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Expert-verified Found in: Page 131 ### 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 # Building Temperature. In Section 3.3 we modeled the temperature inside a building by the initial value problem (13)$$\frac{{{\bf{dT}}}}{{{\bf{dt}}}}{\bf{ = K}}\,\,\left[ {{\bf{M}}\,{\bf{(t) - T}}\,{\bf{(t)}}} \right]{\bf{ + H}}\,{\bf{(t) + U}}\,{\bf{(t),}}\,\,{\bf{T}}\,{\bf{(}}{{\bf{t}}_{\bf{o}}}{\bf{) = }}{{\bf{T}}_{\bf{o}}}$$ , where M is the temperature outside the building, T is the temperature inside the building, H is the additional heating rate, U is the furnace heating or air conditioner cooling rate, K is a positive constant, and $${{\bf{T}}_{\bf{o}}}$$ is the initial temperature at time $${{\bf{t}}_{\bf{o}}}$$ . In a typical model, $${{\bf{t}}_{\bf{o}}}{\bf{ = 0}}$$ (midnight),$${{\bf{T}}_{\bf{o}}}{\bf{ = 6}}{{\bf{5}}^{\bf{o}}}$$, $${\bf{H}}\left( {\bf{t}} \right){\bf{ = 0}}{\bf{.1}}$$, $${\bf{U(t) = 1}}{\bf{.5}}\left[ {{\bf{70 - T(t)}}} \right]$$ and $${\bf{M(t) = 75 - 20cos}}\frac{{{\bf{\pi t}}}}{{{\bf{12}}}}$$ . The constant K is usually between$$\frac{{\bf{1}}}{{\bf{4}}}{\bf{and}}\frac{{\bf{1}}}{{\bf{2}}}$$, depending on such things as insulation. To study the effect of insulating this building, consider the typical building described above and use the improved Euler’s method subroutine with$${\bf{h = }}\frac{{\bf{2}}}{{\bf{3}}}$$ to approximate the solution to (13) on the interval $$0 \le {\bf{t}} \le 24$$ (1 day) for $${\bf{k = 0}}{\bf{.2,}}\,{\bf{0}}{\bf{.4}}$$, and 0.6.

The temperature at midnight when $${\bf{k = 0}}{\bf{.2}}$$ is approx. $${\bf{68}}{\bf{.385}}$$.

The temperature at midnight when $${\bf{k = }}\,{\bf{0}}{\bf{.4}}$$ is approx. $${\bf{67}}{\bf{.050}}$$.

The temperature at midnight when $${\bf{k = 0}}{\bf{.6}}$$ is approx. $${\bf{65}}{\bf{.974}}$$.

See the step by step solution

## Step 1: Important hint.

To get the result apply Euler’s formula.

## Step 2: Find the value of temperature when $${\bf{K = 0}}{\bf{.2}}$$.

The given equation is

$$\begin{array}{c}\frac{{{\bf{dT}}}}{{{\bf{dt}}}}{\bf{ = K}}\,\,\left[ {{\bf{M}}\,{\bf{(t) - T}}\,{\bf{(t)}}} \right]{\bf{ + H}}\,{\bf{(t) + U}}\,{\bf{(t),}}\,\,{\bf{T}}\,{\bf{(}}{{\bf{t}}_{\bf{o}}}{\bf{) = }}{{\bf{T}}_{\bf{o}}}\\\frac{{{\bf{dT}}}}{{{\bf{dt}}}}{\bf{ = K}}\,\,\left[ {{\bf{75 - 20}}\,{\bf{cos}}\frac{{{\bf{\pi t}}}}{{{\bf{12}}}}{\bf{ - T}}\,{\bf{(t)}}} \right]{\bf{ + 0}}{\bf{.1 + 1}}{\bf{.5(70 - T}}\,{\bf{(t))}}\\\frac{{{\bf{dT}}}}{{{\bf{dt}}}}{\bf{ = 75}}\,{\bf{K + 105}}{\bf{.1 - 20}}\,{\bf{K}}\,{\bf{cos}}\,\frac{{{\bf{\pi t}}}}{{{\bf{12}}}}{\bf{ - (K + 1}}{\bf{.5)}}\,{\bf{T}}\,{\bf{(t)}}\\{\bf{T(0) = 65}}\\\frac{{{\bf{dT}}}}{{{\bf{dt}}}}{\bf{ = 120}}{\bf{.1 - 4}}\,{\bf{cos}}\frac{{{\bf{\pi t}}}}{{{\bf{12}}}}{\bf{ - 1}}{\bf{.7}}\,{\bf{T}}\,{\bf{(t)}}\end{array}$$

