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Calculus problem - Mental creativity and flexability needed

  1. Nov 28, 2009 #1
    The spread of a disease in a population is modeled by the differential equation dP/dt=.03P(30-P) where p is less than or equal to 30 is measured in thousands of people and t is greater or equal to 0 is measured in weeks.

    a) Find the general solution of the differential equation.
    b) Find the particular solution of the differential equation given that 5000 people have the disease when first noticed, that is, when t=0, P=5.
    c) Estimate how many people will eventually have the disease. Explain your answer.

    Here's what i did:


    (a)
    dP/[P(30 - P)] = 0.03 dt.
    Now 1/[P(30 - P)] = 1/30[1/P + 1/(30 - P)] so the DE becomes

    [1/P + 1/(30 - P)] dP = 30(0.03) dt and integrating gives
    ln|P| - ln|(30 - P)| = 0.9t + c
    ln(P/(30 - P)) = 0.9t + c
    P/(30 - P) = Ae^(0.9t) where A = e^c
    P = Ae^(0.9t)*(30 - P)
    P[1 + Ae^(0.9t)] = 30Ae^(0.9t)
    so the general solution is
    P(t) = 30Ae^(0.9t)/[1 + Ae^(0.9t)].

    (b)
    When t = 0, P = 5 so
    5 = 30A/[1 + A]
    5[1 + A] = 30A
    25A = 5
    so A = 0.2 and the particular solution is
    P(t) = 6e^(0.9t)/[1 + 0.2e^(0.9t)].

    (c)
    The number eventually having the disease will be the limit of P as t--> infinity.
    Now P(t) = 6e^(0.9t)/[1 + 0.2e^(0.9t)]
    = 6/[e^(-0.9t) + 0.2]
    --> 6/0.2 = 30 as t--> infinity since e^(-0.9t)-->0 as t-->infinity.
    So 30 000 people will eventually have the disease.


    I've tried to do the work, please just let me know if i have anything wrong. Thanks guys.
     
  2. jcsd
  3. Nov 28, 2009 #2
    Looks perfectly fine to me.
     
  4. Nov 29, 2009 #3
    Bumpp - anyone else!?
     
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