Question

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)

Answer 1

Answer:

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)