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Logistic Equation

  1. Mar 30, 2005 #1
    Hi. I'm having some difficulty with a couple of questions, but mainly a problem differentiating an equation:

    dP kP (1-P/K)(1-m/P)
    -- =
    dT

    it says to solve it explicitly but I have no idea how to differentiate explicitly.

    If you could help me out, it'd be greatly appreciated. Also, if you tell me how to insert the the math operations so it looks normal, it'd be appreciated. Thanks again.
     
  2. jcsd
  3. Mar 30, 2005 #2
    [tex] \frac{dP}{dt} = kP(1-\frac{P}{K})(1-\frac{m}{P}) [/tex]

    Well unless you left something out, with respect to time, nothing on the right hand side changes, it is all constant. Derivative of a constant is 0.

    Anyhow if im mistaken, explicit derivative is the fancy way of saying 'take the derivative'.
     
  4. Mar 30, 2005 #3
    no, you are mistaken. I have to find the function that, when you take the derivative of it, gives you that equation. I pretty much have to find what P(t) equals. Thanks anyhow. Oh ya, m, K, and k are constants.
     
  5. Mar 30, 2005 #4
    Gah nevermind P isnt constant. I can't help you.
     
  6. Mar 30, 2005 #5
    ya I was just going to point it out...I'll try and post what I have so far and maybe someone can help me from there
     
  7. Mar 30, 2005 #6

    James R

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    First, use algebra to show that:

    [tex]kP(1-\frac{P}{K})(1-\frac{m}{P}) = -\frac{k}{K}(P - K)(P - m)[/tex]

    The differential equation is then:

    [tex]\frac{1}{(P - K)(P - m)} dP = -k/K dt[/tex]

    Split the integrand on the left hand side using partial fractions, to get:

    [tex][\frac{1}{(K-m)(P-K)} + \frac{1}{(m-K)(P - m)}] dP = -\frac{k}{K} dt[/tex]

    Integrate both sides...

    [tex]\frac{\ln (P-K)}{(K-m)} - \frac{\ln (P - m)}{K - m} = -\frac{k}{K}t + c1[/tex]

    Use log rules to combine the logs...

    [tex]\frac{1}{K-m} \ln [\frac{P-K}{P-m}] = -\frac{k}{K}t + c1[/tex]

    Multiply by (K - m) and take exponentials of both sides...

    [tex]\frac{P-K}{P-m} = A \exp (-k + km/K)t[/tex]

    Note that A, here, is a constant of integration, which must be determined from the boundary conditions. From here, it is simple algebra to get P as the subject. Doing that gives:

    [tex]P(t) = \frac{Am \exp [(-k + km/K)t] - K]}{A \exp[(-k + km/K)t]}[/tex]

    If you know one value of P(t), you can determine A, and you're done.
     
  8. Mar 30, 2005 #7
    what I have so far is:

    [tex] dP (K/(K-P)(P-M)) = kt + c [/tex]

    no idea where I go from here so...help me out please.
     
  9. Mar 30, 2005 #8
    wow...thank you....hah.
     
  10. Mar 30, 2005 #9
    question...why is it -k/K in the first part
     
  11. Mar 31, 2005 #10

    James R

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    It follows from the algebra. Check the first line of my working.

    The way you split the constants between sides of the equation is arbitrary, so my way is not the only way to get to the solution. If you want to leave K on the dP side, as you have done a few posts above this one, that's ok. The working will be a little different, by the answer should be the same.
     
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