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Math Help - 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?

    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 x\,e^{\frac{5x}{2}} + 4e^{\frac{5x}{2}}.


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

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

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


    and 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}}

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


    Substituting into the DE gives...

    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}}

    \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}}

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


    Therefore 5A = 1 and 7A + 5B = 4.

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

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

    5B = \frac{13}{5}

    B = \frac{13}{25}.


    So finally, we have our particular solution

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