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ODE Population problem

  1. Feb 13, 2011 #1
    I know that the rate of change with time of a population is proportional to the square root of t. T=0 is y = 100. Population increases at rate of 20 per month.

    I started out by trying to do dy/dt = p^.5. I am used to the population problems where I use y=Ce^(rt) but am having trouble making the jump to this kind. How do you account for the increase of 20/month?
  2. jcsd
  3. Feb 13, 2011 #2
    The population always increases at a rate of 20 per month, or is that an initial value?
  4. Feb 13, 2011 #3
    Yes, always increases at rate of 20/month and initial population is 100.
  5. Feb 13, 2011 #4
    Than the question makes no sense to me. How can the rate of increase in the population be of 20/month and proportional to the square root of t at the same time?
  6. Feb 13, 2011 #5
    You and me both then. Here is the exact question:
    The time rate of change of a rabbit population P is proportional to the square root of P. At time t = 0 (months) the population numbers 100 rabbits and is increasing at the rate of 20 rabbits per month. How many after one year? The answer I guess is 484.

    How do you set up logistics equations?
  7. Feb 13, 2011 #6
    So 20/month is an initial value.

    The rate of change of P, i.e. it's derivative with respect to P is proportional to the square root of P:

    [tex] \frac{dP(t)}{dt} = \alpha P^{1/2} [/tex]

    where alpha is a constant with respect to. You also know that

    [tex] \frac{dP(t)}{dt} \bigg|_{t=0}} = 20 \ rabbits/month [/tex]

    and that

    [tex]P(0) = 100 \ rabbits[/tex]

    Which is all the info you need to determine P(t)
    Last edited: Feb 13, 2011
  8. Feb 13, 2011 #7
    I appreciate what you wrote but I'm still unsure on where to start. I see that I was wrong to have a p^.5 on the right. Should I go ahead and start separating variables and then moving forward?
  9. Feb 13, 2011 #8
    OOOPSSS, my mistake. I corrected it in the previous post.

    Of course you should. Even in the case where you shouldn't, equations don't bite you know.
  10. Feb 13, 2011 #9
    But how do you incorporate the dp/dt = 20 and P(0) = 100? I've done the sep of variables, no problem there. Sitting on 2p^.5 = (alpha)t + C.
  11. Feb 13, 2011 #10
    Made another attempt at this, can you tell me where I'm going wrong?
    I came up with this equation after sep of variables: 2p^.5 = Bt + C.
    1) With t = 0 and initial population of 100, I came up with a C value of 100.
    2) With t = 1, population = 120, so I came up with a "B" value of 120.
    3) But then when I do time = 12, I don't get anywhere near the 484 rabbits that I should have.
  12. Feb 13, 2011 #11


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    Er, did you solve for p in your equation so you have some p(t) on the left side instead of 2p^.5?
  13. Feb 14, 2011 #12
    This is wrong. What you have is P'(0) = 20. So use the derivative of P to find B. Do you realize that the rate of change in the population is the derivative of P?
    Last edited: Feb 14, 2011
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