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Math Help - non homogeneous second order, should be simple

  1. #1
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    non homogeneous second order, should be simple

    Find the particular solution to:
    y" - 6y' + 9y = 17e^(3t)

    (The homogeneous solution is y_1 = e^(3t), after finding the roots)

    For the particular solution, I try:
    y = Ae^(3t) ....... then y' = 3Ae^(3t)...... y" = 9Ae^(3t)

    Substitute that in.....

    9Ae^(3t) - 6(3Ae^(3t)) + 9(Ae^(3t)) = 17e^(3t)
    (simplified)
    9Ae^(3t) - 18Ae^(3t) + 9Ae^(3t) = 17e^(3t)
    simplify this further, and everything on the left hand side cancels....which means
    0 = 17e^(3t) which is not possible since e is asymptotic at y = 0.

    What do I do!?

    (edit)

    Same thing happens when I tried y = At(e^(3t)). Everything on the left hand side still cancels out after I substitute
    Last edited by plopony; February 24th 2010 at 08:04 AM. Reason: (new attempt)
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  2. #2
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    Quote Originally Posted by plopony View Post
    Find the particular solution to:
    y" - 6y' + 9y = 17e^(3t)

    (The homogeneous solution is y_1 = e^(3t), after finding the roots)

    For the particular solution, I try:
    y = Ae^(3t) ....... then y' = 3Ae^(3t)...... y" = 9Ae^(3t)

    Substitute that in.....

    9Ae^(3t) - 6(3Ae^(3t)) + 9(Ae^(3t)) = 17e^(3t)
    (simplified)
    9Ae^(3t) - 18Ae^(3t) + 9Ae^(3t) = 17e^(3t)
    simplify this further, and everything on the left hand side cancels....which means
    0 = 17e^(3t) which is not possible since e is asymptotic at y = 0.

    What do I do!?

    (edit)

    Same thing happens when I tried y = At(e^(3t)). Everything on the left hand side still cancels out after I substitute
    Since you have two repeated roots of 3 and  3, the complimentary solution is

    y_1 = e^{3t}, y_2 = t e^{3t} so you'll need to try y_p = At^2 e^{3t} .
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  3. #3
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    thank you, it would have been logical for me to keep following the pattern eh? thanks again
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