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Dx/dt = x(t)(1-x(t))

  1. Jan 16, 2013 #1
    1. The problem statement, all variables and given/known data

    Where am I going wrong in solving this differential equation?

    2. Relevant equations

    x' = x(1-x) is what I'm after

    3. The attempt at a solution

    Separating variables: dx / [x(1-x)] = dt
    Partial fraction decomposition on the left side yields: dx [1/x + 1/(1-x)] = dt.
    Integrating both sides with respect to their variable: ln(x) + ln(1-x) = t + C
    Simplifying the left side: ln(x(1-x)) = t + C
    Raising both sides to the e power: x(1-x)=Ket

    Nothing I can really do after this, though. I guess I just define it implicitly like that?
     
  2. jcsd
  3. Jan 16, 2013 #2
    Another question:

    Any idea how I can get a formula for the x for which cos(x)=ex?
     
  4. Jan 17, 2013 #3

    Mark44

    Staff: Mentor

    Your mistake is just above.
    $$ \int \frac{dx}{1 - x} \neq ln(1 - x)$$


     
  5. Jan 17, 2013 #4

    CompuChip

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    Science Advisor
    Homework Helper

    Also, if you get x(1 - x) = f(t) you can definitely solve for x -- if you look more closely it's just a quadratic equation.
    First correct the error pointed out by Mark though :)

    As for your other question cos(x) = ex does not have an algebraic solution.
     
  6. Jan 17, 2013 #5
    Ok. Because one of our hw questions was to solve x'(t) = et - cos(t). Someone asked the professor what to do about the fact that there are infinitely many fixed points, and he told us to just deal with x'(0)=0 but to explain how we know there are infinitely many other solutions.
     
  7. Jan 17, 2013 #6
    Ah, I see. :smile:

    So then we have

    x' = x(1-x)
    ---> dx / [x(1-x)] = dt
    ---> dx/x + dx/(1-x) = dt
    ---> ln(x) - ln(1-x) = t + c
    ---> ln[x/(1-x)] = t + c
    ---> x/(1-x) = Cet
    ---> (1-x)/x = Ke-t
    ---> 1/x - 1 = Ke-t
    ---> 1/x = 1 + Ke-t
    ---> x(t) = 1/(1+Ke-t)
    ---> x(0) = 1(1+K), K = x(0)-1
    ---> x(t) = 1/[1+(X(0)-1)e-t]
     
  8. Jan 17, 2013 #7

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    One way to show it has more than one solution is to show a graph of [itex]y= cos(x)- e^x[/itex] and note that it crosses the x-axis multiple times.
    cos-ex.jpeg
     
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