The differential equation below models the temperature of a 95°C cup of coffee in a 21°C room, where it is known that the coffee cools at a rate of 1°C per minute when its temperature is 71°C. How to solve the differential equation?

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Answer 1 · 2017-12-02T13:52:17

See below.

This is a separable differential equation so it can be arranged as

$\frac{d y}{y - 21} = - \frac{1}{50} d t$ $(dy)/(y-21) = -1/50 dt$

Now, integrating each side

${log}_{e} | y - 21 | = - \frac{1}{50} t + C_{0}$ $log_eabs(y-21) = -1/50 t + C_0$ or

$y - 21 = C_{1} e^{- \frac{t}{50}}$ $y-21 = C_1 e^(-t/50)$

Now, at $t = 0$ $t = 0$ the temperature is $95^{\circ}$ $95^@$ or

$95^{\circ} - 21^{\circ} = C_{1} e^{0} \Rightarrow C_{1} = 74^{\circ}$ $95^@-21^@ = C_1 e^0rArr C_1 = 74^@$ and finally

$y = 21^{\circ} + 74^{\circ} e^{- \frac{t}{50}}$ $y = 21^@ +74^@e^(-t/50)$