The temperatureT(in units of 100 F) of a university classroom on a cold winter day varies with timet(in hours) as$$\begin{gathered} \frac{\mathbf{dT}}{\mathbf{dt}}\mathbf{=}\left\{\mathbf{1}\mathbf{-}\mathbf{T}\mathbf{,}\mathbf{if}\mathbf{heating}\mathbf{units}\text{is}\mathbf{On} \\ \mathbf{-}\mathbf{T}\mathbf{,}\mathbf{if}\mathbf{heating}\mathbf{units}\mathbf{is}\mathbf{OFF}\mathbf{.} \\ \right. \end{gathered}$$

$\mathbf{T}\mathbf{=}\mathbf{0}\mathbf{}$Suppose at 9:00 a.m., the heating unit is ON from 9-10 a.m., OFF from 10-11 a.m., ON again from 11 a.m.–noon, and so on for the rest of the day. How warm will the classroom be at noon? At 5:00 p.m.?

Here heat is on at time 9 a.m, 11a.m and 1 p.m. ${\mathbf{T}}_{\mathbf{0}}$ is the temperature at room.

The differential equation when heat is on.

$\frac{\mathbf{dT}}{\mathbf{dt}}\mathbf{+}\mathbf{T}\mathbf{=}\mathbf{1}$

The integrating factor is ${\mathbf{e}}^{\mathbf{t}}$.then

$$\begin{gathered} \mathbf{T}\mathbf{.}{\mathbf{e}}^{\mathbf{t}}\mathbf{=}\int {\mathbf{e}}^{\mathbf{t}}\mathbf{dt} \\ \mathbf{T}\mathbf{=}\mathbf{1}\mathbf{+}{\mathbf{ce}}^{\mathbf{-}\mathbf{t}} \end{gathered}$$

Apply the initial conditions then

$\mathbf{T}\mathbf{=}\mathbf{1}\mathbf{+}\mathbf{(}{\mathbf{T}}_{\mathbf{o}}\mathbf{-}\mathbf{1}\mathbf{)}{\mathbf{e}}^{\mathbf{-}\mathbf{(}\mathbf{t}\mathbf{-}{\mathbf{t}}_{\mathbf{o}}\mathbf{)}}$

The differential equation when heat is off at time 10 a.m 12 noon and so on.

$\frac{\mathbf{dT}}{\mathbf{dt}}\mathbf{+}\mathbf{T}\mathbf{=}\mathbf{0}$

The integrating factor is${\mathbf{e}}^{\mathbf{t}}$ . And apply the initial conditions then

$\mathbf{T}\mathbf{=}{\mathbf{T}}_{\mathbf{o}}{\mathbf{e}}^{\mathbf{-}\mathbf{(}\mathbf{t}\mathbf{-}{\mathbf{t}}_{\mathbf{0}}\mathbf{)}}$

Let now$\mathbf{t}\mathbf{=}\mathbf{0}$corresponding to 9 a.m. The temperature is 0 degree. And the heat is turned off.

Then$\mathbf{T}\mathbf{=}\mathbf{1}\mathbf{-}{\mathbf{e}}^{\mathbf{-}\mathbf{t}}$.

Now at 10 a.m ,$\mathbf{t}\mathbf{=}\mathbf{1}$then$\mathbf{T}\mathbf{=}\mathbf{(}\mathbf{1}\mathbf{-}{\mathbf{e}}^{\mathbf{-}\mathbf{t}}\mathbf{)} {\mathbf{e}}^{\mathbf{-}\mathbf{(}\mathbf{t}\mathbf{-}\mathbf{1}\mathbf{)}}$

Now at 11 a.m,$\mathbf{t}\mathbf{=}\mathbf{2}$then$\mathbf{T}\mathbf{=}\mathbf{1}\mathbf{+}\mathbf{\left(}\mathbf{\right(}\mathbf{1}\mathbf{-}{\mathbf{e}}^{\mathbf{-}\mathbf{1}}\mathbf{)} {\mathbf{e}}^{\mathbf{-}\mathbf{1}}\mathbf{-}\mathbf{1}\mathbf{)} {\mathbf{e}}^{\mathbf{-}\mathbf{(}\mathbf{t}\mathbf{-}\mathbf{2}\mathbf{)}}$

Proceeding like this for heat is on then$\mathbf{T}\mathbf{(}{\mathbf{t}}_{\mathbf{1}}\mathbf{)}\mathbf{=}\sum _{\mathbf{n}\mathbf{=}\mathbf{1}}^{{\mathbf{t}}_{\mathbf{1}}}{\mathbf{(}\mathbf{-}\mathbf{1}\mathbf{)}}^{\mathbf{n}\mathbf{+}\mathbf{1}}{\mathbf{e}}^{\mathbf{-}\mathbf{n}}$.

And proceeding for heat Is off then $\mathbf{T} \mathbf{(}{\mathbf{t}}_{\mathbf{2}}\mathbf{)}\mathbf{=}\sum _{\mathbf{n}\mathbf{=}\mathbf{1}}^{{\mathbf{t}}_{\mathbf{2}}}{\mathbf{(}\mathbf{-}\mathbf{1}\mathbf{)}}^{\mathbf{n}}{\mathbf{e}}^{\mathbf{-}\mathbf{n}}$

At noon$\mathbf{t}\mathbf{=}\mathbf{3}$then$$\begin{gathered} \mathbf{T}\mathbf{\left(}\mathbf{3}\mathbf{\right)}\mathbf{=}\sum _{\mathbf{n}\mathbf{=}\mathbf{1}}^{\mathbf{3}}{\mathbf{(}\mathbf{-}\mathbf{1}\mathbf{)}}^{\mathbf{n}}{\mathbf{e}}^{\mathbf{-}\mathbf{n}} \\ \mathbf{=}\mathbf{1}\mathbf{-}{\mathbf{e}}^{\mathbf{-}\mathbf{1}}\mathbf{+}{\mathbf{e}}^{\mathbf{-}\mathbf{2}}\mathbf{-}{\mathbf{e}}^{\mathbf{-}\mathbf{3}} \\ \mathbf{=}\mathbf{0}\mathbf{.}\mathbf{718} \end{gathered}$$

Thus, the temperature at this time is 71.8-degree Fahrenheit.

Similarly at 5pmthen $\mathbf{T}\mathbf{\left(}\mathbf{8}\mathbf{\right)}\mathbf{=}\mathbf{26}\mathbf{.}\mathbf{9}$

Thus, the temperature at this time is 26.9-degree Fahrenheit.

Thereforethe temperature is 71.8-degree Fahrenheit at 12 noon and 26.9-degree Fahrenheit at 5pm and so on.