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Method of Variation of Parameters

  1. Jun 13, 2012 #1

    ElijahRockers

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    1. The problem statement, all variables and given/known data

    [itex]y''-2y'+y = \frac{e^x}{1+x^2}[/itex]

    2. Relevant equations

    [itex]u_1 = -\int \frac{y_{2}g(x)}{W}dx[/itex]
    [itex]u_2 = \int \frac{y_{1}g(x)}{W}dx[/itex]
    [itex]g(x) = \frac{e^x}{1+x^2}[/itex]
    W is the wronskian of y1 and y2.

    3. The attempt at a solution

    The characteristic equation for the homogenous solution yields a repeated root of -1, so

    [itex]y_{h} = C_{1}e^{-x} + C_{2}xe^{-x}[/itex]

    When I calculated the Wronskian it simplified to

    [itex]W = e^{-2x}[/itex]

    Plugging in the formula for u1 and simplifying I get

    [itex]u_{1} = -\int \frac{xe^{2x}}{1+x^2}dx[/itex]

    I'm not quite sure how to go about solving this. My best shot at integration by parts didn't really seem to help. I ended up having to integrate ln(x^2 + 1)e^2x which doesn't seem any easier.
     
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  3. Jun 13, 2012 #2

    vela

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    I got W=(1-2x)e-2x.
     
  4. Jun 13, 2012 #3

    ElijahRockers

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    Hmmm, alright.. I can't seem to find my mistake

    if W is
    |y1 y2|
    |y'1 y'2|

    I got
    y1 = e^-x
    y2 = xe^-x
    y'1 = -e^-x
    y'2 = -xe^-x + e^-x = e^-x(1-x) (after factoring off the e^-x)

    and y1y'2 - y'1y2 = e^(-2x)(1-x) + xe^(-2x) = e^(-2x)
     
  5. Jun 13, 2012 #4

    vela

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    Sorry, I made a sign mistake.
     
  6. Jun 13, 2012 #5

    vela

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    Oh, your roots are wrong. They should be +1.
     
  7. Jun 13, 2012 #6

    ElijahRockers

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    O geez... always something little.

    Thanks!
     
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