In Problems 19 – 21, solve the given initial value problem.

$$\begin{gathered} \frac{\mathbf{dx}}{\mathbf{dt}}\mathbf{=}\mathbf{2}\mathbf{x}\mathbf{+}\mathbf{y}\mathbf{-}{\mathbf{e}}^{\mathbf{2}\mathbf{t}}\mathbf{;} \mathbf{x}\left(\mathbf{0}\right)\mathbf{=}\mathbf{1}\mathbf{,} \\ \frac{\mathbf{dy}}{\mathbf{dt}}\mathbf{=}\mathbf{x}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{;} \mathbf{y}\left(\mathbf{0}\right)\mathbf{=}\mathbf{-}\mathbf{1} \end{gathered}$$

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,

$\frac{\mathbf{dx}}{\mathbf{dt}}\mathbf{=}\mathbf{2}\mathbf{x}\mathbf{+}\mathbf{y}\mathbf{-}{\mathbf{e}}^{\mathbf{2}\mathbf{t}}$…… (1)

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

Let us rewrite this system of operators in operator form:

$\left(\mathbf{D}\mathbf{-}\mathbf{2}\right)\left[\mathbf{x}\right]\mathbf{-}\mathbf{y}\mathbf{=}\mathbf{-}{\mathbf{e}}^{\mathbf{2}\mathbf{t}}$ …… (3)

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

Multiply (D-2) on equation (3) and add with equation (4). one gets,

$$\begin{gathered} {\left(\mathbf{D}\mathbf{-}\mathbf{2}\right)}^{\mathbf{2}}\left[\mathbf{x}\right]\mathbf{-}\left(\mathbf{D}\mathbf{-}\mathbf{2}\right)\mathbf{y}\mathbf{+}\left(\mathbf{-}\mathbf{x}\mathbf{+}\left(\mathbf{D}\mathbf{-}\mathbf{2}\right)\left[\mathbf{y}\right]\right)\mathbf{=}\left(\mathbf{D}\mathbf{-}\mathbf{2}\right)\left[\mathbf{-}{\mathbf{e}}^{\mathbf{2}\mathbf{t}}\right] \\ {\left(\mathbf{D}\mathbf{-}\mathbf{2}\right)}^{\mathbf{2}}\left[\mathbf{x}\right]\mathbf{-}\mathbf{x}\mathbf{=}\mathbf{-}\mathbf{2}{\mathbf{e}}^{\mathbf{2}\mathbf{t}}\mathbf{+}\mathbf{2}{\mathbf{e}}^{\mathbf{2}\mathbf{t}} \\ \left({\left(\mathbf{D}\mathbf{-}\mathbf{2}\right)}^{\mathbf{2}}\mathbf{-}\mathbf{1}\right)\left[\mathbf{x}\right]\mathbf{=}\mathbf{0} \\ \left(\mathbf{D}\mathbf{-}\mathbf{4}\mathbf{D}\mathbf{+}\mathbf{4}\mathbf{-}\mathbf{1}\right)\left[\mathbf{x}\right]\mathbf{=}\mathbf{0} \\ \left(\mathbf{D}\mathbf{-}\mathbf{4}\mathbf{D}\mathbf{+}\mathbf{3}\right)\left[\mathbf{x}\right]\mathbf{=}\mathbf{0} \end{gathered}$$

Since the corresponding auxiliary equation is ${\mathbf{r}}^{\mathbf{2}}\mathbf{-}4\mathbf{r}\mathbf{+}3\mathbf{=}\mathbf{0}$. The roots are r=1 and r=3 .

Then, the general solution is $\mathbf{x}\left(\mathbf{t}\right)\mathbf{=}{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{t}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\mathbf{3}\mathbf{t}}$ …… (5)

Substitute the equation (5) in equation (3).

$$\begin{gathered} \left(\mathbf{D}\mathbf{-}\mathbf{2}\right)\left[\mathbf{x}\right]\mathbf{-}\mathbf{y}\mathbf{=}\mathbf{-}{\mathbf{e}}^{\mathbf{2}\mathbf{t}} \\ \mathbf{-}\mathbf{y}\mathbf{=}\mathbf{-}{\mathbf{e}}^{\mathbf{2}\mathbf{t}}\mathbf{-}\left(\mathbf{D}\mathbf{-}\mathbf{2}\right)\left[\mathbf{x}\right] \\ \mathbf{y}\mathbf{=}\left(\mathbf{D}\mathbf{-}\mathbf{2}\right)\left[{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{t}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\mathbf{3}\mathbf{t}}\right]\mathbf{+}{\mathbf{e}}^{\mathbf{2}\mathbf{t}} \\ \mathbf{y}\mathbf{=}\frac{\mathbf{d}}{\mathbf{dt}}\left[{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{t}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\mathbf{3}\mathbf{t}}\right]\mathbf{-}\mathbf{2}{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{t}}\mathbf{-}\mathbf{2}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\mathbf{3}\mathbf{t}}\mathbf{+}{\mathbf{e}}^{\mathbf{2}\mathbf{t}} \\ \mathbf{y}\left(\mathbf{t}\right)\mathbf{=}{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{t}}\mathbf{+}\mathbf{3}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\mathbf{3}\mathbf{t}}\mathbf{-}\mathbf{2}{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{2}\mathbf{t}}\mathbf{-}\mathbf{2}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\mathbf{3}\mathbf{t}}\mathbf{+}{\mathbf{e}}^{\mathbf{2}\mathbf{t}} \\ \mathbf{y}\left(\mathbf{t}\right)\mathbf{=}\mathbf{-}{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{t}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\mathbf{3}\mathbf{t}}\mathbf{+}{\mathbf{e}}^{\mathbf{2}\mathbf{t}} \\ \mathbf{y}\left(\mathbf{t}\right)\mathbf{=}\mathbf{-}{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{t}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\mathbf{3}\mathbf{t}}\mathbf{+}{\mathbf{e}}^{\mathbf{2}\mathbf{t}} \dots \left(6\right) \end{gathered}$$

