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Population Model

  1. Sep 30, 2009 #1
    1. The problem statement, all variables and given/known data

    The differential equation [tex]\frac{dP}{dt} = P(a - bP)[/tex] is a well-known population model. Suppose the DE is changed to:

    [tex]\frac{dP}{dt} = P(aP - b)[/tex]

    Where a and b are positive constants. Discuss what happens to the population P as time t increases.

    2. Relevant equations

    [tex]\frac{dP}{dt} = P(aP - b)[/tex]

    3. The attempt at a solution

    Well I thought the population would increase because in the original equation on the intervals:

    critical points
    [tex] P(t) = 0[/tex]
    [tex] P(t) = a/b[/tex]

    [tex]-\infty < P < 0[/tex] it is decreasing
    [tex]0 < P < a/b[/tex] it is increasing
    [tex]a/b < P < \infty[/tex] it is decreasing

    so in the changed equation:

    critical points
    [tex] P(t) = 0[/tex]
    [tex] P(t) = b/a[/tex]

    [tex]-\infty < P < 0[/tex] it is increasing
    [tex]0 < P < b/a[/tex] it is increasing
    [tex]b/a < P < \infty[/tex] it is increasing
     
  2. jcsd
  3. Sep 30, 2009 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    The derivative is positive (and so the function is increasing) when P(aP- b)> 0. Now, if P< 0, both P and aP-b are negative so P(aP- b)> 0. For 0< P < b/a, P is positive but aP- b is still negative so P(aP-b)< 0 and P is decreasing. Was that a typo?

    From that, it seems to me that what happens to the population "as time t increases" depends upon the initial value. What happens if P(0) is less than b/a? What happens if P(0) is larger than b/a? What happens if P(0)= b/a?
     
  4. Oct 1, 2009 #3
    Ok I got it thanks.
     
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