A building consists of two zones A and B (see Figure 5.5). Only zone A is heated by a furnace, which generates 80,000 Btu/hr. The heat capacity of zone A is per thousand Btu. The time constant for heat transfer between zone A and the outside is 4 hr, between the unheated zone B and the outside is 5 hr, and between the two zones is 2 hr. If the outside temperature stays at , how cold does it eventually get in the unheated zone B?

Elimination Procedure for 2 × 2 Systems:

To find a general solution for the system:

$$\begin{gathered} {\mathbf{L}}_{\mathbf{1}}\left[\mathbf{x}\right]\mathbf{+}{\mathbf{L}}_{\mathbf{2}}\left[\mathbf{y}\right]\mathbf{=}{\mathbf{f}}_{\mathbf{1}}\mathbf{,} \\ {\mathbf{L}}_{\mathbf{3}}\left[\mathbf{x}\right]\mathbf{+}{\mathbf{L}}_{\mathbf{4}}\left[\mathbf{y}\right]\mathbf{=}{\mathbf{f}}_{\mathbf{2}}\mathbf{,} \end{gathered}$$

Where${\mathbf{L}}_{\mathbf{1}}\mathbf{,}{\mathbf{L}}_{\mathbf{2}}\mathbf{,}{\mathbf{L}}_{\mathbf{3}}\mathbf{,}$ and${\mathbf{L}}_{\mathbf{4}}$ are polynomials in $\mathbf{D}\mathbf{=}\frac{\mathbf{d}}{\mathbf{dt}}$:

Vieta’s formulas for finding roots:

For function y(t) to be bounded when$\mathbf{t}\to \mathbf{+}\infty$ we need for both roots of the auxiliary equation to be non-positive if they are reals and, if they are complex, then the real part has to be non-positive. In other words,

1. If ${\mathbf{r}}_{\mathbf{1}}\mathbf{,}{\mathbf{r}}_{\mathbf{2}}\in \mathbf{R}$, then ${\mathbf{r}}_{\mathbf{1}}·{\mathbf{r}}_{\mathbf{2}}⩾\mathbf{0}\mathbf{,}{\mathbf{r}}_{\mathbf{1}}\mathbf{+}{\mathbf{r}}_{\mathbf{2}}⩽\mathbf{0}$,
2. If ${\mathbf{r}}_{\mathbf{1}}\mathbf{,}{\mathbf{r}}_{\mathbf{2}}\mathbf{=}\mathbf{\alpha }\pm \mathbf{\beta i}$, $\mathbf{\beta }\ne 0$, then $\mathbf{\alpha }\mathbf{=}\frac{{\mathbf{r}}_{\mathbf{1}}\mathbf{+}{\mathbf{r}}_{\mathbf{2}}}{\mathbf{2}}⩽0$.

Section 3.3: Heat transfer, it is modeled by the following equation$\frac{\mathbf{dT}}{\mathbf{dt}}\mathbf{=}\mathbf{K}\left(\mathbf{M}\left(\mathbf{t}\right)\mathbf{-}\mathbf{T}\left(\mathbf{t}\right)\right)\mathbf{+}\mathbf{H}\left(\mathbf{t}\right)\mathbf{+}\mathbf{U}\left(\mathbf{t}\right)$ where$\frac{\mathbf{1}}{\mathbf{K}}$ is the time constant for the building given in hours, M(t) is the outside temperature, T(t) is the inside temperature, H(t) is the heating in the building, U(t) is the cooling.

Given that, only zone A is heated by a furnace, which generates 80,000 Btu/hr.

The heat capacity of zone A is$\frac{\mathbf{1}}{\mathbf{4}}{}^{\mathbf{\circ }}\mathbf{F}$per thousand Btu.

The time constant for heat transfer between zone A and the outside is 4 hours, between the unheated zone B and the outside is 5 hours, and between the two zones is 2 hours.

Let x(t) be denoted as the temperature in A at time t and y(t) denoted as the temperature in B at time t.

Using the given information create the system of equation.

Then,$\frac{\mathbf{dx}}{\mathbf{dt}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\left(\mathbf{y}\left(\mathbf{t}\right)\mathbf{-}\mathbf{x}\left(\mathbf{t}\right)\right)\mathbf{-}\frac{\mathbf{x}\left(\mathbf{t}\right)}{\mathbf{4}}\mathbf{+}\mathbf{20}$and$\frac{\mathbf{dy}}{\mathbf{dt}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\left(\mathbf{x}\left(\mathbf{t}\right)\mathbf{-}\mathbf{y}\left(\mathbf{t}\right)\right)\mathbf{-}\frac{\mathbf{y}\left(\mathbf{t}\right)}{\mathbf{5}}$

