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Elimination Procedure for 2 脳 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鈥檚 formulas for finding roots:

For function y(t) to be bounded when$\mathbf{t}鈫�\mathbf{+}鈭�$ 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 ${\mathbf{r}}_{\mathbf{1}}\mathbf{,}{\mathbf{r}}_{\mathbf{2}}鈭�\mathbf{R}$, then ${\mathbf{r}}_{\mathbf{1}}路{\mathbf{r}}_{\mathbf{2}}猢�\mathbf{0}\mathbf{,}{\mathbf{r}}_{\mathbf{1}}\mathbf{+}{\mathbf{r}}_{\mathbf{2}}猢�\mathbf{0}$,
2. If ${\mathbf{r}}_{\mathbf{1}}\mathbf{,}{\mathbf{r}}_{\mathbf{2}}\mathbf{=}\mathbf{伪}卤\mathbf{尾颈}$,$\mathbf{尾}鈮�0$ , then $\mathbf{伪}\mathbf{=}\frac{{\mathbf{r}}_{\mathbf{1}}\mathbf{+}{\mathbf{r}}_{\mathbf{2}}}{\mathbf{2}}猢�0$.

Given that,

$\frac{\mathbf{dx}}{\mathbf{dt}}\mathbf{=}\mathbf{-}\mathbf{x}\mathbf{+}\mathbf{位测}鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�\left(1\right)$

role="math" localid="1664101250446" $\frac{\mathbf{dy}}{\mathbf{dt}}\mathbf{=}\mathbf{x}\mathbf{-}\mathbf{y}鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�\left(2\right)$

Let us rewrite the given system of equations into operator form.

$\left(\mathbf{D}\mathbf{+}\mathbf{1}\right)\left[\mathbf{x}\right]\mathbf{-}\mathbf{位}\left[\mathbf{y}\right]\mathbf{=}\mathbf{0}鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�\left(3\right)$

$\left[\mathbf{x}\right]\mathbf{-}\left(\mathbf{D}\mathbf{+}\mathbf{1}\right)\left[\mathbf{y}\right]\mathbf{=}\mathbf{0}$ 鈥� (4)

Multiply D+1 on equation (4). Then, subtract with equation (3).

$$\begin{gathered} \left(\mathbf{D}\mathbf{+}\mathbf{1}\right)\left[\mathbf{x}\right]\mathbf{-}\mathbf{位}\left[\mathbf{y}\right]\mathbf{-}\left(\mathbf{D}\mathbf{+}\mathbf{1}\right)\left[\mathbf{x}\right]\mathbf{+}{\left(\mathbf{D}\mathbf{+}\mathbf{1}\right)}^{\mathbf{2}}\left[\mathbf{y}\right]\mathbf{=}\mathbf{0} \\ {\left(\mathbf{D}\mathbf{+}\mathbf{1}\right)}^{\mathbf{2}}\left[\mathbf{y}\right]\mathbf{-}\mathbf{位}\left[\mathbf{y}\right]\mathbf{=}\mathbf{0} \\ \left({\left(\mathbf{D}\mathbf{+}\mathbf{1}\right)}^{\mathbf{2}}\mathbf{-}\mathbf{位}\right)\left[\mathbf{y}\right]\mathbf{=}\mathbf{0} \\ \left({\mathbf{D}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathbf{D}\mathbf{+}\mathbf{1}\mathbf{-}\mathbf{位}\right)\left[\mathbf{y}\right]\mathbf{=}\mathbf{0} \\ \left({\left(\mathbf{D}\mathbf{+}\mathbf{1}\right)}^{\mathbf{2}}\mathbf{-}\mathbf{位}\right)\left[\mathbf{y}\right]\mathbf{=}\mathbf{0}鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈰�\left(5\right) \end{gathered}$$

Since the auxiliary equation to the corresponding homogeneous equation is .

${\left(\mathbf{r}\mathbf{+}\mathbf{1}\right)}^{\mathbf{2}}\mathbf{-}\mathbf{位}\mathbf{=}\mathbf{0}$

Then,

$$\begin{gathered} {\left(\mathbf{r}\mathbf{+}\mathbf{1}\right)}^{\mathbf{2}}\mathbf{-}\mathbf{位}\mathbf{=}\mathbf{0} \\ {\left(\mathbf{r}\mathbf{+}\mathbf{1}\right)}^{\mathbf{2}}\mathbf{=}\mathbf{位} \\ \mathbf{r}\mathbf{+}\mathbf{1}\mathbf{=}卤\sqrt{\mathbf{位}} \\ \mathbf{r}\mathbf{=}\mathbf{-}\mathbf{1}卤\sqrt{\mathbf{位}} \end{gathered}$$

So, the roots are$\mathbf{r}\mathbf{=}\mathbf{-}\mathbf{1}\mathbf{+}\sqrt{\mathbf{位}}$ and $\mathbf{r}\mathbf{=}\mathbf{-}\mathbf{1}\mathbf{-}\sqrt{\mathbf{位}}$.