Now apply improved Euler’s method subroutine with $${\bf{h = }}\frac{{\bf{2}}}{{\bf{3}}} \approx 0.6667$$ and $${\bf{N = 36}}$$.

$$\begin{array}{c}{\bf{f}}\,{\bf{(t,T) = 120}}{\bf{.1 - 4}}\,{\bf{cos}}\,\,\frac{{{\bf{\pi t}}}}{{{\bf{12}}}}{\bf{ - 1}}{\bf{.7}}\,{\bf{T}}\,{\bf{(t)}}\\{\bf{F = f}}\,{\bf{(t,T)}}\\{\bf{ = 120}}{\bf{.1 - 4}}\,{\bf{cos}}\,\,\frac{{{\bf{\pi t}}}}{{{\bf{12}}}}{\bf{ - 1}}{\bf{.7T}}\,{\bf{(t)}}\\{\bf{G = f}}\,{\bf{(t + h,T + h}}\,{\bf{F)}}\end{array}$$

Apply initial conditions$${\bf{t = }}{{\bf{t}}_{\bf{o}}}{\bf{ = 0,T = }}{{\bf{T}}_{\bf{o}}}{\bf{ = 65}}$$.

$$\begin{array}{c}{\bf{F(0,65) = 5}}{\bf{.6}}\\{\bf{G(0,65) = 0}}{\bf{.6862}}\\{\bf{t = }}{{\bf{t}}_{\bf{0}}}{\bf{ + h = 0}}{\bf{.6667}}\\{\bf{T = }}{{\bf{T}}_{\bf{o}}}{\bf{ + h}}\frac{{{\bf{F + G}}}}{{\bf{2}}}{\bf{ = 66}}{\bf{.638}}\end{array}$$

Therefore at 0.6667h after midnight which is 12.40 AM, the temperature is approx. 66.638.

Now apply the same procedure for the period of 24h.

## Step 3: Get the value for 36 steps.

Since there are 36 steps so by construct a table to get the required result.

 Time $${{\bf{t}}_{\bf{0}}}$$ $${{\bf{T}}_{\bf{o}}}$$ Midnight 0 65 12:40 AM 0.667 66.638 2:00 AM 2 68.073 4:00 AM 4 69.073 6:00 AM 6 70.301 8:00 AM 8 71.484 10:00 AM 10 72.437 12:00 PM 12.001 72.909 2:00 PM 14.001 72.775 4:00 PM 16.001 72.071 6:00 PM 18.001 70.985 8:00 PM 20.001 69.809 10:00 PM 22.001 68.857 MIDNIGHT 24.001 68.385

## Step 4: Determine the value of temperature when $${\bf{K = 0}}{\bf{.4}}$$.

The given equation is

$$\begin{array}{l}{\bf{T(0) = 65}}\\\frac{{{\bf{dT}}}}{{{\bf{dt}}}}{\bf{ = }}135.{\bf{1 - }}8{\bf{cos}}\frac{{{\bf{\pi t}}}}{{{\bf{12}}}}{\bf{ - 1}}.9{\bf{T(t)}}\end{array}$$

Now apply improved Euler’s method subroutine with $${\bf{h = }}\frac{{\bf{2}}}{{\bf{3}}} \approx 0.6667$$ and $${\bf{N = 36}}$$.

$$\begin{array}{c}{\bf{f(t,T) = 135}}{\bf{.1 - 8cos}}\frac{{{\bf{\pi t}}}}{{{\bf{12}}}}{\bf{ - 1}}{\bf{.9T(t)}}\\{\bf{F = f(t,T) = 135}}{\bf{.1 - 8cos}}\frac{{{\bf{\pi t}}}}{{{\bf{12}}}}{\bf{ - 1}}{\bf{.9T(t)}}\\{\bf{G = f(t + h,T + hF)}}\\{\bf{ = 135}}{\bf{.1 - 8cos}}\frac{{{\bf{\pi (t + 0}}{\bf{.6667)}}}}{{{\bf{12}}}}{\bf{ - 1}}{\bf{.9(T + 0}}{\bf{.6667(135}}{\bf{.1 - 8cos}}\frac{{{\bf{\pi t}}}}{{{\bf{12}}}}{\bf{ - 1}}{\bf{.9T))}}\end{array}$$

Apply initial conditions $${\bf{t = }}{{\bf{t}}_{\bf{o}}}{\bf{ = 0,T = }}{{\bf{T}}_{\bf{o}}}{\bf{ = 65}}$$.

$$\begin{array}{c}{\bf{F(0,65) = 3}}{\bf{.6}}\\{\bf{G(0,65) = - 0}}{\bf{.838678}}\\{\bf{t = }}{{\bf{t}}_{\bf{0}}}{\bf{ + h = 0}}{\bf{.6667}}\\{\bf{T = }}{{\bf{T}}_{\bf{o}}}{\bf{ + h}}\frac{{{\bf{F + G}}}}{{\bf{2}}}{\bf{ = 65}}{\bf{.920}}\end{array}$$

Thus, at 0.6667h after midnight which is 12.40 AM, the temperature is approx. 65.920.

