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Why doesn't this diff eq. have two solutions?

  1. Sep 28, 2011 #1
    Say you have the diff eq. x'=2x(x-13); x(0)=20. After separating and integrating we get,

    ln|(x-13)/x|=26t+C

    From here, we raise e to the power of each side to get rid of the natural log. Does this get rid of the absolute value signs? If so, why? If not, why is there only one solution (as dictated by wolfram alpha)?
     
  2. jcsd
  3. Sep 29, 2011 #2

    HallsofIvy

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    Wolfram alpha gives a solution. It does not necessarily mean there are not other solutions.

    As for this particular problem, you seem to be forgetting the "initial value" part of the intial value problem. If you take the exponential of both sides you get
    [tex]\frac{x- 13}{x}= e^{26t+ C}= e^Ce^{26t}= C'e^{26t}[/tex]
    where [itex]C'= e^C[/itex]

    Yes, C' can be either positive or negative but to satisfy x(0)= 20, we must have
    [tex]\frac{20- 13}{20}= \frac{7}{20}=C'[/tex]
    so there is only one solution.
     
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