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Separating an ODE

  1. May 21, 2007 #1
    The rate of growth of the population x of a country is proportional to the product of x and (M-x), where M is a constant. Assume that x can be regarded as a continuous function of the time t with a continuous derivative at all times and show that if x=M/2 when t=50 then
    [tex] x=\frac{M}{1+\exp^{-Mk(t-50)}} \\[/tex].
    My attempt:
    [tex]\frac{dx}{dt} \propto x(M-x) [/tex] Therefore
    [tex] \int\frac{1}{x(M-x)}dx = k\int dt [/tex] Solving the first integral, using partial fractions, we get:
    [tex] \int \frac{1}{x(M-x)} dx = \frac{-1}{M} \int \frac{1}{x} dx + \frac{1}{M} \int \frac{1}{m-x} dx \\ [/tex]
    Solving the 2nd integral on the l.h.s., let u=M-x => -du=dx, therefore [tex]\int \frac{1}{x(M-x)} dx = \frac{-1}{M} \ln{x} + \frac{-1}{M} \int \frac{1}{u}du \\ [/tex] =-1/M(ln(x) + ln(m-x)) = -1/M*ln(x(M-x)) = [tex] k\int dt[/tex].
    Therefore ln(x(M-x)) = -M*k*t - M*c. => [tex] x(M-x) = \exp{-M(kt+c)} [/tex].
    As you can see x is not separated, but I cannot find the mistake/s that gave rise to this. Thanks for helping.
    Last edited: May 21, 2007
  2. jcsd
  3. May 21, 2007 #2
    Recheck the integration by partial fractions.
  4. May 21, 2007 #3
    I now get x/(M-x) =[tex] \exp{-M(kt+c) [/tex]
  5. May 21, 2007 #4
    What you wrote as

    [tex] \int \frac{1}{x(M-x)} dx = \frac{-1}{M} \int \frac{1}{x} dx + \frac{1}{M} \int\frac{1}{m-x} dx \\ [/tex]

    should have been...

    [tex] \int \frac{1}{x(M-x)} dx = \frac{1}{M} \int \frac{1}{x} dx + \frac{1}{M} \int \frac{1}{M-x} dx \\ [/tex]

    i.e.No minus signs.

    After integrating, it should read,

    [tex]\frac{1}{M}\left[\ln{x} - \ln{\left(M-x\right)}\right] = k\int{dt}[/tex]
    Last edited: May 21, 2007
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