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Thread: y'' + 4y' + 5y = xe^x

  1. #1
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    y'' + 4y' + 5y = xe^x

    Hi I have some problem with the following d.e

    y'' + 4y' + 5y = xe^x.

    I will solve it as much as I can


    Yh

    fives yh = c1 cosx + c2 sinx)*e^-2x


    Yp

    hard to know what to choose, but I looked at an example

    Yp = (A+Bx)e^x (I don't really know why we take it)
    Yp' = (A + B + Bx)e^x
    Yp'' = (A + 2b + Bx)e^x

    Right lets put it into the d.e


    ((A + 2b + Bx)e^x) + 4((A + B + Bx)e^x) + 5((A+Bx)e^x) = x^x

    I dont whats happenening but i dont get the real, constants, I get B = -1/4 etc
    would be nice if someone could correct my solution, and help me

    Thanks
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  2. #2
    Super Member Random Variable's Avatar
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    You make that choice for $\displaystyle y_{p} $ because any derivative of $\displaystyle xe^{x} $ will have that form.

    $\displaystyle y_{p} = (A+Bx)e^{x} $

    $\displaystyle y'_{p} = Be^{x} + (A+Bx)e^{x} = (A+B+Bx)e^{x}$

    $\displaystyle y''_{p} = Be^{x} + (A+B+Bx) e^{x} = (A+2B+Bx)e^{x}$

    So that part was correct.


    $\displaystyle (A+2B+Bx)e^{x} +4(A+B+Bx)e^{x} + 5(A+Bx)e^{x} = xe^{x} $

    $\displaystyle (A+2B+4A+4B+5A)e^{x} + (B+4B+5B)xe^{x} = xe^{x} $

    so $\displaystyle 10A + 6B = 0 $ and $\displaystyle 10B = 1 $

    $\displaystyle B = \frac{1}{10} $

    $\displaystyle A = -\frac{6}{100} = -\frac{3}{50} $
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  3. #3
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    Thank you very much
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