Now apply the same procedure for the period of 24h.

## Step 5: Get the value for 36 steps.

Since there are 36 steps so by construct a table to get the required result.

 Time $${{\bf{t}}_{\bf{0}}}$$ $${{\bf{T}}_{\bf{o}}}$$ Midnight 0 65 12:40 AM 0.667 65.92 2:00 AM 2 67.01 4:00 AM 4 68.565 6:00 AM 6 70.561 8:00 AM 8 72.667 10:00 AM 10 74.349 12:00 PM 12.001 75.161 2:00 PM 14.001 74.885 4:00 PM 16.001 73.597 6:00 PM 18.001 71.641 8:00 PM 20.001 69.541 10:00 PM 22.001 67.861 MIDNIGHT 24.001 67.05

## Step 6: Evaluate the value of temperature when $${\bf{K = 0}}{\bf{.6}}$$.

The given equation is at

$$\begin{array}{c}{\bf{T(0) = 65}}\\\frac{{{\bf{dT}}}}{{{\bf{dt}}}}{\bf{ = 150}}{\bf{.1 - 12cos}}\frac{{{\bf{\pi t}}}}{{{\bf{12}}}}{\bf{ - 2}}{\bf{.1T(t)}}\end{array}$$

Now apply improved Euler’s method subroutine with $${\bf{h = }}\frac{{\bf{2}}}{{\bf{3}}} \approx 0.6667$$ and $${\bf{N = 36}}$$.

$$\begin{array}{c}{\bf{f(t,T) = 150}}{\bf{.1 - 12cos}}\frac{{{\bf{\pi t}}}}{{{\bf{12}}}}{\bf{ - 2}}{\bf{.1T(t)}}\\{\bf{F = f(t,T) = 150}}{\bf{.1 - 12cos}}\frac{{{\bf{\pi t}}}}{{{\bf{12}}}}{\bf{ - 2}}{\bf{.1T(t)}}\\{\bf{G = f(t + h,T + hF)}}\\{\bf{ = 150}}{\bf{.1 - 12cos}}\frac{{{\bf{\pi (t + 0}}{\bf{.6667)}}}}{{{\bf{12}}}}{\bf{ - 2}}{\bf{.1(T + 0}}{\bf{.6667(150}}{\bf{.1 - 12cos}}\frac{{{\bf{\pi t}}}}{{{\bf{12}}}}{\bf{ - 2}}{\bf{.1T))}}\end{array}$$

Apply initial conditions $${\bf{t = }}{{\bf{t}}_{\bf{o}}}{\bf{ = 0,T = }}{{\bf{T}}_{\bf{o}}}{\bf{ = 65}}$$.

$$\begin{array}{c}{\bf{F(0,65) = 1}}{\bf{.6}}\\{\bf{G(0,65) = - 0}}{\bf{.1483}}\\{\bf{t = }}{{\bf{t}}_{\bf{0}}}{\bf{ + h = 0}}{\bf{.6667}}\\{\bf{T = }}{{\bf{T}}_{\bf{o}}}{\bf{ + h}}\frac{{{\bf{F + G}}}}{{\bf{2}}}{\bf{ = 65}}{\bf{.381}}\end{array}$$

Hence, at 0.6667h after midnight which is 12.40 AM, the temperature is approx. $${\bf{68}}{\bf{.38}}1$$.

Now apply the same procedure for the period of 24h.

## Step 7: Find the value for 36 steps.

Since there are 36 steps so by construct a table to get the required result.

 Time $${{\bf{t}}_{\bf{0}}}$$ $${{\bf{T}}_{\bf{o}}}$$ Midnight 0 65 12:40 AM 0.667 65.381 2:00 AM 2 66.199 4:00 AM 4 68.13 6:00 AM 6 70.825 8:00 AM 8 73.668 10:00 AM 10 75.919 12:00 PM 12.001 76.978 2:00 PM 14.001 76.563 4:00 PM 16.001 74.784 6:00 PM 18.001 72.119 8:00 PM 20.001 69.282 10:00 PM 22.001 67.032 MIDNIGHT 24.001 65.974

Hence, this is the required result. ### Want to see more solutions like these? 