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Y''' + 8y = 2x - 5 + 8 e^(-2x); y(0) = -5, y'(0) = 3, y''(0) = -4

  1. Feb 20, 2014 #1

    s3a

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    1. The problem statement, all variables and given/known data
    The problem:
    y''' + 8y = 2x - 5 + 8 e^(-2x); y(0) = -5, y'(0) = 3, y''(0) = -4

    Correct answer:
    y = -23/12 e^(-2x) + e^x (-59/24 cos(sqrt(3) x) + 17 sqrt(3)/72 sin(sqrt(3) x)) + 1/4 x - 5/8 + 2/3 x e^(-2x)

    2. Relevant equations
    Method of undetermined coefficients.

    3. The attempt at a solution
    My work is attached as the TheProblemAndMyWork.pdf file. The work is typed (not handwritten), so it looks nice.

    What I'm having trouble with is finding the values for c_1, c_2 and c_3. I entered the system of equations from my work into Wolfram Alpha, and the answer is wrong! I re-did this problem so many times, and I keep ending up in the same situation. I suppose, I keep making the exact same mistake.

    Could someone please tell me what I am doing wrong?

    Any help would be GREATLY appreciated!
     

    Attached Files:

  2. jcsd
  3. Feb 20, 2014 #2

    vela

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    The constant term you get when you calculated f''(0) should be -8/3, not 8.
     
  4. Feb 20, 2014 #3

    Ray Vickson

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    It is a quite a bit easier if you first get rid of the x and constant terms on the right. You can do this by looking at ##z(x) = y(x) + ax + b## for some constants ##a, b##, then adjusting ##a## and ##b## until you get ##z''' + 8z = 8 e^{ -2x}##. Then the "undetermined coefficient" equations simplify down quite a lot. Of course you need to determine the initial conditions on z(t) and use them instead.
     
  5. Feb 22, 2014 #4

    s3a

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    Thanks guys, I get it now!
     
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