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Population Modeling using DE's

  1. Oct 28, 2006 #1
    Hi,

    I was wondering if anyone out here on a Friday night could help me understand population modeling. Here is what I have as a problem (this is pretty simple because my goal here is to understand the thinking behind the madness :rolleyes: )

    -
    The population of a certain community is known to increase at a rate proportional to the number of people present at any time. If the population has doubled in 5 years, how long will it take to triple, to quadruple?
    -

    So I understand that the rate = dy/dt and proportional translates to "something" = "some constant" times "something", or (dy/dt)=Ky, or in this case, dP/dt = kP.

    Solving this DE I get the equation P(t)= P(initial)e^(kt). So all I am told is that the population doubles in 5 years. So what can I do? I can't assume an arbitrary number as the initial population can I? So I have one equation with three unknowns, P(initial), P(t), and k.

    Any insight would be great. Thanks!!
     
  2. jcsd
  3. Oct 28, 2006 #2
    you know that [tex] \frac{dP}{dt} = kP [/tex]. The solution is:

    [tex] P(t) = P_{0}e^{kt} [/tex]

    You also know that [tex] 2P_{0}= P_{0}e^{5k} [/tex]. What can you do from here? Solve for k:

    [tex] e^{5k} = 2 [/tex]

    [tex] k = \frac{\ln 2}{5} [/tex].



    Then solve the equations for t:

    [tex] 3P_{0} = P_{0}e^{(\frac{\ln 2}{5})t} [/tex]
    [tex] 4P_{0} = P_{0}e^{(\frac{\ln 2}{5})t} [/tex]
     
    Last edited: Oct 28, 2006
  4. Oct 28, 2006 #3
    Oh man... So here the [tex] P_{0} [/tex]'s will cancel and all I have to do is take the natural logs of both sides to get t? man... I think I was making it a lot harder than it really was. Thank you for taking the time to show me this.
     
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