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Another change of variable problem

  1. Jul 24, 2005 #1
    I'm also attempting another problem...

    A population has a periodic growth rate r(t) = A[1 + sin(t/(2*pi))], but otherwise follows the logistic population model with carrying capacity K. There is no threshold and the initial population is Yo = Y knot = K/2.

    a. Modify the basic logistic equation for this population.
    b. Use a change of variable z(t) = y/K to find a initial value equation in z(t). Then Solve for z(t).

    So far I have:
    Using the logistic growth equation: dy/dt = r(1-y/k)y
    Leading to:
    integral of [dy/((1-y/k)y)] = rt +c
    substituting for r:
    integral of [dy/((1-y/k)y)] = t*A[1 + sin(t/(2*pi))] + c

    I can't really follow how to use the change of variables well so part b is the most confusing part.
  2. jcsd
  3. Jul 25, 2005 #2
    as a friend of mine likes to say, "Classic mistake."
    if r = r(t) you can't integrate it as if it were a constant then
    plug in the value. you need:

    [tex] \int \ldots dy = C + \int r(t) dt [/tex]

    There shouldn't be anything tricky to a change of variables.
    (except, of course, deciding what varibles to use.)
    It's a plug-and-chug operation.

    if y = k z then
    y' = ?

    now, just replace every occurence of y with (kz) and every
    occurence of y' with ?.
  4. Jul 25, 2005 #3
    Thanks for the point about r(t), so I fixed that part, but I'm still confused as how to go about the rest...
    If I substitute y with kz and y' with ?, then how do I integrate the following:
    Because I could integrate it if the "?" were a "dz" right?
  5. Jul 25, 2005 #4
    I guess I was unclear. it's your job to find out what
    to put in in place of the ?.

    What I thought i had hinted at
    was if y = kz then y' = kz', because
    k is a constant.
  6. Jul 25, 2005 #5
    Sorry, looking back that was pretty clear...
    So then it would be the integral of:
    dz/(1-z)z = ln(1-z) + ln(z)
    Then set this equal to:
    C + Integral( r(t))
    and just solve for z?
    Last edited: Jul 25, 2005
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