Then, the general solution of y is $\mathbf{y}\left(\mathbf{t}\right)\mathbf{=}{\mathbf{Ae}}^{\left(\mathbf{-}\mathbf{1}\mathbf{+}\sqrt{\mathbf{位}}\right)\mathbf{t}}\mathbf{+}{\mathbf{Be}}^{\left(\mathbf{-}\mathbf{1}\mathbf{-}\sqrt{\mathbf{位}}\right)\mathbf{t}}鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�\left(2\right)$

Now substitute the equation (6) in equation (4).

$$\begin{gathered} \left[\mathbf{x}\right]\mathbf{-}\left(\mathbf{D}\mathbf{+}\mathbf{1}\right)\left[\mathbf{y}\right]\mathbf{=}\mathbf{0} \\ \mathbf{x}\mathbf{=}\left(\mathbf{D}\mathbf{+}\mathbf{1}\right)\left[\mathbf{y}\right] \\ \mathbf{x}\mathbf{=}\left(\mathbf{D}\mathbf{+}\mathbf{1}\right)\left[{\mathbf{Ae}}^{\left(\mathbf{-}\mathbf{1}\mathbf{+}\sqrt{\mathbf{位}}\right)\mathbf{t}}\mathbf{+}{\mathbf{Be}}^{\left(\mathbf{-}\mathbf{1}\mathbf{-}\sqrt{\mathbf{位}}\right)\mathbf{t}}\right] \\ \mathbf{=}\left(\mathbf{-}\mathbf{1}\mathbf{+}\sqrt{\mathbf{位}}\right){\mathbf{Ae}}^{\left(\mathbf{-}\mathbf{1}\mathbf{+}\sqrt{\mathbf{位}}\right)\mathbf{t}}\mathbf{+}\left(\mathbf{-}\mathbf{1}\mathbf{-}\sqrt{\mathbf{位}}\right){\mathbf{Be}}^{\left(\mathbf{-}\mathbf{1}\mathbf{-}\sqrt{\mathbf{位}}\right)\mathbf{t}}\mathbf{+}{\mathbf{Ae}}^{\left(\mathbf{-}\mathbf{1}\mathbf{+}\sqrt{\mathbf{位}}\right)\mathbf{t}}\mathbf{+}{\mathbf{Be}}^{\left(\mathbf{-}\mathbf{1}\mathbf{-}\sqrt{\mathbf{位}}\right)\mathbf{t}} \\ \mathbf{=}\mathbf{A}\sqrt{\mathbf{位}}{\mathbf{e}}^{\left(\mathbf{-}\mathbf{1}\mathbf{+}\sqrt{\mathbf{位}}\right)\mathbf{t}}\mathbf{-}\mathbf{B}\sqrt{\mathbf{位}}{\mathbf{e}}^{\left(\mathbf{-}\mathbf{1}\mathbf{-}\sqrt{\mathbf{位}}\right)\mathbf{t}} \\ \mathbf{x}\left(\mathbf{t}\right)\mathbf{=}\mathbf{A}\sqrt{\mathbf{位}}{\mathbf{e}}^{\left(\mathbf{-}\mathbf{1}\mathbf{+}\sqrt{\mathbf{位}}\right)\mathbf{t}}\mathbf{-}\mathbf{B}\sqrt{\mathbf{位}}{\mathbf{e}}^{\left(\mathbf{-}\mathbf{1}\mathbf{-}\sqrt{\mathbf{位}}\right)\mathbf{t}}鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�\left(7\right) \end{gathered}$$

Since x and y do not approach the $鈭�$. When, $\mathbf{t}鈫�\mathbf{+}鈭�$. So, the relation between them is taken as below.

$\left(\mathbf{-}\mathbf{1}\mathbf{+}\sqrt{\mathbf{位}}\right)\left(\mathbf{-}\mathbf{1}\mathbf{-}\sqrt{\mathbf{位}}\right)猢�\mathbf{0}$

Let us rearrange the terms. To get,

$$\begin{gathered} \left(\sqrt{\mathbf{位}}\mathbf{-}\mathbf{1}\right)\left(\sqrt{\mathbf{位}}\mathbf{+}\mathbf{1}\right)猢�\mathbf{0} \\ \left(\mathbf{位}\mathbf{-}\mathbf{1}\right)猢�\mathbf{0} \\ \mathbf{位}猢�\mathbf{1} \end{gathered}$$

Therefore, the values of$\mathbf{位}$ for which the solution of x(t) and y(t) remains bounded are $\mathbf{位}猢�\mathbf{1}$.