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Math Help - 2OLE (Simple)

  1. #1
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    2OLE (Simple)

    y'' - 2 y' + 1 y = -18 e^{1 t}

    Find a particular solution to this equation:

    How exactly am I supposed to answer this?

    My homologous equation renders repeated roots:

    [Math] r_1=e^x, r_2=xe^x[/tex]

    And when solve this with undetermined coefficients, I get:
    <br />
Y_p=Ae^t<br />

    but since that e^x and xe^x already exist in the homologous solution, I should get:

    <br />
Y_p=\alpha T^2e^t

    Is this the correct answer? I cant solve for alpha, because I have no initial values missing. My assignment wont accept this format. What am I missing?
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Vamz View Post
    y'' - 2 y' + 1 y = -18 e^{1 t}

    Find a particular solution to this equation:

    How exactly am I supposed to answer this?

    My homologous equation renders repeated roots:

    [Math] r_1=e^x, r_2=xe^x[/tex]
    Shouldn't this be
    y = Ae^t + Bte^t

    Quote Originally Posted by Vamz View Post
    And when solve this with undetermined coefficients, I get:
    <br />
Y_p=Ae^t<br />

    but since that e^x and xe^x already exist in the homologous solution, I should get:

    <br />
Y_p=\alpha T^2e^t

    Is this the correct answer? I cant solve for alpha, because I have no initial values missing. My assignment wont accept this format. What am I missing?
    T is not the same variable as t. So your particular solution should be y_p = \alpha t^2e^t. Put your particular solution into the differential equation. I get \alpha = -9. So your overall solution will be
    y(t) = Ae^t + Bte^t - 9t^2 e^t.

    -Dan
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