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Thread: Second order linear ODE

  1. #1
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    Second order linear ODE

    Just wondering how would you find the particular solution to the following differential equation. Is there a quick technique or way of doing this?

    $\displaystyle 2y''-3y'-5y=(x+4)e^{5x/2}$
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  2. #2
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    Inspired guesswork.

    Note that the right hand side can be expanded as $\displaystyle x\,e^{\frac{5x}{2}} + 4e^{\frac{5x}{2}}$.


    So an appropriate particular solution could be $\displaystyle y_p = Ax\,e^{\frac{5x}{2}} + Be^{\frac{5x}{2}}$.

    This means $\displaystyle y_p' = Ae^{\frac{5x}{2}} + \frac{5}{2}Ax\,e^{\frac{5x}{2}} + \frac{5}{2}Be^{\frac{5x}{2}}$

    $\displaystyle = \left(A + \frac{5}{2}B\right)e^{\frac{5x}{2}} + \frac{5}{2}Ax\,e^{\frac{5x}{2}}$


    and $\displaystyle y_p'' = \frac{5}{2}\left(A + \frac{5}{2}B\right)e^{\frac{5x}{2}} + \frac{5}{2}Ae^{\frac{5x}{2}} + \frac{25}{4}Ax\,e^{\frac{5x}{2}}$

    $\displaystyle = \left(5A + \frac{25}{4}B\right)e^{\frac{5x}{2}} + \frac{25}{4}Ax\,e^{\frac{5x}{2}}$.


    Substituting into the DE gives...

    $\displaystyle 2\left[\left(5A + \frac{25}{4}B\right)e^{\frac{5x}{2}} + \frac{25}{4}Ax\,e^{\frac{5x}{2}}\right] - 3\left[\left(A + \frac{5}{2}B\right)e^{\frac{5x}{2}} + \frac{5}{2}Ax\,e^{\frac{5x}{2}}\right] - 5\left(Ax\,e^{\frac{5x}{2}} + Be^{\frac{5x}{2}}\right) = x\,e^{\frac{5x}{2}} + 4e^{\frac{5x}{2}}$

    $\displaystyle \left(10A + \frac{25}{2}B\right)e^{\frac{5x}{2}} + \frac{25}{2}Ax\,e^{\frac{5x}{2}} + \left(-3A - \frac{15}{2}B\right)e^{\frac{5x}{2}} - \frac{15}{2}Ax\,e^{\frac{5x}{2}} - 5Ax\,e^{\frac{5x}{2}} - 5Be^{\frac{5x}{2}} = x\,e^{\frac{5x}{2}} + 4e^{\frac{5x}{2}}$

    $\displaystyle 5Ax\,e^{\frac{5x}{2}} + \left(7A + 5B\right)e^{\frac{5x}{2}} = x\,e^{\frac{5x}{2}} + 4e^{\frac{5x}{2}}$.


    Therefore $\displaystyle 5A = 1$ and $\displaystyle 7A + 5B = 4$.

    This gives $\displaystyle A = \frac{1}{5}$

    and $\displaystyle \frac{7}{5} + 5B = 4$

    $\displaystyle 5B = \frac{13}{5}$

    $\displaystyle B = \frac{13}{25}$.


    So finally, we have our particular solution

    $\displaystyle y_p = \frac{1}{5}x\,e^{\frac{5x}{2}} + \frac{13}{25}e^{\frac{5x}{2}}$.
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