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Math Help - Solving a fifth order differential equation?

  1. #1
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    Question Solving a fifth order differential equation?

    Solve: y'''''-3y''''-y'''+23y''-64y'+60y=x^4 + e^2x + e^3x + sinx; when x=0, y=a, y'=a2, y''=a3, y'''=a4, y''''=a5


    I wanted to approach this the way I would a second order, ie. y= yc+yp. would that be the way to go? if so, how would i factor the left side? thanks in advance
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  2. #2
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    Since they're constant coefficient, I assume that process should be fine.

    Assume a solution of the form y = e^{rx}, then for the solution to the homogeneous equation, you would have

    r^5e^{rx} - 3r^4e^{rx} - r^3e^{rx} + 23r^2e^{rx} - 64r\,e^{rx} + 60e^{rx} = 0

    r^5 - 3r^4 - r^3 + 23r^2 - 64r + 60 = 0.


    Now use the factor theorem and long division to factorise this, giving

    (r - 2)^2(r + 3)(r^2 - 2r + 5) = 0.

    Therefore r = 2 (multiplicity 2), r = -3, r = 1 - 2i, 1 + 2i.


    So the homogeneous solution is

    y_c = Ae^{2x} + Bx\,e^{2x} + Ce^{-3x} + De^{x}(E\cos{2x} + F\sin{2x}).
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  3. #3
    A Plied Mathematician
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    Try the factors of 60. Use synthetic division to check.

    [EDIT]: ProveIt beat me to it.
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  4. #4
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    Thanks a lot you guys! This was a huge help!
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