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Variation of parameters (Kinda having trouble with the integral)

  1. Oct 26, 2009 #1
    1. The problem statement, all variables and given/known data

    Solve the problem: 4y'' - y = 8e^(.5t)/(2 + e^(.5t))


    2. Relevant equations

    Particular solution of Y = X*integral(inverse of X multiplied by G)

    Finding eigenvalues and eigenvectors


    3. The attempt at a solution

    This might be a little too messy for anyone to make sense of, but

    I found the eigenvalues first. 4x^2 - 1 = 0, so eigenvalues are +/- 1/2.
    Solution for the homogenous is therefore C_1e^(.5t) + C_2e^(-.5t)
    The matrix corresponding to the equation X' = AX is:
    [0 1]
    [.25 0]

    The eigenvector for value 1/2 is [1 ] and for -1/2 is [ 1 ]
    [.5] [-.5]

    So my matrix for the theorem Y = X*integral(inverse of X multiplied by G)
    is:
    [e^(.5t) e^(-.5t)]
    [.5e^(.5t) -.5e^(-.5t)]

    So far so good, I think.

    The inverse of this is [.5e^(-.5t) e^(-.5t)]
    [.5e^(.5t) -e^(.5t) ]

    G is [8e^(.5t)/(2 + e^(.5t))]
    [ 0 ]

    So plugging that into my integral equation thing, I get the integral of

    [ 4/(2 + e^(.5t)) ]
    [4e^t/(2 + e^(.5t))]

    This is where I am stuck and don't know how to integrate.

    I'd appreciate any help or advice you can guys can give! Also, please let me know if this problem can be solved a way different from the one I tried to use. Thanks a lot!
     
  2. jcsd
  3. Oct 26, 2009 #2

    LCKurtz

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    Science Advisor
    Homework Helper
    Gold Member

    I haven't checked over your work, but for an integral of the form:

    [tex] \int \frac 1 {a+e^{bt}}\, dt[/tex]

    try letting [itex]u = a + e^{bt},\ du =be^{bt}dt,\ e^{bt} = u-a[/itex]

    should give you something to work with. Same idea for the other integral.
     
  4. Oct 27, 2009 #3
    Thank you so much! I got the first one, and I think I can get the second one.
     
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