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Thread: Second order non-homogeneous ODE

  1. #1
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    Second order non-homogeneous ODE

    The question:
    $\displaystyle y'' - y = e^{-x}$

    My attempt:
    Finding homogeneous solution:
    $\displaystyle \lambda^2 - 1 = 0$
    $\displaystyle \lambda = 1, -1$

    So we have $\displaystyle y = Ae^x + Be^{-x}$

    Finding a particular solution, I tried $\displaystyle y = Ce^{ax}$, but noticed that this is also a solution to the homogeneous equation. So instead, I went with $\displaystyle y = Cxe^{ax}$.

    Finding derivatives:

    $\displaystyle y' = Caxe^{ax} + Ce^{ax}$
    $\displaystyle y'' = Ca^2xe^{ax} + 2Cae^{ax}$

    Substituting into the ODE:

    $\displaystyle Ca^2xe^{ax} + 2Cae^{ax} - Cxe^{ax} = e^{-x}$

    $\displaystyle Ca^2xe^{ax} + Cxe^{ax} = e^{-x}$

    Clearly a = -1, so:

    $\displaystyle Cxe^{-x} + Cxe^{-x} = e^{-x}$

    $\displaystyle 2Cxe^{-x} = e^{-x}$

    So we want $\displaystyle 2Cx = 1$

    $\displaystyle C =\frac{1}{2x}$

    Substituting into the general solution:
    $\displaystyle y = \frac{1}{2x}xe^{-x}$

    $\displaystyle y = \frac{1}{2}e^{-x}$

    Adding homogeneous and particular solutions:

    $\displaystyle y = Ae^x + Be^{-x} + \frac{1}{2}e^{-x}$

    $\displaystyle y = e^{-x}(B + \frac{1}{2}) + Ae^x$


    However, the solution is:

    $\displaystyle y = (A - \frac{x}{2})e^{-x} + Be^x$

    I'm not sure where that x came from. I'm sure I'm missing something obvious here. Any help would be great!
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  2. #2
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    Your particular solution has too many constants. You can try $\displaystyle \displaystyle y = Cx\,e^{-x}$.
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  3. #3
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    Ahh, thank you. I'll try that after I have some dinner. :P
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