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Math Help - Ode

  1. #1
    Newbie
    Joined
    Dec 2007
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    16

    Ode

    Find a general sol'n for the following ODE:

    y''-6y'+9y=2+te^{4t}

    So I use the auxiliary eq:

    m^2 - 6m + 9 = 0

    m = 3.

    So, the homogeneous sol'n, y_c, is:

    y_c = c_1e^{3t}

    How'd I complete it?
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  2. #2
    Senior Member
    Joined
    Dec 2007
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    Melbourne
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    428
    Good job so far.

    What needs to be done next is find the parts of the solution that are linearly independent to e^{3t}. This is done by substituting y= ue^{3t}. This will net you both the 2nd part of the homogeneous solution and the particular solution.

    y = ue^{3t}
    y' = u'e^{3t}+3ue^{3t}
    y'' = u''e^{3t}+6u'e^{3t}+9ue^{3t}

    So our equation becomes:

    (u''e^{3t}+6u'e^{3t}+9ue^{3t})-6(u'e^{3t}+3ue^{3t})+9ue^{3t} = 2+te^{4t}
    u''e^{3t} = 2+te^{4t}

    notice that the u term has cancelled out. This will always happen when we make this substitution. In this case the u' term has also been cancelled but this was just lucky.

    u'' = 2e^{-3t}+te^t
    u' = \int 2e^{-3t}+te^tdt
    = -\frac {2}{3}e^{-3t}+te^t-e^t+c_2 (I have integrated by parts)
    u = \int -\frac {2}{3}e^{-3t}+te^t-e^t+c_2 dt
    = \frac {2}{9}e^{-3t}+te^t-e^t-e^t+c_2t+c_1
    The reason for reusing c_1 will soon be apparent
    u =  \frac {2}{9}e^{-3t}+te^t-2e^t+c_2t+c_1

    y = ue^{3t}
    = (\frac {2}{9}e^{-3t}+te^t-2e^t+c_2t+c_1)e^{3t}
    = \frac {2}{9}+te^{4t}-2e^{4t}+c_2te^{3t}+c_1e^{3t}
    Last edited by badgerigar; December 10th 2007 at 12:42 AM. Reason: fixed typo
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