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18 December, 07:13

Substitute yequalse Superscript rx into the given differential equation to determine all values of the constant r for which yequalse Superscript rx is a solution of the equation. y double prime plus 4 y prime minus 12 y equals 0

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  1. 18 December, 07:35
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    y (x) = C_1·e^{2x} + C_2·e^{-6x}

    Step-by-step explanation:

    From Exercise we have the differential equation

    y''+4y'-12=0.

    This is a characteristic differential equation and we are solved as follows:

    y''+4y'-12=0

    m²+4m-12=0

    m_{1,2}=/frac{-4±/sqrt{16+48}}{2}

    m_{1,2}=/frac{-4±/sqrt{64}}{2}

    m_{1,2}=/frac{-4±8}{2}

    m_1=2

    m_2=-6

    The general solution of this differential equation is in the form

    y (x) = C_1·e^{m_1 ·x} + C_2·e^{m_2 ·x}

    Therefore, we get

    y (x) = C_1·e^{2x} + C_2·e^{-6x}
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