Question

The differential equation below models the temperature of a 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

just separate the variables as usual

dy/dt = -1/50 (y-17)
dy/(y-17) = -1/50 dt
ln(y-17) = -1/50 t + C
y = c e^(-t/50) + 17
y(0) = 87, so c=70

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

Answer 2

I also have a little variation of this question but why do you take the ln(y-17)?

Answer 3

You take the ln because you integrate it and have log base e