In Problems 3 – 18, use the elimination method to find a general solution for the given linear system, where differentiation is with respect to t.

$$\begin{gathered} \mathbf{x}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{5}\mathbf{x}\mathbf{-}\mathbf{4}\mathbf{y}\mathbf{=}\mathbf{0}\mathbf{,} \\ \mathbf{-}\mathbf{x}\mathbf{+}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{=}\mathbf{0} \end{gathered}$$

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

c. (Shortcut) If possible, use the system to derive an equation that involves y(t) but not its derivatives. [Otherwise, go to step (d).] Substitute the found expression for x(t) into this equation to get a formula for y(t). The expressions for x(t), and y(t) give the desired general solution.

d. Eliminate x from the system and solve for y(t). [Solving for y(t) gives more constants----in fact, twice as many as needed.]

Given that,

$\mathbf{x}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{5}\mathbf{x}\mathbf{-}\mathbf{4}\mathbf{y}\mathbf{=}\mathbf{0}$ … (1)

$-\mathrm{x}+\mathrm{y}\text{'}\text{'}+2\mathrm{y}=0$ … (2)

Let us rewrite the system in operator form,

$\left({\mathrm{D}}^2+5\right)\left[\mathrm{x}\right]-4\left[\mathrm{y}\right]=0$ … (3)

$-\mathrm{x}+\left({\mathrm{D}}^2+2\right)\left[\mathrm{y}\right]=0$ … (4)

Multiply$\left({\mathbf{D}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\right)$ on equation (3) and 4 on equation (4) then add equation (3) and (4) together one gets,

$$\begin{gathered} \left({\mathbf{D}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\right)\left({\mathbf{D}}^{\mathbf{2}}\mathbf{+}\mathbf{5}\right)\left[\mathbf{x}\right]\mathbf{-}\mathbf{4}\left({\mathbf{D}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\right)\left[\mathbf{y}\right]\mathbf{+}\left(\mathbf{-}\mathbf{4}\mathbf{x}\mathbf{+}\mathbf{4}\left({\mathbf{D}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\right)\left[\mathbf{y}\right]\right)\mathbf{=}\mathbf{0} \\ \left({\mathbf{D}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\right)\left({\mathbf{D}}^{\mathbf{2}}\mathbf{+}\mathbf{5}\right)\left[\mathbf{x}\right]\mathbf{-}\mathbf{4}\mathbf{x}\mathbf{=}\mathbf{0} \\ \left({\mathbf{D}}^{\mathbf{4}}\mathbf{+}\mathbf{7}{\mathbf{D}}^{\mathbf{2}}\mathbf{+}\mathbf{6}\right)\left[\mathbf{x}\right]\mathbf{=}\mathbf{0} \\ \left({\mathbf{D}}^{\mathbf{4}}\mathbf{+}\mathbf{7}{\mathbf{D}}^{\mathbf{2}}\mathbf{+}\mathbf{6}\right)\left[\mathbf{x}\right]\mathbf{=}\mathbf{0} \dots \left(5\right) \end{gathered}$$

Since the corresponding auxiliary equation is ${\mathbf{r}}^{\mathbf{4}}\mathbf{+}\mathbf{7}{\mathbf{r}}^{\mathbf{2}}\mathbf{+}\mathbf{6}\mathbf{=}\mathbf{0}$. The roots are$\mathbf{r}\mathbf{=}\mathbf{\pm }\mathbf{i}$ and $\mathbf{r}\mathbf{=}\mathbf{\pm }\sqrt{\mathbf{6}}\mathbf{i}$.

Then, the general solution of x is;

$\mathrm{x}\left(\mathrm{t}\right)={\mathrm{c}}_1\mathrm{cost}+{\mathrm{c}}_2\mathrm{sint}+{\mathrm{c}}_3\cos \sqrt{6}\mathrm{t}+{\mathrm{c}}_4\sin \sqrt{6}\mathrm{t}$ … (6)

Now, take equation (3).

$$\begin{gathered} \left({\mathbf{D}}^{\mathbf{2}}\mathbf{+}\mathbf{5}\right)\left[\mathbf{x}\right]\mathbf{-}\mathbf{4}\left[\mathbf{y}\right]\mathbf{=}\mathbf{0} \\ 4\left[\mathbf{y}\right]\mathbf{=}\left({\mathbf{D}}^{\mathbf{2}}\mathbf{+}\mathbf{5}\right)\left[\mathbf{x}\right] \\ \mathbf{=}\left({\mathbf{D}}^{\mathbf{2}}\mathbf{+}\mathbf{5}\right)\left[{\mathbf{c}}_{\mathbf{1}}\mathbf{cost}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}\mathbf{sint}\mathbf{+}{\mathbf{c}}_{\mathbf{3}}\mathbf{cos}\sqrt{\mathbf{6}}\mathbf{t}\mathbf{+}{\mathbf{c}}_{\mathbf{4}}\mathbf{sin}\sqrt{\mathbf{6}}\mathbf{t}\right] \end{gathered}$$

Now derivate the x value.

$$\begin{gathered} \mathbf{D}\left[\mathbf{x}\left(\mathbf{t}\right)\right]\mathbf{=}\mathbf{-}{\mathbf{c}}_{\mathbf{1}}\mathbf{sint}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}\mathbf{cost}\mathbf{-}\sqrt{\mathbf{6}}{\mathbf{c}}_{\mathbf{3}}\mathbf{sin}\sqrt{\mathbf{6}}\mathbf{t}\mathbf{+}\sqrt{\mathbf{6}}{\mathbf{c}}_{\mathbf{4}}\mathbf{cos}\sqrt{\mathbf{6}}\mathbf{t} \\ {\mathbf{D}}^{\mathbf{2}}\left[\mathbf{x}\left(\mathbf{t}\right)\right]\mathbf{=}\mathbf{-}{\mathbf{c}}_{\mathbf{1}}\mathbf{cost}\mathbf{-}{\mathbf{c}}_{\mathbf{2}}\mathbf{sint}\mathbf{-}\mathbf{6}{\mathbf{c}}_{\mathbf{3}}\mathbf{cos}\sqrt{\mathbf{6}}\mathbf{t}\mathbf{-}\mathbf{6}{\mathbf{c}}_{\mathbf{4}}\mathbf{sin}\sqrt{\mathbf{6}}\mathbf{t} \end{gathered}$$

Substitute in the above equation