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Math Help - Diff Eq. Non-homogeneous const coeffs, particular solution is particularly hard

  1. #1
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    Diff Eq. Non-homogeneous const coeffs, particular solution is particularly hard

    y'' + 3y' + 2y = 5e^{-2x} <-- problem

    General Solution
    r^2 + 3r + 2 = 0
     (r + 1)(r + 2) = 0
     r = -1, -2
     y = c1*e^{-1x} + c2*e^{-2x} + yparticular

    Particular Solution
    yparticular = A*e^{-2x}
    yparticular' = -2A*e^{-2x}
    yparticular'' = 4A*e^{-2x}
    4A*e^{-2x} -6A*e^{-2x}+2A*e^{-2x} = 5*e^{-2x}
    6A-6A = 5
    0 = 5

    The conclusion I arrive at after figuring out the particular solution cannot be true and I am lost as to what this means and or how to solve the problem. Can anyone help clear this up for me?

    Thanks in advance
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  2. #2
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    Quote Originally Posted by vitaemage View Post
    y'' + 3y' + 2y = 5e^{-2x} <-- problem

    General Solution
    r^2 + 3r + 2 = 0
     (r + 1)(r + 2) = 0
     r = -1, -2
     y = c1*e^{-1x} + c2*e^{-2x} + yparticular

    Particular Solution
    yparticular = A*e^{-2x}
    yparticular' = -2A*e^{-2x}
    yparticular'' = 4A*e^{-2x}
    4A*e^{-2x} -6A*e^{-2x}+2A*e^{-2x} = 5*e^{-2x}
    6A-6A = 5
    0 = 5

    The conclusion I arrive at after figuring out the particular solution cannot be true and I am lost as to what this means and or how to solve the problem. Can anyone help clear this up for me?

    Thanks in advance
    You should know to try y_p = a x e^{-2x}.

    This 'problem' occurs whenever \lambda = \lambda_1 is one of the solutions to the auxillary equation and e^{\lambda_1 x} is one of the terms in the right hand side.
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