The differential equation below models the temperature of an 87°C cup of coffee in a 19°C room, where it is known that the coffee cools at a rate of 1°C per minute when its temperature is 69°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 − 19)
dy/dt= − 1/50(y − 19)
dy/(y-19) = -1/50 dt
ln(y-19) = -1/50 t + C
y-19 = ce^(-t/50)
y = 19 + ce^(-t/50)