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Math Help - Differential Equations

  1. #1
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    Differential Equations

    Hello!

    Q: A population of animals live on an isolated island such that at any time, the number of births per unit time is proportional to the population and the number of deaths per unit time is proportional to the square of the population. If the population at time t is x and the population is changing at the greatest rate when x = 1, show that

    \frac{dx}{dt} = kx (2 - x)

    where k is positive constant.

    (I have done the showing step, but I don't know how to solve the following parts.)

    i) Find the general solution of the differential equation in terms of k.
    ii) Show that there is a limit to the size of the population.


    Pleas help! Thank you!
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  2. #2
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    Quote Originally Posted by Tangera View Post
    Hello!

    Q: A population of animals live on an isolated island such that at any time, the number of births per unit time is proportional to the population and the number of deaths per unit time is proportional to the square of the population. If the population at time t is x and the population is changing at the greatest rate when x = 1, show that

    \frac{dx}{dt} = kx (2 - x)

    where k is positive constant.

    (I have done the showing step, but I don't know how to solve the following parts.)

    i) Find the general solution of the differential equation in terms of k.
    ii) Show that there is a limit to the size of the population.


    Pleas help! Thank you!
    http://www.math.neu.edu/~gilmore/U343su05files/logistic.pdf
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  3. #3
    MHF Contributor chisigma's Avatar
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    Unhappy

    In order to simplify the problem we write the equation as...

    \frac{dx}{dt}= - k\cdot x\cdot (x-2) , k>0 (1)

    Before the 'attack' some useful considerations...

    a) the (1) has two 'stationary solutions' : x=0 and x=2

    b) for  x>2 is x'(t)<0, for 0<x<2 is x'(t)>0. That means that if we set for x(0)=x_{0} it is...

    \lim_{t \rightarrow \infty} x(t)= 2 , x_{0}>0

    c)  x'(t) is independent from t so that if x_{0}(t) is solution od (1), x_{0}(t-t_{0}) with t_{0}>0 arbitrary is also solution of (1)

    The (1) can be solved with standard sepation of variable and the general solution is...

    x= \frac{2}{1-2\cdot c\cdot e^{-kt}} (2)

    ... where the constant c is determined by the 'initial condition'. Some problem exist however when is x_{0}=0

    Kind regards

    \chi \sigma
    Last edited by chisigma; August 7th 2009 at 04:29 AM. Reason: error in (2).... sorry!...
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