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.

$$\begin{gathered} \frac{\mathbf{dx}}{\mathbf{dt}}\mathbf{=}\mathbf{x}\mathbf{-}\mathbf{4}\mathbf{y} \\ \frac{\mathbf{dy}}{\mathbf{dt}}\mathbf{=}\mathbf{x}\mathbf{+}\mathbf{y} \end{gathered}$$

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

Given that,

$\frac{\mathbf{dx}}{\mathbf{dt}}\mathbf{=}\mathbf{x}\mathbf{-}\mathbf{4}\mathbf{y}$ … (1)

$\frac{\mathbf{dy}}{\mathbf{dt}}\mathbf{=}\mathbf{x}\mathbf{+}\mathbf{y}$ … (2)

Let us rewrite the system in operator form,

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

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

Multiply (D-1) on both sides of equation (2) then subtract equation (1) and (2) together one gets,

$$\begin{gathered} \left(\mathbf{D}\mathbf{-}\mathbf{1}\right)\left[\mathbf{x}\right]\mathbf{+}\mathbf{4}\mathbf{y}\mathbf{+}\left(\mathbf{-}\left(\mathbf{D}\mathbf{-}\mathbf{1}\right)\mathbf{x}\mathbf{+}{\left(\mathbf{D}\mathbf{-}\mathbf{1}\right)}^{\mathbf{2}}\left[\mathbf{y}\right]\right)\mathbf{=}\mathbf{0} \\ \mathbf{4}\mathbf{y}\mathbf{+}{\left(\mathbf{D}\mathbf{-}\mathbf{1}\right)}^{\mathbf{2}}\left[\mathbf{y}\right]\mathbf{=}\mathbf{0} \\ \left({\left(\mathbf{D}\mathbf{-}\mathbf{1}\right)}^{\mathbf{2}}\mathbf{+}\mathbf{4}\right)\left[\mathbf{y}\right]\mathbf{=}\mathbf{0} \\ \left({\mathbf{D}}^{\mathbf{2}}\mathbf{-}\mathbf{2}\mathbf{D}\mathbf{+}\mathbf{5}\right)\left[\mathbf{y}\right]\mathbf{=}\mathbf{0} \\ \left({\mathbf{D}}^{\mathbf{2}}\mathbf{-}\mathbf{2}\mathbf{D}\mathbf{+}\mathbf{5}\right)\left[\mathbf{y}\right]\mathbf{=}\mathbf{0} \dots \left(5\right) \end{gathered}$$

Since the corresponding auxiliary equation is ${\mathbf{r}}^{\mathbf{2}}\mathbf{-}\mathbf{2}\mathbf{r}\mathbf{+}\mathbf{5}\mathbf{=}\mathbf{0}$. The roots are$\mathbf{r}\mathbf{=}\mathbf{1}\mathbf{-}\mathbf{2}\mathbf{i}$ and $\mathbf{r}\mathbf{=}\mathbf{1}\mathbf{+}\mathbf{2}\mathbf{i}$.

Then, the general solution of y is:

$\mathbf{y}\left(\mathbf{t}\right)\mathbf{=}{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{t}}\mathbf{cos}\mathbf{2}\mathbf{t}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\mathbf{t}}\mathbf{sin}\mathbf{2}\mathbf{t}$ … (4)

Now, take equation (4).

$$\begin{gathered} \mathbf{-}\mathbf{x}\mathbf{+}\left(\mathbf{D}\mathbf{-}\mathbf{1}\right)\left[\mathbf{y}\right]\mathbf{=}\mathbf{0} \\ \mathbf{x}\left(\mathbf{t}\right)\mathbf{=}\left(\mathbf{D}\mathbf{-}\mathbf{1}\right)\left[\mathbf{y}\left(\mathbf{t}\right)\right] \\ \mathbf{=}\left(\mathbf{D}\mathbf{-}\mathbf{1}\right)\left[{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{t}}\mathbf{cos}\mathbf{2}\mathbf{t}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\mathbf{t}}\mathbf{sin}\mathbf{2}\mathbf{t}\right] \\ \mathbf{=}{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{t}}\mathbf{cos}\mathbf{2}\mathbf{t}\mathbf{-}\mathbf{2}{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{t}}\mathbf{sin}\mathbf{2}\mathbf{t}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\mathbf{t}}\mathbf{sin}\mathbf{2}\mathbf{t}\mathbf{+}\mathbf{2}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\mathbf{t}}\mathbf{cos}\mathbf{2}\mathbf{t}\mathbf{-}{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{t}}\mathbf{cos}\mathbf{2}\mathbf{t}\mathbf{-}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\mathbf{t}}\mathbf{sin}\mathbf{2}\mathbf{t} \\ \mathbf{x}\left(\mathbf{t}\right)\mathbf{=}\mathbf{-}\mathbf{2}{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{t}}\mathbf{sin}\mathbf{2}\mathbf{t}\mathbf{+}\mathbf{2}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\mathbf{t}}\mathbf{cos}\mathbf{2}\mathbf{t} \end{gathered}$$

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