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Differential equations - logistic law problem

  1. Feb 7, 2007 #1
    1. A population grows accordin to the logistic law, with a limiting population of 5 x 10^8 individuals. When the population is low, it doubles every 40 minutes. What will the population be after 2 hours if initially it is (a) 10^8, and (b) 10^9?

    P=P0e^a(t-t0)
    N=a/b
    p(t) = ap0/[bp0 + (a-bp0)e^-a(t-t0)]


    I have actually solved the problem, but I want to confirm that I am doing it right so here is my solution, i only showed (a) since a and b are the same.

    p=p0e^a(t-t0) (i used this since it gives you info about a low population)
    2=1e^a(40-0)
    2=e^40a
    40a = ln 2
    a = 1/40 ln 2

    then i used N = a/b since N is given to be 5 x 10^8
    5 x 10^8 = a/b
    5 x 10^8 = (1/40 ln 2) / b
    b = (1/40 ln 2)/(5 x 10^8)

    then i used the logistic model
    p(t) = ap0/[bp0 + (a-bp0)e^-a(t-t0)]
    p(120)= (1/40 ln 2)(10^8)/[((1/40 ln 2)/(5x10^8)) (10^8) + (1/40 ln 2 - ((1/40 ln 2)/(5x10^8)) (10^8) )e^(-1/40 ln 2)(120-0)] (sorry i know this looks confusing bare with me)
    P(120) = (10^8/40 ln 2) / [1/8 ln 2 + (1/40 ln 2- (1/8 ln 2))e^-3 ln 2)]
    P(120) = (10^8/40 ln 2) / [1/8 ln 2 + (1/40 ln 2 - (1/8 ln 2))(1/8)] (e^-3 ln 2 = 1/8, is this correct??)
    P(120) = (10^8/40 ln 2) / [1/8 ln 2 + 1/320 ln 2 - 1/64 ln 2]
    then i got rid of all the ln 2's
    P(120) = (10^8/40) / [1/8 + 1/320 - 1/64]
    P(120) = (10^8/40) / 9/80
    P(120) = (10^8/40)(80/9)
    P(120) = (2 x 10^8)/9 = 22,222,222


    i really hope this is right.
    and i did the same thing for (b) but changed 10^8 to 10^9 and got 22,792,022
     
  2. jcsd
  3. Feb 8, 2007 #2
    anyone???
    i am not sure if this is right at all, any feedback would be great
     
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