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Rodent Population as a differential equation.

  1. Jan 12, 2014 #1
    1. The problem statement, all variables and given/known data

    For those who can not see the screen shot here is the question...
    Suppose the population P of rodents satisfies the diff eq dP/dt = kP^2.
    Initially there are P(0) = 2 rodents, and their number is increasing at the rate of dP/dt = 1 rodent per month when P = 10. How long does it take for the population to reach 105 rodents.

    2. Relevant equations

    3. The attempt at a solution

    First I found k by substituting in 10 for p and 1 for dp/dt.
    1 = k * 10^2
    k = 1/100

    Then to solve the differential equation I integrated both sides with respect to t.

    ∫dp/dt * dt = ∫0.01 * p^2 dt
    p = 0.01p^2 * t + C

    I solved for C and found C = 2.

    Solving for t gives
    t = 100(p - 2) / p^2

    I then plunged in 105 for p and the answer was wrong.
    Last edited: Jan 12, 2014
  2. jcsd
  3. Jan 12, 2014 #2
    It would be helpful to see the question.
  4. Jan 12, 2014 #3


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    At the moment the question has not been posted or is not visible to me.
  5. Jan 12, 2014 #4
    I can see the screen shot of the question uploaded. I will rewrite the question and update the original post.
  6. Jan 12, 2014 #5


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    Homework Helper

    Integrating ∫0.01 * p^2 dt to get 0.01p^2*t+C is incorrect. That's assuming p is a constant. You need to separate the ODE first.
  7. Jan 13, 2014 #6


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    This is incorrect. You are treating p, on the right, as if it were a constant but it is a function of t.
    Instead, write it as [tex]\int \frac{dp}{p^2}dp= \int .01dt [/tex]
    [tex]-\frac{1}{p}= .01t+ C[/tex]
    [tex]p(t)= -\frac{1}{.01t+ C}[/tex]
    [tex]p(0)= -\frac{1}{C}= 2[/tex]

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