The above equations can be rewritten as,

$$\begin{gathered} \frac{\mathbf{dx}}{\mathbf{dt}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\left(\mathbf{y}\left(\mathbf{t}\right)\mathbf{-}\mathbf{x}\left(\mathbf{t}\right)\right)\mathbf{-}\frac{\mathbf{x}\left(\mathbf{t}\right)}{\mathbf{4}}\mathbf{+}\mathbf{20} \\ \mathbf{4}\frac{\mathbf{dx}}{\mathbf{dt}}\mathbf{=}\mathbf{2}\left(\mathbf{y}\left(\mathbf{t}\right)\mathbf{-}\mathbf{x}\left(\mathbf{t}\right)\right)\mathbf{-}\mathbf{x}\left(\mathbf{t}\right)\mathbf{+}\mathbf{80} \\ \mathbf{=}\mathbf{2}\mathbf{y}\left(\mathbf{t}\right)\mathbf{-}\mathbf{2}\mathbf{x}\left(\mathbf{t}\right)\mathbf{+}\mathbf{80}\mathbf{-}\mathbf{x}\left(\mathbf{t}\right) \\ \mathbf{=}\mathbf{2}\mathbf{y}\left(\mathbf{t}\right)\mathbf{-}\mathbf{3}\mathbf{x}\left(\mathbf{t}\right)\mathbf{+}\mathbf{80} \\ \mathbf{4}\frac{\mathbf{dx}}{\mathbf{dt}}\mathbf{+}\mathbf{3}\mathbf{x}\left(\mathbf{t}\right)\mathbf{-}2\mathbf{y}\left(\mathbf{t}\right)\mathbf{=}8\mathbf{0} \dots \left(1\right) \end{gathered}$$

$$\begin{gathered} \frac{\mathbf{dy}}{\mathbf{dt}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\left(\mathbf{x}\left(\mathbf{t}\right)\mathbf{-}\mathbf{y}\left(\mathbf{t}\right)\right)\mathbf{-}\frac{\mathbf{y}\left(\mathbf{t}\right)}{\mathbf{5}} \\ 10\frac{\mathbf{dy}}{\mathbf{dt}}\mathbf{=}5\left(\mathbf{x}\left(\mathbf{t}\right)\mathbf{-}\mathbf{y}\left(\mathbf{t}\right)\right)\mathbf{-}2\mathbf{y}\left(\mathbf{t}\right) \\ \mathbf{=}5\mathbf{x}\left(\mathbf{t}\right)\mathbf{-}5\mathbf{y}\left(\mathbf{t}\right)\mathbf{-}2\mathbf{y}\left(\mathbf{t}\right) \\ \mathbf{=}5\mathbf{x}\left(\mathbf{t}\right)\mathbf{-}7\mathbf{y}\left(\mathbf{t}\right) \\ 10\frac{\mathbf{dy}}{\mathbf{dt}}\mathbf{-}5\mathbf{x}\left(\mathbf{t}\right)\mathbf{+}7\mathbf{y}\left(\mathbf{t}\right)\mathbf{=}0 \dots \left(2\right) \end{gathered}$$

Rewrite the system in operator form:

$\left(\mathbf{4}\mathbf{D}\mathbf{+}\mathbf{3}\right)\left[\mathbf{x}\right]\mathbf{-}2\mathbf{y}\mathbf{=}80$ …… (3)

$\mathbf{-}5\mathbf{x}\mathbf{+}\left(10\mathbf{D}\mathbf{+}7\right)\left[\mathbf{y}\right]\mathbf{=}0$…… (4)

Multiply 5 on equation (3) and multiply 4D+3 on equation (4). Then, add them together to get.

$$\begin{gathered} \mathbf{5}(\mathbf{4}\mathbf{D}\mathbf{+}\mathbf{3})\left[\mathbf{x}\right]\mathbf{-}\mathbf{10}\mathbf{y}\mathbf{-}\mathbf{5}(\mathbf{4}\mathbf{D}\mathbf{+}\mathbf{3})\mathbf{x}\mathbf{+}(\mathbf{4}\mathbf{D}\mathbf{+}\mathbf{3})(\mathbf{10}\mathbf{D}\mathbf{+}\mathbf{7})\left[\mathbf{y}\right]\mathbf{=}\mathbf{400} \\ (\mathbf{4}\mathbf{D}\mathbf{+}\mathbf{3})(\mathbf{10}\mathbf{D}\mathbf{+}\mathbf{7})\left[\mathbf{y}\right]\mathbf{-}\mathbf{10}\mathbf{y}\mathbf{=}\mathbf{400} \\ (\mathbf{40}{\mathbf{D}}^{\mathbf{2}}\mathbf{+}\mathbf{58}\mathbf{D}\mathbf{+}\mathbf{21}\mathbf{-}\mathbf{10})\left[\mathbf{y}\right]\mathbf{=}\mathbf{400} \\ (\mathbf{40}{\mathbf{D}}^{\mathbf{2}}\mathbf{+}\mathbf{58}\mathbf{D}\mathbf{+}\mathbf{11})\left[\mathbf{y}\right]\mathbf{=}\mathbf{400} \\ (\mathbf{40}{\mathbf{D}}^{\mathbf{2}}\mathbf{+}\mathbf{58}\mathbf{D}\mathbf{+}\mathbf{11})\left[\mathbf{y}\right]\mathbf{=}\mathbf{400} \dots \left(5\right) \end{gathered}$$

Since the auxiliary equation to the corresponding homogeneous equation is .

