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$\mathbf{D}\mathbf{=}\frac{\mathbf{d}}{\mathbf{dt}}$

The differential for only x(t) is $\left({\mathbf{L}}_{\mathbf{1}}\mathbf{×}{\mathbf{L}}_{\mathbf{4}}\mathbf{-}{\mathbf{L}}_{\mathbf{2}}\mathbf{×}{\mathbf{L}}_{\mathbf{3}}\right)\left[\mathbf{x}\right]\mathbf{=}{\mathbf{L}}_{\mathbf{4}}\left[{\mathbf{g}}_{\mathbf{1}}\left(\mathbf{t}\right)\right]\mathbf{-}{\mathbf{L}}_{\mathbf{1}}\left[{\mathbf{g}}_{\mathbf{2}}\left(\mathbf{t}\right)\right]$.

Given that,

$\mathbf{x}\mathbf{\text{'}}\mathbf{+}\mathbf{y}\mathbf{\text{'}}\mathbf{-}\mathbf{x}\mathbf{=}\mathbf{5}                     …\left(1\right)$

$\mathbf{x}\mathbf{\text{'}}\mathbf{+}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{y}\mathbf{=}\mathbf{1}                …\left(2\right)$

Let us rewrite this system of operators in operator form:

$\left(\mathbf{D}\mathbf{-}\mathbf{1}\right)\left[\mathbf{x}\right]\mathbf{+}\mathbf{D}\left[\mathbf{y}\right]\mathbf{=}\mathbf{5}                 …\left(3\right)$

$\mathbf{D}\left[\mathbf{x}\right]\mathbf{+}\left(\mathbf{D}\mathbf{+}\mathbf{1}\right)\left[\mathbf{y}\right]\mathbf{=}\mathbf{1}                   …\left(4\right)$

For the system of the equation of the form is shown above.

The differential for only x(t) for equations (3) and (4). We get,

$$\begin{gathered} \left(\left(\mathbf{D}\mathbf{-}\mathbf{1}\right)\left(\mathbf{D}\mathbf{+}\mathbf{1}\right)\mathbf{-}\mathbf{D}\left(\mathbf{D}\right)\right)\left[\mathbf{x}\right]\mathbf{=}\left(\mathbf{D}\mathbf{+}\mathbf{1}\right)\mathbf{5}\mathbf{-}\mathbf{D} \\ \left({\mathbf{D}}^{\mathbf{2}}\mathbf{-}\mathbf{1}\mathbf{-}{\mathbf{D}}^{\mathbf{2}}\right)\left[\mathbf{x}\right]\mathbf{=}\mathbf{5} \\ \mathbf{-}\left[\mathbf{x}\right]\mathbf{=}\mathbf{5} \\ \mathbf{x}\left(\mathbf{t}\right)\mathbf{=}\mathbf{-}\mathbf{5} \end{gathered}$$

Then, the general solution is$\mathbf{x}\left(\mathbf{t}\right)\mathbf{=}\mathbf{-}\mathbf{5}                …\left(5\right)$

Substitute equation (5) in equation (1).

Since$\mathbf{x}\left(\mathbf{t}\right)\mathbf{=}\mathbf{-}\mathbf{5}$ and $\mathbf{x}\mathbf{\text{'}}\left(\mathbf{t}\right)\mathbf{=}\mathbf{0}$.

Then,

$$\begin{gathered} \mathbf{x}\mathbf{\text{'}}\mathbf{+}\mathbf{y}\mathbf{\text{'}}\mathbf{-}\mathbf{x}\mathbf{=}\mathbf{5} \\ \mathbf{0}\mathbf{+}\mathbf{y}\mathbf{\text{'}}\mathbf{-}\left(\mathbf{-}\mathbf{5}\right)\mathbf{=}\mathbf{5} \\ \mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{5}\mathbf{=}\mathbf{5} \\ \mathbf{y}\mathbf{\text{'}}\left(\mathbf{t}\right)\mathbf{=}\mathbf{0} \end{gathered}$$

Substitute the values in equation (2).

$$\begin{gathered} \mathbf{x}\mathbf{\text{'}}\mathbf{+}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{y}\mathbf{=}\mathbf{1} \\ \mathbf{0}\mathbf{+}\mathbf{0}\mathbf{+}\mathbf{y}\mathbf{=}\mathbf{1} \\ \mathbf{y}\left(\mathbf{t}\right)\mathbf{=}\mathbf{1} \end{gathered}$$

Therefore, the solutions for the given linear system are$\mathbf{x}\left(\mathbf{t}\right)\mathbf{=}\mathbf{-}\mathbf{5}$ and $\mathbf{y}\left(\mathbf{t}\right)\mathbf{=}1$.