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Math Help - Variation of Parameters

  1. #1
    sc8nt4u
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    Variation of Parameters

    I must use Variation of Parameters to solve the following eqn:

    dy/dt − 2ty = t

    I don't know where to move the -2t so that I can solve for the homogeneous solution to find y1 and y2.
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by sc8nt4u View Post
    I must use Variation of Parameters to solve the following eqn:

    dy/dt − 2ty = t

    I don't know where to move the -2t so that I can solve for the homogeneous solution to find y1 and y2.
    For starters, since this is a first order differential equation, there is only one homogeneous solution, not two. I'm not sure you can do a variation of parameters on a first order equation (at least, my text doesn't discuss it.)

    -Dan
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  3. #3
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    Hello, sc8nt4u!

    I agree with Dan; Variation of Parameters doesn't apply here.


    I must use Variation of Parameters to solve: . \frac{dy}{dt} - 2ty \:=\:t


    Integrating factor: . I \:=\:e^{\int(-2t)dt} \:=\:e^{-t^2}

    We have: . e^{-t^2}\frac{dy}{d1} -2te^{-t^2}y \:=\:te^{-t^2}

    . . which is: . \frac{d}{dt}\left(e^{-t^2}y\right) \:=\:te^{-t^2}

    Integrate: . e^{-t^2}y \;= \;-\frac{1}{2}e^{-t^2} + C

    Multiply by e^{t^2}\!:\;\;y \;= \;Ce^{t^2} - \frac{1}{2}

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