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2nd order ODE solution bases /wronskain question

  1. Mar 31, 2012 #1
    hello

    this question from my coarse notes has been giving me some trouble so i thought i would ask for some help on here,

    http://img88.imageshack.us/img88/9764/asfar.jpg [Broken]

    i understand that since the bases are bases of the same solutions then they are just a multiple of each other, but i'm not sure how you would show it using the wronskain.

    i first tried starting by saying since both are two different bases for the solutions then
    [itex]k\phi_{1}=\psi_{1}[/itex]

    [itex]k\phi_{2}=\psi_{2}[/itex]

    then doing the wronskian, but it gives k2

    it seems like there is something straight forward that i am not seeing

    anyone know what it might be?
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Mar 31, 2012 #2

    LCKurtz

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    No, they aren't just multiples of each other. For example, the equation ##y''-y=0## could have ##y=A\cosh(x)+B\sinh(x)## or ##y=Ce^x+De^{-x}##.
    What you do know is that each function in the second basis is a linear combination of the functions in the first basis. Calciulate the Wronskian for the second basis and use that fact to relate it to the Wronskian in the first basis.
     
    Last edited by a moderator: May 5, 2017
  4. Apr 2, 2012 #3
    Hey,

    Thanks for your reply!

    Sorry I had an assignment I had to work on before I started studying math again,

    So I did what you said just now using

    [itex]c_1\phi_{1} +c_2\phi_{2}=\psi_{1}[/itex]

    [itex]c_3\phi_{1} +c_4\phi_{2}=\psi_{1}[/itex]

    I got the required result with k = c1c4-c2c3

    It seems good to me thanks!

    Does it look right to you?
     
  5. Apr 2, 2012 #4

    LCKurtz

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    Yes, that looks to be correct.
     
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