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Expert-verified Found in: Page 1 ### Fundamentals Of Differential Equations And Boundary Value Problems

Book edition 9th
Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider
Pages 616 pages
ISBN 9780321977069 # The temperature T (in units of 100 F) of a university classroom on a cold winter day varies with time t (in hours) as $\frac{\mathbf{dT}}{\mathbf{dt}}{\mathbf{=}}\left\{\mathbf{1}\mathbf{-}\mathbf{T}\mathbf{,}\mathbf{if}\mathbf{heating}\mathbf{units}\text{is}\mathbf{On}\phantom{\rule{0ex}{0ex}}\mathbf{-}\mathbf{T}\mathbf{,}\mathbf{if}\mathbf{heating}\mathbf{units}\mathbf{is}\mathbf{OFF}\mathbf{.}\phantom{\rule{0ex}{0ex}}\right\$${\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.?

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

See the step by step solution

## Step 1: Find the solution when heat is on

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

$\mathbf{T}\mathbf{.}{\mathbf{e}}^{\mathbf{t}}\mathbf{=}\int {\mathbf{e}}^{\mathbf{t}}\mathbf{dt}\phantom{\rule{0ex}{0ex}}\mathbf{T}\mathbf{=}\mathbf{1}\mathbf{+}{\mathbf{ce}}^{\mathbf{-}\mathbf{t}}\phantom{\rule{0ex}{0ex}}$

Apply the initial conditions then

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

## Step 2: Determine the solution when heat is off

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

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

## Step 3: Find temperature

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 $\mathbf{T}\mathbf{=}\mathbf{\left(}\mathbf{1}\mathbf{-}{\mathbf{e}}^{\mathbf{-}\mathbf{t}}\mathbf{\right)} {\mathbf{e}}^{\mathbf{-}\mathbf{\left(}\mathbf{t}\mathbf{-}\mathbf{1}\mathbf{\right)}}$

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

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

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

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

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

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

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

Therefore the temperature is 71.8-degree Fahrenheit at 12 noon and 26.9-degree Fahrenheit at 5pm and so on. ### Want to see more solutions like these? 