Differential Equation Problem?

Math Word Problem (differential equation)?

• The differential equation below models the temperature of a 95°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 95°C.) dy/dt = -(y-17)/50 I have absolutely no idea where to start from this lol, Thnx.

• Answer:

  dy/dt = - (y-17)/50 dy/(y-17) = -dt/50 integrate both sides ln ly-17l = -t/50 + C y - 17 = e^(-t/50 + C) y - 17 = e^C e^(-t/50) since e^C is an arbitrary constant, let e^C = k, where k is also an arbitrary constant. y - 17 = k e^(-t/50) given at t = 0, the temperature is 95 degrees C 95 - 17 = k e^(-0/50) y - 17 = 78 e^(-t/50) y = 78 e^(-t/50) + 17 the equation above tells the temperature of the coffee at any time.