Wai Fung 發問於 科學及數學數學 · 1 十年前

differential equation

y''+3y'+2y=10e^3x+4x^2

Find the original equation of "y"?

1 個解答

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  • 1 十年前
    最愛解答

    A particular solution is

    y = Ae^3x + Bx^2 + Cx + D

    y' = 3Ae^3x + 2Bx + C

    y'' = 9Ae^3x + 2B

    y''+3y'+2y

    =20Ae^3x + 2Bx^2 + (6B + 2C)x + 2B + 3C + 2D

    By comparing coefficient, we have

    A = 0.5,

    B =2,

    C= -6,

    D= 7

    Therefore, the particular equation is

    0.5 e^3x + 2x^2 - 6x + 7

    2010-11-15 15:39:00 補充:

    For the general equation,

    The characteristic equation is

    r^2 + 3r + 2 =0

    r = -1, -2

    which is Ae^-x + B e^-2x

    The general equation is therefore

    Ae^-x + B e^-2x + 0.5 e^3x + 2x^2 - 6x + 7

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