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DIFFEQ - Modeling Problem

  1. Sep 6, 2005 #1
    Ok, I have this modeling problem that I cannot figure out how to setup.
    The problem is given:

    The population of mosquitoes in a certain area increases at a rate proportional to the current population, and in the absence of other factors, the population doubles each week. There are 200,000 mosquitoes in the area initially, and predators (birds, bats, so forth) eat 20,000 mosquitoes/day. Determine the population of mosquitoes in the area at any time.


    I don't understand what "absense of other factors" means. I'll put down what I have.

    Let P(t) = the population of mosquitoes.
    P(0)=200,000

    dP/dt = 2*P(t) mosquitoes/week - 20,000 mosquitoes/day

    (now simply convert week to days)
    dP/dt = 2*P(t) mosquitoes/(7*day) - 20,000 mosquitoes/day

    So yeah, that's the equation I have to model it. I have NO idea if that is right not, because like I said, I don't really understand what the question is asking.
    Now, when I solve this problem I get a different answer then what the book has.
    The book gives:
    P=201,977.31-1977.31*e^(ln(2))*t, 0<=t<=t_f ~=6.6745 weeks

    So, if anybody could point me in a direction of what I am doing wrong with the modeling... or maybe, just maybe my model is fine and I'm solving the DIFFEQ problem wrong...

    Anyways... thanks in advance, I appreciate any help.
     
  2. jcsd
  3. Sep 6, 2005 #2

    arildno

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    Why do you think the proportionality factor is 2??
    Let us consider the exponential growth scenario, that is, with no predators present.
    Let time be measured in weeks.
    We then have the diff.eq:
    [tex]\frac{dp}{dt}=kp[/tex] where we regard k as momentarily an unkown constant.
    This equation is easy to solve with an intial value [tex]p(0)=p_{0}\neq{0}[/tex]:
    [tex]p(t)=p_{0}e^{kt}[/tex]
    Now, we invoke the condition that after 1 week (t=1), the population is doubled ([tex]p(1)=2p_{0}[/tex]):
    [tex]2p_{0}=p_{0}e^{k}\to{k}=ln(2)[/tex]

    Thus, when we are to formulate the associated problem with predators, we get (with time still measured in weeks):
    [tex]\frac{dp}{dt}=ln(2)p-7*a, a=20.000[/tex]
     
  4. Sep 6, 2005 #3
    Ok, so the basic idea for problems like this is you hit condition A with some constant (say C from the integrating factor/solving process) and you hit the next condition with B (say with the constant K).


    Yeah, that doubling process makes more sense. I was really just tossing variables and stuff in there, hoping it would work like magic without understanding it. That makes a hell of a lot more sense with what you are saying, and I'll play around with it when I get home tonight. Thanks man, I appreciate the help.
     
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