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The classic second order differential equation

  1. Sep 28, 2013 #1
    If we have [itex]\frac{1}{X(x)}[/itex] [itex]\frac{d^2 X}{dx^2}[/itex]=-κ^2, the literature is saying that the solution must be: e^(±iκx), but am always getting e^(±k^2x).

    Isn't the approach is to decently integrate twice and then raise the ln by the exponential? Where am I going wrong? Thanks
     
    Last edited: Sep 28, 2013
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  3. Sep 28, 2013 #2

    arildno

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    What do you mean with "integrating twice"?
    Show your steps.
     
  4. Sep 28, 2013 #3

    jambaugh

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    The devil is in the details... could you show your integration work?

    Keep in mind that to integrate you need a differential instead of derivative format. letting [itex]Y=\frac{dX}{dx}[/itex] your equation is of the form:
    [tex]\frac{1}{X}\frac{dY}{dx} = -k^2[/tex] so
    [tex]dY = -k^2 X dx[/tex]
    Integrating doesn't work out directly:
    [tex] Y(x)=k^2 \int X(x)dx[/tex]
    You can try different integration by parts and substitutions... but eventually you come back to the problem of finding an integration factor for the original equation:

    [tex] YdY = -k^2 XYdx = -k^2 XdX[/tex]
    [tex] Y^2/2 = -k^2 X^2/2[/tex]
    [tex]Y = -ik X[/tex]
    [tex]dX = -ikX dx, dX/X = -ik dx[/tex]
    [tex] X = e^{-ikx}[/tex]
    (here I'm sloppy with the integration constants, you need to work them in.)
     
  5. Sep 28, 2013 #4
    Jambaugh, thank you for the reply. But that was derivation with second order. You worked it out with first order, right?
     
  6. Sep 28, 2013 #5
    One more question, if for x=0, X=0 then X=sinkx according to your result? Or would it be isinkx? Am just confused with the imaginary number..
     
  7. Sep 28, 2013 #6

    arildno

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    Show your work, and we'll tell you where you are mistaken.
     
  8. Sep 29, 2013 #7

    LCKurtz

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    It is just a constant coefficient equation ##X''+k^2X=0## with characteristic equation ##r^2+k^2=0##. So ##r=\pm ik##.
     
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