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Thread: Solve the IVP

  1. #1
    Junior Member
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    Solve the IVP

    Solve the IVP:
    4y''(x)-4y'(x)-3y(x)=0
    y'(0)=5
    y(0)=1

    So far I have y = (-1/2) and y=(3/2)
    y(x) = (c1e^(3/2)(x))+(c^2e^(-1/2)(x))

    c1= -1
    c2 = 3
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  2. #2
    MHF Contributor
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    Re: Solve the IVP

    $4y^{\prime \prime} - 4 y^\prime - 3y = 0$

    characteristic equation is

    $4s^2 - 4s -3 = 0$

    with roots $s=-\dfrac 1 2,~\dfrac 3 2$

    and thus solution

    $y(x) = c_1 e^{-x/2}+c_2 e^{3x/2}$

    $y^\prime(x) = -\dfrac {c_1}{2} e^{-x/2} + \dfrac {3c_2}{2} e^{3x/2}$

    $y^\prime(0) = -\dfrac{c_1}{2} + \dfrac{3c_2}{2} = 5$

    $y(0) = c_1 + c_2 = 1$

    solving we get

    $c_1 = -\dfrac 7 4, ~ c_2 = \dfrac {11}{4}$

    $y(x) = -\dfrac 7 4 e^{-x/2} + \dfrac {11}{4} e^{3x/2}$
    Thanks from topsquark and HallsofIvy
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