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29 August, 17:36

If y = e^5t is a solution to the differential equation d^2 y/dt^2 - 13 dy/dt + ky = 0, find the value of the constant k and the general form y = Ae^5t + Be^at of the solution to the above equation, (i. e. find a). (Use constants A, B, etc., for any constants in your solution formula.)

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  1. 29 August, 19:38
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    k = - 12/5

    A = 125/12

    B = - 325/12

    a = 5

    Step-by-step explanation:

    y = e^5t

    Dy/dt = 5e^5t

    d2y/dt2 = 25e^5t

    Inputting the values of dy/dt and d2y/dt2 into the equation above, we have:

    25e^5t - 13e^5t + 5k (e^5t) = 0

    12e^5t + 5k (e^5t) = 0

    e^5t (12 + 5k) = 0

    12 + 5k = 0

    k = - 12/5

    The equation becomes,

    d2y/dt2 - 13dy/dt - 12/5y = o

    So rearranging the equation,

    -5/12d2y/dt2 + 65/12dy/dt + y = 0

    y = 5/12 (25e^5t) - 65/12 (5e^5t)

    y = 125/12e^5t - 325/12e^5t

    Therefore,

    k = - 12/5

    A = 125/12

    B = - 325/12

    a = 5
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