Verify that the solution to the initial value problem

$$\begin{gathered} \mathbf{x}\mathbf{\text{'}}\mathbf{=}\mathbf{5}\mathbf{x}\mathbf{-}\mathbf{3}\mathbf{y}\mathbf{-}\mathbf{2}\mathbf{;} \mathbf{x}\left(\mathbf{0}\right)\mathbf{=}\mathbf{2}\mathbf{,} \\ \mathbf{y}\mathbf{\text{'}}\mathbf{=}\mathbf{4}\mathbf{x}\mathbf{-}\mathbf{3}\mathbf{y}\mathbf{-}\mathbf{1}\mathbf{;} \mathbf{y}\left(\mathbf{0}\right)\mathbf{=}\mathbf{0} \end{gathered}$$

Satisfies $\left|x\left(t\right)\right|\mathbf{+}\left|y\left(t\right)\right|\mathbf{\to }\mathbf{+}\mathbf{\infty }$as$\mathbf{t}\mathbf{\to }\mathbf{+}\mathbf{\infty }$

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,

$$\begin{gathered} x\text{'}=5x-3y-2 ......\mathbf{\left(}\mathbf{1}\mathbf{\right)} \\ y\text{'}=4x-3y-1 ......\mathbf{\left(}\mathbf{2}\mathbf{\right)} \end{gathered}$$

To verify:$\left|x\left(t\right)\right|+\left|y\left(t\right)\right|\to +\infty$ as $t\to +\infty$.

Let us rewrite this system of operators in operator form:

$$\begin{gathered} \left(D-5\right)\left[x\right]+3y=-2 ......\mathbf{\left(}\mathbf{3}\mathbf{\right)} \\ 4x-\left(D+3\right)\left[y\right]=1 ......\mathbf{\left(}\mathbf{4}\mathbf{\right)} \end{gathered}$$

Multiply 4 on equation (3) and D-5 on equation (4). Then, add them together to get,

$$\begin{gathered} \left(4\left(D-5\right)\left[x\right]+12y\right)-\left(4\left(D-5\right)x-\left(D-5\right)\left(D+3\right)\left[y\right]\right)=-8-\left(D-5\right)1 \\ \left(D-5\right)\left(D+3\right)\left[y\right]-12y=-8+5 \\ \left(D^2-2D-15+12\right)\left[y\right]=-3 \\ \left(D^2-2D-3\right)\left[y\right]=-3 \end{gathered}$$

$\left(D^2-2D-3\right)\left[y\right]=-3 ......\mathbf{\left(}\mathbf{5}\mathbf{\right)}$

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

Then, the homogeneous solution is${\mathbf{y}}_{\mathbf{h}}\left(\mathbf{t}\right)\mathbf{=}{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{-}\mathbf{t}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\mathbf{3}\mathbf{t}} ......\mathbf{\left(}\mathbf{6}\mathbf{\right)}$

Let us take the undetermined coefficients and assume that${\mathbf{y}}_{\mathbf{p}}\left(\mathbf{t}\right)\mathbf{=}\mathbf{C} ......\mathbf{\left(}\mathbf{7}\mathbf{\right)}$

Now derivate the equation (7)

$D\left[y_p\left(t\right)\right]=0$

Substitute the derivation in equation (5).

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

So, $y_p\left(t\right)=1$.

Then,$y\left(t\right)=c_1{e}^{-t}+c_2{e}^{3t}+1 ......\mathbf{\left(}\mathbf{8}\mathbf{\right)}$

Substitute equation (8) in equation (4).

$$\begin{gathered} 4x-\left(D+3\right)\left[y\right]=1 \\ 4x=1+\left(D+3\right)\left[y\right] \\ 4x=1+\left(D+3\right)\left[c_1{e}^{-t}+c_2{e}^{3t}+1\right] \end{gathered}$$

$$\begin{gathered} 4x=1-c_1{e}^{-t}+3c_2{e}^{3t}+3c_1{e}^{-t}+3c_2{e}^{3t}+3 \\ =2c_1{e}^{-t}+6c_2{e}^{3t}+4 \\ x=\frac{2c_1{e}^{-t}+6c_2{e}^{3t}+4}{4} \\ =\frac{1}{2}{c}_1{e}^{-t}+\frac{3}{2}{c}_2{e}^{3t}+1 \end{gathered}$$

$x\left(t\right)=\frac{1}{2}{c}_1{e}^{-t}+\frac{3}{2}{c}_2{e}^{3t}+1 ......\mathbf{\left(}\mathbf{9}\mathbf{\right)}$

Given, $x\left(0\right)=2, y\left(0\right)=0$.

Now substitute the values in equations (8) and (9).

$$\begin{gathered} x\left(t\right)=\frac{1}{2}{c}_1{e}^{-t}+\frac{3}{2}{c}_2{e}^{3t}+1 \\ x\left(0\right)=\frac{1}{2}{c}_1{e}^0+\frac{3}{2}{c}_2{e}^{3\left(0\right)}+1 \\ 1=\frac{1}{2}{c}_1+\frac{3}{2}{c}_2 \\ {c}_1+3c_2=2 ......\mathbf{\left(}\mathbf{10}\mathbf{\right)} \end{gathered}$$

$$\begin{gathered} y\left(t\right)=c_1{e}^{-t}+c_2{e}^{3t}+1 \\ y\left(0\right)=c_1{e}^{-0}+c_2{e}^{3\left(0\right)}+1 \\ -1=c_1+c_2 \\ {c}_1+c_2=-1 ......\mathbf{\left(}\mathbf{11}\mathbf{\right)} \end{gathered}$$

First, solve the equations (10) and (11).

$$\begin{gathered} c_1+3c_2-c_1-c_2=2+1 \\ 2c_2=3 \\ {c}_2=\frac{3}{2} \end{gathered}$$

Then,

$$\begin{gathered} 3c_1+3c_2-c_1-3c_2=-3-2 \\ 2c_1=-5 \\ {c}_1=-\frac{5}{2} \end{gathered}$$

Now substitute the values of c in equations (8) and (9).

$$\begin{gathered} x\left(t\right)=-\frac{5}{4}{e}^{-t}+\frac{9}{4}{c}_2{e}^{3t}+1 \\ y\left(t\right)=-\frac{5}{2}{e}^{-t}+\frac{3}{2}{e}^{3t}+1 \end{gathered}$$

Now calculate the limits:

$$\begin{gathered} \underset{t\to \infty }{lim}\left(\left|x\left(t\right)\right|+\left|y\left(t\right)\right|\right)=\underset{t\to \infty }{lim}\left(\left|-\frac{5}{4}{e}^{-t}+\frac{9}{4}{c}_2{e}^{3t}+1\right|+\left|-\frac{5}{2}{e}^{-t}+\frac{3}{2}{e}^{3t}+1\right|\right) \\ =\underset{t\to \infty }{lim}\left|-\frac{5}{4}{e}^{-t}+\frac{9}{4}{c}_2{e}^{3t}+1\right|+\underset{t\to \infty }{lim}\left|-\frac{5}{2}{e}^{-t}+\frac{3}{2}{e}^{3t}+1\right| \\ =\left|-0+\infty +1\right|+\left|-0+\infty +1\right| \\ \underset{t\to \infty }{lim}\left(\left|x\left(t\right)\right|+\left|y\left(t\right)\right|\right)=\infty \end{gathered}$$

So, the solution is founded.