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.

To get the result apply Euler’s formula.

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.

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

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.

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

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.

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

Hence, this is the required result.