91Ó°ÊÓ

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

Given that,

$$\begin{gathered} \mathbf{x}\mathbf{\text{'}}\mathbf{=}\mathbf{x}\mathbf{-}\mathbf{y}                    …\left(1\right) \\ \mathbf{y}\mathbf{\text{'}}\mathbf{=}\mathbf{y}\mathbf{-}\mathbf{4}\mathbf{x}                 …\left(2\right) \end{gathered}$$

Let us rewrite this system of operators in operator form:

$$\begin{gathered} \left(\mathbf{D}\mathbf{-}\mathbf{1}\right)\left[\mathbf{x}\right]\mathbf{+}\mathbf{y}\mathbf{=}\mathbf{0}                        …\left(3\right) \\ \mathbf{4}\mathbf{x}\mathbf{+}\left(\mathbf{D}\mathbf{-}\mathbf{1}\right)\left[\mathbf{y}\right]\mathbf{=}\mathbf{0}                        …\left(4\right) \end{gathered}$$

Multiply (D-1) on equation (3) and subtract with equation (4). one gets,

$$\begin{gathered} {\left(\mathbf{D}\mathbf{-}\mathbf{1}\right)}^{\mathbf{2}}\left[\mathbf{x}\right]\mathbf{-}\mathbf{4}\mathbf{x}\mathbf{=}\mathbf{0} \\ \left({\mathbf{D}}^{\mathbf{2}}\mathbf{-}\mathbf{2}\mathbf{D}\mathbf{+}\mathbf{1}\right)\left[\mathbf{x}\right]\mathbf{-}\mathbf{4}\mathbf{x}\mathbf{=}\mathbf{0} \\ \left({\mathbf{D}}^{\mathbf{2}}\mathbf{-}\mathbf{2}\mathbf{D}\mathbf{-}\mathbf{3}\right)\left[\mathbf{x}\right]\mathbf{=}\mathbf{0} \end{gathered}$$

Since the corresponding auxiliary equation is ${\mathbf{r}}^{\mathbf{2}}\mathbf{-}\mathbf{2}\mathbf{r}\mathbf{-}\mathbf{3}\mathbf{=}\mathbf{0}$. The roots are$\mathbf{r}\mathbf{=}\mathbf{-}\mathbf{1}$ and $\mathbf{r}\mathbf{=}3$.

Then, the general solution is$\mathbf{x}\left(\mathbf{t}\right)\mathbf{=}{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{-}\mathbf{t}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\mathbf{3}\mathbf{t}}                …\left(5\right)$

Substitute the equation (5) in equation (3).

$$\begin{gathered} \left(\mathbf{D}\mathbf{-}\mathbf{1}\right)\left[{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{-}\mathbf{t}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\mathbf{3}\mathbf{t}}\right]\mathbf{y}\mathbf{=}\mathbf{0} \\ \mathbf{y}\mathbf{=}\left(\mathbf{1}\mathbf{-}\mathbf{D}\right)\left[{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{-}\mathbf{t}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\mathbf{3}\mathbf{t}}\right] \\ \mathbf{y}\mathbf{=}{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{-}\mathbf{t}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\mathbf{3}\mathbf{t}}\mathbf{-}\frac{\mathbf{d}}{\mathbf{dt}}\left[{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{-}\mathbf{t}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\mathbf{3}\mathbf{t}}\right] \\ \mathbf{=}{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{-}\mathbf{t}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\mathbf{3}\mathbf{t}}\mathbf{-}\left(\mathbf{-}{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{-}\mathbf{t}}\mathbf{+}\mathbf{3}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\mathbf{3}\mathbf{t}}\right) \\ \mathbf{=}\mathbf{2}{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{-}\mathbf{t}}\mathbf{-}\mathbf{2}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\mathbf{3}\mathbf{t}} \end{gathered}$$

Thus, the solutions for the given linear system are$\mathbf{x}\left(\mathbf{t}\right)\mathbf{=}{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{-}\mathbf{t}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{3\mathbf{t}}$ and $\mathbf{y}\left(\mathbf{t}\right)\mathbf{=}\mathbf{2}{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{-}\mathbf{t}}\mathbf{-}\mathbf{2}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\mathbf{3}\mathbf{t}}$.