The differential equation below models the temperature of a 95°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 70°C. Solve the differential equation to find an expression for the temperature of the coffee at time t. dy dt = - 1 51 (y - 19)

y (t) = 19 + 76*e^ ( - (1/51) * t)

Step-by-step explanation:

We have the following differential equation model:

dy / dt = - (1/51) * (y - 19)

The initial conditions are: y (0) = 95 and y ' (70) = 1

We solve the differential equation, separating variables:

1 / (and - 19) * dy = - (1/51) * dt

we integrate on both sides:

ln (y - 19) = - (1/51) * t + C

e ^ ln (y - 19) = e ^ ( - (1/51) * t + C)

y - 19 = C1 * e ^ ( - (1/51) * t)

y = 19 + C1 * e ^ ( - (1/51) * t)

replacing, and (0) = 95

95 = 19 + C1 * e ^ ( - (1/51) * 0)

95 = 19 + C1 * e ^ 0

95 = 19 + C1

C1 = 95 - 19 = 76

Therefore the equation would be:

y (t) = 19 + 76 * e ^ ( - (1/51) * t)