Question

The differential equation below models the temperature of an 88°C cup of coffee in a 24°C room, where it is known that the coffee cools at a rate of 1°C per minute when its temperature is 74°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 88°C.) dy dt = − 1 50 (y − 24)

Answer 1

Answer:

$y = 24+64\cdot e^{-150\cdot t}$

Explanation:

Let solve the differential equation by separating corresponding variables:

$\int\limits^t_0\, dt = -\frac{1}{150} \int\limits^y_{y_{o}} \frac{dy}{y-24}$

The solution of this equation is:

$t = -\frac{1}{150}\cdot (\ln|y-24|-\ln |y_{o}-24|)$

The explicit form of the temperature as a function of time is:

$\ln |y-24|=-150\cdot t + \ln |y_{o}-24|$

$y-24 = C\cdot e^{-150\cdot t}$

The value of the integration constant is:

$C = 64$

The complete expression is:

$y = 24+64\cdot e^{-150\cdot t}$