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Initial Value Problem

  1. Jan 22, 2017 #1

    Drakkith

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    Staff: Mentor

    1. The problem statement, all variables and given/known data
    Solve the initial value problem:
    ##\frac{dx}{dt} = x(2-x)##, ##x(0) = 1##
    for ##x(t=ln2)##.

    2. Relevant equations


    3. The attempt at a solution

    I moved the right side to the left and multiplied both sides by dt to get:
    ##\frac{dx}{x(2-x)} = dt##

    Integrating gave me:
    ##\frac{ln|x|}{2} - \frac{ln|x-2|}{2} = t+C##

    Then:
    ##ln|x| - ln|x-2| = 2t + 2C##
    ##ln|\frac{x}{x-2}| = 2t + 2C##
    ##\frac{x}{x-2} = e^{2t}e^{2c}##
    ##\frac{1}{1-\frac{2}{x}} = Ke^{2t}##

    Manipulating for a while, I end up with:
    ##x=\frac{2}{Ke^{2t}}##

    Since ##x(0) = 1##, I set the left side to 1 and solve for K, winding up with ##k=3##.
    However, when trying to solve for ##x(ln2)## I end up with ##\frac{2}{11}##, which isn't one of my possible answers.

    Does my process look even remotely correct?
     
  2. jcsd
  3. Jan 22, 2017 #2

    LCKurtz

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    So you could say ##\frac x {x-2} = Ke^{2t}## Put ##t=0,~x=1## there to get ##\frac x {x-2} = -e^{2t}##. Now put ##t=\ln 2## in that, then solve for ##x##. See if that fixes it.
     
  4. Jan 23, 2017 #3

    Ray Vickson

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    For ##t## near 0 you have ##x## near 1, and so ##x/(x-2) = e^{2C} e^{2t}## is impossible unless you let ##C## be a complex number. However, ##\ln(|x|/|x-2|) = \ln(x/(2-x))##, and so having ##x/(2-x) = e^{2C} e^{2t}## is OK. Of course, when you re-wrote ##e^{2C}## as ##K## and then forgot the origin of ##K##, you were then able to have a valid formula!
     
  5. Jan 23, 2017 #4

    LCKurtz

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    @Drakkith: Ray is quite correct and I didn't notice you had been sloppy with the absolute value signs. A better way for you to have written it would be like this:$$
    ln\left | \frac{x}{x-2}\right | = 2t + 2C$$ $$
    \left|\frac{x}{x-2}\right | =e^{2t + 2C}=e^{2C}e^{2t}$$ $$
    \frac{x}{x-2} =\pm e^{2C}e^{2t} = Ke^{2t}$$
     
  6. Jan 23, 2017 #5

    Drakkith

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    Yes, that seemed to work. I got x=8/5, which is one of the possible answers. Thanks guys!
     
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