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Math Help - Population growth model

  1. #1
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    Population growth model

    The question

    The simple population growth model, \frac{dy}{dt} = ky (because of limitation on resources, pollution, ...) is unsatisfactory over a 'long' period. We might look at

    \frac{dy}{dt} = k(y)y ...........(1)

    (which of course ignores seasonal and other variations with time). Mathematically one of the simplest assumptions we can make is that k(y) decreases linearly as y increases. In this case (1) may be written in the form

    \frac{dy}{dt} = k(1 - \frac{y}{K})y, ...........(2)
    where k and K are constants.

    a) i) Equation (2) has two constant (stationary) solutions. What are they?

    I'm not sure how to calculate this. Do I have to solve the ODE? Or is there an obvious method? Thanks.
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  2. #2
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    Hmm, I think I get it now. I just need find values of 'y' to make the RHS zero?
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  3. #3
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    Quote Originally Posted by Glitch View Post


    \frac{dy}{dt} = k(1 - \frac{y}{K})y, ...........(2)
    where k and K are constants.

    a) i) Equation (2) has two constant (stationary) solutions. What are they?

    I'm not sure how to calculate this. Do I have to solve the ODE?
    This is worth solving, separate, decompose with partial fractions and integrate.

    Then look at solutions for \pm K
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  4. #4
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    Ok, I tried to do that, and I got this:

    \frac{1}{y(1-\frac{y}{K})}dy = k dt
    \frac{A}{y} + \frac{B}{(1 - \frac{y}{K})} = \frac{1}{y(1-\frac{y}{K})}}

    So we get:
    A = 1
    B = \frac{1}{K}

    \int{\frac{1}{y} + \frac{1}{K(1-\frac{y}{K})} = kt + C

    ln(y) + \int{\frac{1}{K - y}} = kt + C

    ln(\frac{y}{K - y}) = kt + C

    Take the exponential of each side:

    \frac{y}{K - y} = e^{kt + C}

    y = e^{kt + C}(K - y)

    Is this right so far? Thanks.
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  5. #5
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    However, to answer the question asked, "i) Equation (2) has two constant (stationary) solutions. What are they?", a "stationary solution" is a constant solution, dy/dt= 0 so, yes, just set the right side of the equation equal to 0: solve k\left(1- \frac{y}{K}\right)y= 0.
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