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Math Help - non-homogeneous equation, method of undetermined coefficients

  1. #1
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    non-homogeneous equation, method of undetermined coefficients

    Find the solution of:

    y" + 3y' = 72sin(3t) + 36cos(3t)
    where y(0) = 6 and y'(0) = 9

    I first found the solution to the homogeneous eq:

    the roots (R^2 + 3R = 0) are R = 0, -3
    which gives the family of solutions:
    y = a(1) + be^(-3t)
    and y' = -3be^(-3t)

    using the initial conditions (maybe Im not supposed to use them here?)
    I find a = 9, b = -3

    For the inhomogeneous eq:

    I try (guess)
    y = Asin(3t) + Bsin(3t)
    y' = 3Acos(3t) - 3Bsin(3t)
    y" = -9Asin(3t) - 9Bcos(3t)

    substitute those values into the original equation (left hand side) I find

    sin3t(-9A-9B) + cos3t(-9B-9A) = 72sin3t + 36cos3t

    therefore
    -9A - 9B = 72
    -9B - 9A = 36
    giving B = -6, A = -2


    Therefore I get the solution:

    y = 9 - 3e^(-3t) - 2sin3t - 6cos3t

    What did I do wrong (this answer is incorrect)
    Thanks
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  2. #2
    Member Mauritzvdworm's Avatar
    Joined
    Aug 2009
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    Pretoria
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    using the initial conditions (maybe Im not supposed to use them here?)
    I find a = 9, b = -3
    don't use the initial conditions here, use it at the end

    y(x)=-2\sin(3t)-6\cos(3t)+a+be^{-3t}

    so

    y(0)=0=-6+a+b so a=6-b and
    y^{\prime}(t)=-6\cos(3t)+18\sin(3t)-3be^{-3t} and
    y^{\prime}(0)=9=-6-3b so b=-5 and a=11

    then finally you end up with

    y(x)=-2\sin(3t)-6\cos(3t)+11-5e^{-3t}
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