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Thread: Ode

  1. #1
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    Dec 2007
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    Ode

    Find a general sol'n for the following ODE:

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

    So I use the auxiliary eq:

    m^2 - 6m + 9 = 0

    m = 3.

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

    $\displaystyle y_c = c_1e^{3t}$

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

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

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

    So our equation becomes:

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

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

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