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.

$$\begin{gathered} \frac{\mathbf{dx}}{\mathbf{dt}}\mathbf{+}\mathbf{x}\mathbf{+}\frac{\mathbf{dy}}{\mathbf{dt}}\mathbf{=}{\mathbf{e}}^{\mathbf{4}\mathbf{t}}\mathbf{,} \\ \mathbf{2}\mathbf{x}\mathbf{+}\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{y}}{{\mathbf{dt}}^{\mathbf{2}}}\mathbf{=}\mathbf{0} \end{gathered}$$

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

1. Make sure that the system is written in operator form.
   
   
2. Eliminate one of the variables, say, y, and solve the resulting equation for x(t). If the system is degenerating, stop! A separate analysis is required to determine whether or not there are solutions.

Given that,

$\frac{\mathbf{dx}}{\mathbf{dt}}\mathbf{+}\mathbf{x}\mathbf{+}\frac{\mathbf{dy}}{\mathbf{dt}}\mathbf{=}{\mathbf{e}}^{\mathbf{4}\mathbf{t}}$ … (1)

$\mathbf{2}\mathbf{x}\mathbf{+}\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{y}}{{\mathbf{dt}}^{\mathbf{2}}}\mathbf{=}\mathbf{0}$ … (2)

Let us rewrite the system in operator form,

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

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

Multiply D on both sides of equation (3) then subtract equation (3) and (4) together one gets,

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

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

Then, the homogeneous solution of u is;

${\mathbf{x}}_{\mathbf{h}}\left(\mathbf{t}\right)\mathbf{=}{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{-}\mathbf{2}\mathbf{t}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\mathbf{t}}$ … (6)

Let us take the undetermined coefficients and assume that,

${\mathbf{x}}_{\mathbf{p}}\left(\mathbf{t}\right)\mathbf{=}{\mathbf{Ae}}^{\mathbf{4}\mathbf{t}}$ … (7)

Now find the derivate the equation (7).

$$\begin{gathered} \mathbf{D}\left[{\mathbf{x}}_{\mathbf{p}}\left(\mathbf{t}\right)\right]\mathbf{=}\mathbf{4}{\mathbf{Ae}}^{\mathbf{4}\mathbf{t}} \\ {\mathbf{D}}^{\mathbf{2}}\left[{\mathbf{x}}_{\mathbf{p}}\left(\mathbf{t}\right)\right]\mathbf{=}\mathbf{16}{\mathbf{Ae}}^{\mathbf{4}\mathbf{t}} \end{gathered}$$

Substitute the derivation in equation (5).

$$\begin{gathered} \left({\mathbf{D}}^{\mathbf{2}}\mathbf{+}\mathbf{D}\mathbf{-}\mathbf{2}\right)\left[{\mathbf{Ae}}^{\mathbf{4}\mathbf{t}}\right]\mathbf{=}\mathbf{4}{\mathbf{e}}^{\mathbf{4}\mathbf{t}} \\ \mathbf{16}{\mathbf{Ae}}^{\mathbf{4}\mathbf{t}}\mathbf{+}\mathbf{4}{\mathbf{Ae}}^{\mathbf{4}\mathbf{t}}\mathbf{-}\mathbf{2}{\mathbf{Ae}}^{\mathbf{4}\mathbf{t}}\mathbf{=}\mathbf{4}{\mathbf{e}}^{\mathbf{4}\mathbf{t}} \\ \mathbf{18}{\mathbf{Ae}}^{\mathbf{4}\mathbf{t}}\mathbf{=}\mathbf{4}{\mathbf{e}}^{\mathbf{4}\mathbf{t}} \end{gathered}$$

Now, equalize the like terms.

$$\begin{gathered} \mathbf{18}\mathbf{A}\mathbf{=}\mathbf{4} \\ \mathbf{A}\mathbf{=}\frac{\mathbf{4}}{\mathbf{18}} \\ \mathbf{A}\mathbf{=}\frac{\mathbf{2}}{\mathbf{9}} \end{gathered}$$

