Respuesta :
Answer:
[tex]y(t) = 68e^{\frac{-t}{50}}+21[/tex]
Step-by-step explanation:
Consider the differential equation [tex]\frac{dy}{dt}= \frac{-1}{50}(y-21)[/tex]. NOte that this fullfills the condition that the coffee cools down 1°C when y=71°C. We have the initial condition that y(0) = 89°C. Note that this is a separable differential equation, since
[tex]\frac{dy}{(y-21)}= \frac{-dt}{50}[/tex] which leads to
[tex]\int\frac{dy}{(y-21)}= \int \frac{-dt}{50}[/tex]. This gives us the following result
[tex]\text{ln}\left|y-21\right| = \frac{-t}{50}+C[/tex] where C is a constant. Then, using the exponential function we have
[tex]y-21 = e^{\text{ln}(y-21)} = Ae^{\frac{-t}{50}}, where A = e^{C}[/tex] which is another constant.
Given the initial condition, we have that when t=0, y = 89, then
[tex]89-21 = Ae^{\frac{0}{50}} = A = 68[/tex]
Then, the final solution is
[tex]y(t) = 68e^{\frac{-t}{50}}+21[/tex]
