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First Order D.E

  1. Jun 12, 2015 #1

    AAO

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    1. The problem statement, all variables and given/known data

    A carrier of an infectious disease joins a herd of 500 initially uninfected cattle. At any
    instant in time, the rate at which the disease spreads dx/dt is known to be proportional to
    the product of:
    (i) the number of infected cattle x(t); and
    (ii) the number of uninfected cattle.
    If the number of cattle infected after 4 days is 50, how many will have been infected after
    6 days?

    2. Relevant equations

    3. The attempt at a solution
    I interpreted the problem as the following D.E:
    dx/dt=k*x*(500-x); where k is a constant to be determined

    And the initial condition: x(0)=0
    And the given information: x(4)=50

    The equation looks to be separable:
    dx/(500*k*x-k*x^(2)) = dt

    but I cannot integrate the left hand side.

    Is my interpreted D.E correct? If yes how to integrate the left hand side?

    Appreciate your help.
     
    Last edited: Jun 12, 2015
  2. jcsd
  3. Jun 12, 2015 #2

    Orodruin

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    Your initial condition here is saying that 500 kettle are infected at t=0 ...
     
  4. Jun 12, 2015 #3

    AAO

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    Thanks I corrected it. Any hints about the solution to this DE?

    I tried to use Matlab:
    >> syms x(t) k
    >> x(t) = dsolve(diff(x,t) == k*x(t)*(500-x(t)) , x(0)==0)

    But I got
    x(t) = 0

    !!
     
  5. Jun 12, 2015 #4

    Orodruin

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    Try splitting the integrand into two by partial fraction decomposition.

    Edit: Also, your initial condition is not x(0) = 0. Obviously, if there are no infected cattle, there will be no spread of the disease.
     
  6. Jun 12, 2015 #5

    RUber

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    How many infected cattle are there at t=0?
    If it is zero, transmission rate will surely be zero.
     
  7. Jun 12, 2015 #6

    Ray Vickson

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    x(0) is not zero. The problem said that one infected cow joined the others, so (presumably), x(0) = 1.
     
  8. Jun 13, 2015 #7

    AAO

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    Many Thanks All for your great help.

    I managed to get the answer assuming that x(0)=1, and performing the integration using partial fraction.

    The final answer is:
    500
    ---------------------------
    exp(log(499) - t) + 1

    I also confirmed this using Matlab.

    Once again, thank you very much.
     
  9. Jun 13, 2015 #8

    LCKurtz

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    Shouldn't your original DE have been$$\frac {dx}{dt}= kx(501-x),~x(0) = 1$$
     
  10. Jun 13, 2015 #9

    AAO

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    I believe you are right LCKurtz, I should have done this. Thanks a lot for the hint.
     
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