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Solve using Undetermined Coefficients

  1. Oct 27, 2011 #1
    [itex]y''(t) - \frac{2}{t^2}y(t) = 3 - \frac{1}{t^2}[/itex]

    In this problem I had to solve two ways: Variation of Parameters and Undetermined Coefficients. I solved it using Variation of Parameters and got the correct answer for the particular solution in the back of the book being [itex]y_p(t) = t^2ln|t| + \frac{1}{2}[/itex]

    I can't seem to get the same answer when I'm solving using Undetermined Coefficients. I let my assumption be [itex]y_p(t) = At^2 + Bt + C[/itex] but I only end up with [itex]C=\frac{1}{2}[/itex] while [itex]A=0[/itex] and [itex]B=0[/itex] when I go back and substitute into the original to find their values. What am I doing wrong here? Is my assumption incorrect? I somehow have to get a [itex]ln|t|[/itex] to match up with a [itex]t^2[/itex] as an A value then I should be able to get the same particular solution.
     
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  3. Oct 27, 2011 #2

    LCKurtz

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    If you multiply that equation through by t2 you get
    [tex]t^2y''(t) - y(t) = 3t^2 - 1[/tex]

    It isn't a constant coefficient equation for which you would use undetermined coefficients. One way of working it is to look for a solution in the form y = tr to get a characteristic equation. Alternatively you can change the variable from t to x by x = ln(t) to make a constant coefficient DE out of it. This type of equation is called an Euler equation. Surely your text has something about these.
     
  4. Oct 28, 2011 #3
    Yeah it does, I think I just misheard the teacher in him saying use undetermined coefficients when I think he meant use 2 methods of variation of parameters to get the particular solution. At least I hope he said that as that's what I did.
     
  5. Oct 28, 2011 #4

    lurflurf

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    Undetermined coefficients works fine with variable coefficients, it is just harder to guess the form, and you guessed wrong. A better guess would be

    [itex]y_p(t) = At^2 \log(t) + Bt + C[/itex]

    If it is hard to guess the form undetermined coefficients becomes a hard to use method.
     
  6. Oct 28, 2011 #5
    Treating it as an Euler equation would be better. Or maybe try a series solution.
     
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