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Solutions of 2nd order DEs

  1. Jul 24, 2009 #1
    1. The problem statement, all variables and given/known data
    I was curious if anyone could help me prove the equivalence between the two forms of solutions to second order ODEs, one being the linear combination of two solutions and the other being the phase-shifted sin/cos function.


    2. Relevant equations
    [tex]\frac{d^{2}x}{dt^{2}}+\frac{k}{m}x=0[/tex]

    [tex]x(t)=Asin(\omega t+\phi)[/tex]

    [tex]x(t)=C_{1}cos(\sqrt{\frac{k}{m}}t)+C_{2}sin(\sqrt{\frac{k}{m}}t[/tex]

    [tex]A[/tex] and [tex]\phi[/tex] should be treated as constants. I know that this is the differential equation for a harmonic oscillator, but I figured since it's more of how the mathematics behind it work, it belongs in this forum. If the moderator believes it should be moved, then by all means please move it.

    3. The attempt at a solution
    I have no idea where to start the proof for this. The second solution I wrote can be simply verified from the characteristic equation for the differential equation.

    Thanks.

    EDIT: I found my error. Oops. I wrote it as r^2 + (k/m)r. Thats what I get for doing math at 3AM, lol.
     
    Last edited: Jul 24, 2009
  2. jcsd
  3. Jul 24, 2009 #2

    Pengwuino

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    You're missing the second exponential in the 2nd equation.
     
  4. Jul 24, 2009 #3
    Hm? Which exponential? I got roots of 0 and -k/m for that differential equation's characteristic equation. I did forget the negative sign to the -k/m though.
     
  5. Jul 24, 2009 #4

    Pengwuino

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    Check the roots again, the characteristic equation should be [tex]\lambda ^2 + \frac{k}{m} = 0[/tex]. Assuming k>0, you have imaginary roots giving a positive and negative exponential solution.
     
  6. Jul 24, 2009 #5
    I don't think the second solution satisfies the equation.

    ---
     
    Last edited by a moderator: Aug 6, 2009
  7. Jul 24, 2009 #6

    HallsofIvy

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    Then you've either got the wrong characteristic equation or you have solved it incorrectly. Neither 0 nor -k/m satisfy the characteristic equation for this d.e.

    Please show your work.
     
  8. Jul 24, 2009 #7
    Got it. Thanks.
     
  9. Jul 24, 2009 #8

    HallsofIvy

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    Oh, that's not fair! Show your work so we can all laugh at your mistake and feel superior!

    (Some silly little arithmetic or algebra mistake.)
     
  10. Jul 24, 2009 #9
    Haha no i just added an r for the term (k/m)x. I took DE my sr year in high school, so most of it's begun to leave me. I have a feeling it's gonna come back bite me in the butt when I take circuits... sigh.
     
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