'c' until you need it.

    Also, those quantities for logarithms, aren't they just:


    So we have:


    So you take exponentials of both sides and get:


    This is where you're having problems right, isolating P.

    Suppose I have:


    Well, multiplying by 1-x gives me:


    soooooo . . . ax=k-kx
    You got it right?
  7. May 22, 2005 #6
    Sorry for the late reply. So are you saying that

    P = (10^4 * e^(1/100 (t+c))) / (10^6 + e^(1/100 (t+c)))
  8. May 22, 2005 #7


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    That's it. However, if you would divide the numerator and denominator by the numerator, then the limit as t goes to infinity is clear.
  9. May 22, 2005 #8
    Um, is it 0? After I simplify, I get

    P = 1 / (10^6 + e^(1/100 (t+c))) * (10^4 * e^(1/100 (t+c)))

    Or did I mess up somewhere?
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