Given: $\mathbf{x}\left(\mathbf{0}\right)\mathbf{=}\mathbf{1}\mathbf{,} \mathbf{y}\left(\mathbf{0}\right)\mathbf{=}\mathbf{-}\mathbf{1}$.

Now substitute the values in equations (5) and (6).

$$\begin{gathered} \mathbf{x}\left(\mathbf{t}\right)\mathbf{=}{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{t}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\mathbf{3}\mathbf{t}} \\ \mathbf{x}\left(\mathbf{0}\right)\mathbf{=}{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\left(\mathbf{0}\right)}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\mathbf{3}\left(\mathbf{0}\right)} \end{gathered}$$

${\mathbf{c}}_{\mathbf{1}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}\mathbf{=}1$ …… (7)

$$\begin{gathered} \mathbf{y}\left(\mathbf{t}\right)\mathbf{=}\mathbf{-}{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{t}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\mathbf{3}\mathbf{t}}\mathbf{+}{\mathbf{e}}^{\mathbf{2}\mathbf{t}} \\ \mathbf{y}\left(\mathbf{0}\right)\mathbf{=}\mathbf{-}{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{0}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\mathbf{3}\left(\mathbf{0}\right)}\mathbf{+}{\mathbf{e}}^{\mathbf{2}\left(\mathbf{0}\right)} \\ \mathbf{-}\mathbf{1}\mathbf{=}\mathbf{-}{\mathbf{c}}_{\mathbf{1}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}\mathbf{+}\mathbf{1} \\ \mathbf{-}{\mathbf{c}}_{\mathbf{1}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}\mathbf{=}\mathbf{-}\mathbf{2} \dots \left(8\right) \end{gathered}$$

Solve the equations (7) and (8).

$$\begin{gathered} {\mathbf{c}}_{\mathbf{1}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}\mathbf{-}{\mathbf{c}}_{\mathbf{1}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}\mathbf{=}\mathbf{1}\mathbf{-}\mathbf{2} \\ \mathbf{2}{\mathbf{c}}_{\mathbf{2}}\mathbf{=}\mathbf{-}\mathbf{1} \\ {\mathbf{c}}_{\mathbf{2}}\mathbf{=}\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}} \end{gathered}$$

Then,

$$\begin{gathered} {\mathbf{c}}_{\mathbf{1}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}\mathbf{=}\mathbf{1} \\ {\mathbf{c}}_{\mathbf{1}}\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{=}\mathbf{1} \\ {\mathbf{c}}_{\mathbf{1}}\mathbf{=}\frac{\mathbf{3}}{\mathbf{2}} \end{gathered}$$

Now substitute the values of c in equations (5) and (6).

$\mathbf{x}\left(\mathbf{t}\right)\mathbf{=}\frac{\mathbf{3}}{\mathbf{2}}{\mathbf{e}}^{\mathbf{t}}\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{e}}^{\mathbf{3}\mathbf{t}}$

$\mathbf{y}\left(\mathbf{t}\right)\mathbf{=}\mathbf{-}\frac{\mathbf{3}}{\mathbf{2}}{\mathbf{e}}^{\mathbf{t}}\mathbf{+}{\mathbf{e}}^{\mathbf{2}\mathbf{t}}\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{e}}^{\mathbf{3}\mathbf{t}}$

Thus, the required solutions are;

$\mathbf{x}\left(\mathbf{t}\right)\mathbf{=}\frac{\mathbf{3}}{\mathbf{2}}{\mathbf{e}}^{\mathbf{t}}\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{e}}^{\mathbf{3}\mathbf{t}}$

$\mathbf{y}\left(\mathbf{t}\right)\mathbf{=}\mathbf{-}\frac{\mathbf{3}}{\mathbf{2}}{\mathbf{e}}^{\mathbf{t}}\mathbf{+}{\mathbf{e}}^{\mathbf{2}\mathbf{t}}\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{e}}^{\mathbf{3}\mathbf{t}}$