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15 November, 00:40

the resquest of the y'+6y=e^4t, y (0) = 2

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  1. 15 November, 03:45
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    This is a linear differential equation of first order. Solve this by integrating the coefficient of the y term and then raising e to the integrated coefficient to find the integrating factor, i. e. the integrating factor for this problem is e^ (6x).

    Multiplying both sides of the equation by the integrating factor:

    (y') e^ (6x) + 6ye^ (6x) = e^ (12x)

    The left side is the derivative of ye^ (6x), hence

    d/dx[ye^ (6x) ] = e^ (12x)

    Integrating

    ye^ (6x) = (1/12) e^ (12x) + c where c is a constant

    y = (1/12) e^ (6x) + ce^ (-6x)

    Use the initial condition y (0) = - 8 to find c:

    -8 = (1/12) + c

    c=-97/12

    Hence

    y = (1/12) e^ (6x) - (97/12) e^ (-6x)
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