$40{\mathbf{r}}^{\mathbf{2}}\mathbf{+}58\mathbf{r}\mathbf{+}11\mathbf{=}\mathbf{0}$

Then,

$$\begin{gathered} \mathbf{r}\mathbf{=}\frac{\mathbf{-}58\pm \sqrt{{\left(58\right)}^{\mathbf{2}}\mathbf{-}44\mathbf{\times }40}}{\mathbf{2}\mathbf{\times }40} \\ \mathbf{=}\frac{\mathbf{-}29\pm \sqrt{401}}{40} \end{gathered}$$

So, the roots are$\mathbf{r}\mathbf{=}\frac{\mathbf{-}29\mathbf{+}\sqrt{401}}{40}$and $\mathbf{r}\mathbf{=}\frac{\mathbf{-}29\mathbf{-}\sqrt{401}}{40}$.

Then, the general solution of y is ${\mathbf{y}}_{\mathbf{h}}\left(\mathbf{t}\right)\mathbf{=}{\mathbf{Ae}}^{\left(\frac{\mathbf{-}29\mathbf{+}\sqrt{401}}{40}\right)\mathbf{t}}\mathbf{+}{\mathbf{Be}}^{\left(\frac{\mathbf{-}29\mathbf{-}\sqrt{401}}{40}\right)\mathbf{t}}$ …… (6)

Let us assume that, ${\mathbf{y}}_{\mathbf{p}}\left(\mathbf{t}\right)\mathbf{=}\mathbf{C}$ …… (7)

Substitute the equation (7) in equation (5).

$$\begin{gathered} \left(\mathbf{40}{\mathbf{D}}^{\mathbf{2}}\mathbf{+}\mathbf{58}\mathbf{D}\mathbf{+}\mathbf{11}\right)\left[\mathbf{y}\right]\mathbf{=}\mathbf{400} \\ \left(40{\mathbf{D}}^{\mathbf{2}}\mathbf{+}58\mathbf{D}\mathbf{+}11\right)\left[\mathbf{C}\right]\mathbf{=}400 \\ 11\mathbf{C}\mathbf{=}\mathbf{4}00 \\ \mathbf{C}\mathbf{=}\frac{400}{11} \end{gathered}$$

Substitute the value of C in equation (7).

$$\begin{gathered} \mathbf{y}\left(\mathbf{t}\right)\mathbf{=}{\mathbf{y}}_{\mathbf{h}}\left(\mathbf{t}\right)\mathbf{+}{\mathbf{y}}_{\mathbf{p}}\left(\mathbf{t}\right) \\ \mathbf{=}{\mathbf{Ae}}^{\left(\frac{\mathbf{-}29\mathbf{+}\sqrt{401}}{40}\right)\mathbf{t}}\mathbf{+}{\mathbf{Be}}^{\left(\frac{\mathbf{-}29\mathbf{-}\sqrt{401}}{40}\right)\mathbf{t}}\mathbf{+}\frac{400}{11} \end{gathered}$$

So, the general solution is $\mathbf{y}\left(\mathbf{t}\right)\mathbf{=}{\mathbf{Ae}}^{\left(\frac{\mathbf{-}29\mathbf{+}\sqrt{401}}{40}\right)\mathbf{t}}\mathbf{+}{\mathbf{Be}}^{\left(\frac{\mathbf{-}29\mathbf{-}\sqrt{401}}{40}\right)\mathbf{t}}\mathbf{+}\frac{400}{11}$…… (8)

To find: $\underset{t\to \infty }{lim}x$.

Implement the limits on equation (8).

role="math" localid="1664044965414" $$\begin{gathered} \underset{\mathbf{t}\mathbf{\to }\mathbf{\infty }}{\mathbf{lim}}\mathbf{y}\left(\mathbf{t}\right)\mathbf{=}\underset{\mathbf{t}\mathbf{\to }\mathbf{\infty }}{\mathbf{l}\mathbf{i}\mathbf{m}}\left[{\mathbf{Ae}}^{\left(\frac{\mathbf{-}29\mathbf{+}\sqrt{401}}{40}\right)\mathbf{t}}\mathbf{+}{\mathbf{Be}}^{\left(\frac{\mathbf{-}29\mathbf{-}\sqrt{401}}{40}\right)\mathbf{t}}\mathbf{+}\frac{400}{11}\right] \\ \mathbf{=}\frac{400}{11} \end{gathered}$$

Therefore, the solution is founded.