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4 June, 21:14

The differential equation below models the temperature of an 87°C cup of coffee in a 17°C room, where it is known that the coffee cools at a rate of 1°C per minute when its temperature is 67°C. Solve the differential equation to find an expression for the temperature of the coffee at time t. (Let y be the temperature of the cup of coffee in °C, and let t be the time in minutes, with t = 0 corresponding to the time when the temperature was 87°C.) dy dt = - 1 50 (y - 17)

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  1. 5 June, 00:43
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    y (t) = 17+70e^ (-t/50)

    Step-by-step explanation: The model of equation given is:

    dy/dt = - 1/50 (y-17)

    Integrate both sides

    dy / (y-17) = - 1/50 dt

    This lead to

    ln (y-17) = - 1/50 t + C

    Log both sides:

    y = c e^ (-t/50) + 17 ... (1)

    and let t be the time in minutes, with t = 0 corresponding to the time when the temperature was 87°C.

    87 = c e^0 + 17

    e^0 = 1

    Therefore c = 87-17 = 70

    y (0) = 87, so c=70

    Substitute c into equation 1. This lead to:

    y (t) = 17+70e^ (-t/50)
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