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Second Order ODE Using Laplace Transforms

  1. Feb 16, 2009 #1
    Little homework problem that I'm beating myself up over...

    Solve:

    [tex]xy'' - 2y' + xy = -2\cos x[/tex]

    Using the method of Laplace transforms....

    So I do some jiggling and get to:

    [tex](1+s^{2})\frac{dF}{ds}+4sF(s) = \frac{2s}{s^{2} + 1}[/tex]

    To which I find the following solution:

    [tex]F(s) = \frac{C}{(1 + s^{2})^{2}} + \frac{s^{2}}{(1 + s^{2})^{2}}[/tex]

    But I'm supposed to get to:

    [tex]F(s) = \frac{C}{(1 + s^{2})^{2}} + \frac{s^{2}-1}{(1 + s^{2})^{2}}[/tex]

    But the method I'm using to solve the first order ODE (variation of parameters/constants) won't seem to lead to the answer I need. It almost looks as though I should include an aribtrary constant where they're expecting me to get -1, but it seems that any constant value will be a solution to the first order DE obtained after applying the transform. So I don't really understand how they've deduced that it should be -1...

    We've been given the answer to the problem,

    [tex]y = \frac{C}{2}(\sin x - x\cos x) + x\cos x[/tex]

    Where the result is obtained easily using a table of Laplace Transforms.
     
  2. jcsd
  3. Feb 16, 2009 #2

    Tom Mattson

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    Those are equivalent. To see this, start from the second expression.

    [tex]F(s) = \frac{C}{(1 + s^{2})^{2}} + \frac{s^{2}-1}{(1 + s^{2})^{2}}[/tex]

    [tex]F(s) = \frac{C+s^2-1}{(1 + s^{2})^{2}}[/tex]

    [tex]F(s) = \frac{s^2+(C-1)}{(1 + s^{2})^{2}}[/tex]

    Since C is an arbitrary constant, so is C-1.
     
  4. Feb 16, 2009 #3
    So the justification is altering the constant to give a nice form for the inverse Laplace transform?

    Thanks very much!
     
  5. Feb 16, 2009 #4

    Tom Mattson

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    Yes, you can use either C or C-1. It doesn't matter.
     
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