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Homework Help: Nonhomogeneous Differential Equation

  1. Mar 19, 2006 #1

    Zem

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    I got a particular solution y_p(x) that is different from what the book has.

    y'' + 9y = 2cos3x + 3sin3x

    Characteristic equation: r^2 + 9 = 0
    (r+3i)(r-3i) = 0
    y_c = c_1cos3x + c_2sin3x
    y_p = Acos3x + Bsin3x (not linearly independent, so I'll try another y_p)
    y_p = Axcos3x + Bxsin3x
    y'_p = Acos3x - 3Axsin3x + Bsin3x + 3Bxcos3x
    y''_p = -3Asin3x - 3Asin3x - 9Axcos3x + 3Bcos3x + 3Bcos3x - 9Bxsin3x
    x = 0
    y(x) = Acos3x + Bsin3x - 6Asin3x + 6Bcos3x + Acos3x + Bsin3x - 6Asin3x + 6Bcos3x = 2cos3x + 3sin3x

    Acos3x + 6Bcos3x = 2cos3x
    A + 6B = 1
    Bsin3x - 6Asin3x = 3sin3x
    -12A + 2B = 3

    My ti89 says A = -8/37 , B = 15/74
    y_p = (15/37sin3x - 8/37xcos3x)
    But the book's answer is y_p = 1/6(2xsin3x - 3xcos3x)

    What have I missed?
     
  2. jcsd
  3. Mar 19, 2006 #2

    TD

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    Homework Helper

    Your derivatives look right but I don't think you substituted them well in the equation, where is the factor 9 of y_p? You need to do y''_p+9y_p = ...
     
  4. Mar 19, 2006 #3

    Zem

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    And I should not have included y'_p (oops).
    Here is my new solution.
    y(x) = 6Bcos3x - 6Asin3x + 9Acos3x + 9Bsin3x = 2cos3x + 3sin3x
    9A + 6B = 2
    -2A + 3B = 1
    [ 9 6 2 ]
    [ -2 3 1 ]~

    [ 1 (2/3) (2/9) ](2)
    [ -2 3 1 ]~

    [ 1 (2/3) (2/9) ] (13)
    [ 0 (13/3) (13/9) ](-2) ~

    [1 0 0 ]
    [ 0 3 1 ]
    A = 0, B= 1/3
    The book says A = 1/2, B = 1/3. So I'm still a little off here, but can't see how. Thanks!
     
  5. Mar 19, 2006 #4

    Zem

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    Actually, I did not include y_p is that x = 0, so y_p = A(0)cos3x + B(0)sin3x. I included my original try for y_p in my above answer. But you helped me find the solution. Thanks so much!

    6Bcos3x = 2cos3x
    3B = 1
    -6Asin3x = 3sin3x
    -2A = 1
    y_p(x) = 1/6(2xsin3x - 3xcos3x)
     
  6. Mar 19, 2006 #5

    TD

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    Nice :smile:
     
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