So, ${\mathbf{x}}_{\mathbf{p}}\left(\mathbf{t}\right)\mathbf{=}\frac{2}{9}{\mathbf{e}}^{\mathbf{4}\mathbf{t}}$ … (8)

Use equations (6) and (8) to get,

$$\begin{gathered} \mathbf{x}\left(\mathbf{t}\right)\mathbf{=}{\mathbf{x}}_{\mathbf{h}}\left(\mathbf{t}\right)\mathbf{+}{\mathbf{x}}_{\mathbf{p}}\left(\mathbf{t}\right) \\ \mathbf{x}\left(\mathbf{t}\right)\mathbf{=}{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{-}\mathbf{2}\mathbf{t}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\mathbf{t}}\mathbf{+}\frac{2}{9}{\mathbf{e}}^{\mathbf{4}\mathbf{t}} \dots \left(9\right) \end{gathered}$$

Now, take equation (3).

$$\begin{gathered} \left(\mathbf{D}\mathbf{+}\mathbf{1}\right)\left[\mathbf{x}\right]\mathbf{+}\mathbf{D}\left[\mathbf{y}\right]\mathbf{=}{\mathbf{e}}^{\mathbf{4}\mathbf{t}} \\ \mathbf{D}\left[\mathbf{y}\right]\mathbf{=}{\mathbf{e}}^{\mathbf{4}\mathbf{t}}\mathbf{-}\left(\mathbf{D}\mathbf{+}\mathbf{1}\right)\left[\mathbf{x}\right] \\ \mathbf{=}{\mathbf{e}}^{\mathbf{4}\mathbf{t}}\mathbf{-}\left(\mathbf{D}\mathbf{+}\mathbf{1}\right)\left[{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{-}\mathbf{2}\mathbf{t}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\mathbf{t}}\mathbf{+}\frac{\mathbf{2}}{\mathbf{9}}{\mathbf{e}}^{\mathbf{4}\mathbf{t}}\right] \\ \mathbf{=}{\mathbf{e}}^{\mathbf{4}\mathbf{t}}\mathbf{+}\mathbf{2}{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{-}\mathbf{2}\mathbf{t}}\mathbf{-}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\mathbf{t}}\mathbf{-}\frac{\mathbf{8}}{\mathbf{9}}{\mathbf{e}}^{\mathbf{4}\mathbf{t}}\mathbf{-}{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{-}\mathbf{2}\mathbf{t}}\mathbf{-}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\mathbf{t}}\mathbf{-}\frac{\mathbf{2}}{\mathbf{9}}{\mathbf{e}}^{\mathbf{4}\mathbf{t}} \\ \mathbf{=}{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{-}\mathbf{2}\mathbf{t}}\mathbf{-}\mathbf{2}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\mathbf{t}}\mathbf{-}\frac{\mathbf{1}}{\mathbf{9}}{\mathbf{e}}^{\mathbf{4}\mathbf{t}} \\ \mathbf{y}\left(\mathbf{t}\right)\mathbf{=}\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{-}\mathbf{2}\mathbf{t}}\mathbf{-}\mathbf{2}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\mathbf{t}}\mathbf{-}\frac{\mathbf{1}}{\mathbf{36}}{\mathbf{e}}^{\mathbf{4}\mathbf{t}}\mathbf{+}{\mathbf{c}}_{\mathbf{3}} \end{gathered}$$

Thus, the solutions for the given linear system are$\mathbf{x}\left(\mathbf{t}\right)\mathbf{=}{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{-}\mathbf{2}\mathbf{t}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\mathbf{t}}\mathbf{+}\frac{2}{9}{\mathbf{e}}^{\mathbf{4}\mathbf{t}}$ and $\mathbf{y}\left(\mathbf{t}\right)\mathbf{=}\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{-}\mathbf{2}\mathbf{t}}\mathbf{-}\mathbf{2}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\mathbf{t}}\mathbf{-}\frac{\mathbf{1}}{\mathbf{36}}{\mathbf{e}}^{\mathbf{4}\mathbf{t}}\mathbf{+}{\mathbf{c}}_{\mathbf{3}}$.