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

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

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,

$\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{x}}{{\mathbf{dt}}^{\mathbf{2}}}\mathbf{=}\mathbf{y}$…… (1)

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

Let us rewrite this system of operators in operator form:

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

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

Multiply D² on equation (3) and add with equation (4).

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

Since the corresponding auxiliary equation is ${\mathbf{r}}^4\mathbf{-}1\mathbf{=}\mathbf{0}$. The roots are$\mathbf{r}\mathbf{=}\mathbf{-}\mathbf{1}\mathbf{,}\mathbf{r}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{r}\mathbf{=}\mathbf{-}\mathbf{i}$

and $\mathbf{r}\mathbf{=}\mathbf{i}$.

Then, the general solution is $\mathbf{x}\left(\mathbf{t}\right)\mathbf{=}{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{-}\mathbf{t}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\mathbf{t}}\mathbf{+}{\mathbf{c}}_{\mathbf{3}}\mathbf{cost}\mathbf{+}{\mathbf{c}}_{\mathbf{4}}\mathbf{sint}$ …… (5)

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

$$\begin{gathered} {\mathbf{D}}^{\mathbf{2}}\left[\mathbf{x}\right]\mathbf{-}\mathbf{y}\mathbf{=}\mathbf{0} \\ \mathbf{-}\mathbf{y}\mathbf{=}\mathbf{-}{\mathbf{D}}^{\mathbf{2}}\left[\mathbf{x}\right] \\ \mathbf{y}\mathbf{=}{\mathbf{D}}^{\mathbf{2}}\left[\mathbf{x}\right] \\ \mathbf{y}\mathbf{=}\frac{{\mathbf{d}}^{\mathbf{2}}}{{\mathbf{dt}}^{\mathbf{2}}}\left[{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{-}\mathbf{t}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\mathbf{t}}\mathbf{+}{\mathbf{c}}_{\mathbf{3}}\mathbf{cost}\mathbf{+}{\mathbf{c}}_{\mathbf{4}}\mathbf{sint}\right] \\ \mathbf{y}\mathbf{=}{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{-}\mathbf{t}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\mathbf{t}}\mathbf{-}{\mathbf{c}}_{\mathbf{3}}\mathbf{cost}\mathbf{-}{\mathbf{c}}_{\mathbf{4}}\mathbf{sint} \\ \mathbf{y}\left(\mathbf{t}\right)\mathbf{=}{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{-}\mathbf{t}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\mathbf{t}}\mathbf{-}{\mathbf{c}}_{\mathbf{3}}\mathbf{cost}\mathbf{-}{\mathbf{c}}_{\mathbf{4}}\mathbf{sint} \dots \left(6\right) \end{gathered}$$

Given,$\mathbf{x}\left(\mathbf{0}\right)\mathbf{=}\mathbf{3}\mathbf{,} \mathbf{x}\mathbf{\text{'}}\left(\mathbf{0}\right)\mathbf{=}\mathbf{1}\mathbf{,} \mathbf{y}\left(\mathbf{0}\right)\mathbf{=}\mathbf{1}\mathbf{,}$and $\mathbf{y}\mathbf{\text{'}}\left(\mathbf{0}\right)\mathbf{=}\mathbf{-}\mathbf{1}$.

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

$$\begin{gathered} \mathbf{x}\left(\mathbf{t}\right)\mathbf{=}{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{-}\mathbf{t}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\mathbf{t}}\mathbf{+}{\mathbf{c}}_{\mathbf{3}}\mathbf{cost}\mathbf{+}{\mathbf{c}}_{\mathbf{4}}\mathbf{sint} \\ \mathbf{x}\left(\mathbf{0}\right)\mathbf{=}{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{-}\left(\mathbf{0}\right)}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\left(\mathbf{0}\right)}\mathbf{+}{\mathbf{c}}_{\mathbf{3}}\mathbf{cos}\left(\mathbf{0}\right)\mathbf{+}{\mathbf{c}}_{\mathbf{4}}\mathbf{sin}\left(\mathbf{0}\right) \\ \mathbf{3}\mathbf{=}{\mathbf{c}}_{\mathbf{1}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}\mathbf{+}{\mathbf{c}}_{\mathbf{3}} \\ {\mathbf{c}}_{\mathbf{1}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}\mathbf{+}{\mathbf{c}}_{\mathbf{3}}\mathbf{=}\mathbf{3} \dots \left(7\right) \end{gathered}$$

$$\begin{gathered} \mathbf{x}\mathbf{\text{'}}\left(\mathbf{t}\right)\mathbf{=}\mathbf{-}{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{-}\mathbf{t}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\mathbf{t}}\mathbf{-}{\mathbf{c}}_{\mathbf{3}}\mathbf{sint}\mathbf{+}{\mathbf{c}}_{\mathbf{4}}\mathbf{cost} \\ \mathbf{x}\mathbf{\text{'}}\left(\mathbf{0}\right)\mathbf{=}\mathbf{-}{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{-}\left(\mathbf{0}\right)}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\left(\mathbf{0}\right)}\mathbf{-}{\mathbf{c}}_{\mathbf{3}}\mathbf{sin}\left(\mathbf{0}\right)\mathbf{+}{\mathbf{c}}_{\mathbf{4}}\mathbf{cos}\left(\mathbf{0}\right) \\ \mathbf{1}\mathbf{=}\mathbf{-}{\mathbf{c}}_{\mathbf{1}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}\mathbf{+}{\mathbf{c}}_{\mathbf{4}} \\ \mathbf{-}{\mathbf{c}}_{\mathbf{1}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}\mathbf{+}{\mathbf{c}}_{\mathbf{4}}\mathbf{=}\mathbf{1} \dots \left(8\right) \end{gathered}$$

$$\begin{gathered} \mathbf{y}\left(\mathbf{t}\right)\mathbf{=}{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{-}\mathbf{t}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\mathbf{t}}\mathbf{-}{\mathbf{c}}_{\mathbf{3}}\mathbf{cost}\mathbf{-}{\mathbf{c}}_{\mathbf{4}}\mathbf{sint} \\ \mathbf{y}\left(\mathbf{0}\right)\mathbf{=}{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{-}\left(\mathbf{0}\right)}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\left(\mathbf{0}\right)}\mathbf{-}{\mathbf{c}}_{\mathbf{3}}\mathbf{cos}\left(\mathbf{0}\right)\mathbf{-}{\mathbf{c}}_{\mathbf{4}}\mathbf{sin}\left(\mathbf{0}\right) \\ \mathbf{1}\mathbf{=}{\mathbf{c}}_{\mathbf{1}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}\mathbf{-}{\mathbf{c}}_{\mathbf{3}} \\ {\mathbf{c}}_{\mathbf{1}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}\mathbf{-}{\mathbf{c}}_{\mathbf{3}}\mathbf{=}1 \dots \left(9\right) \end{gathered}$$

$$\begin{gathered} \mathbf{y}\mathbf{\text{'}}\left(\mathbf{t}\right)\mathbf{=}\mathbf{-}{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{-}\mathbf{t}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\mathbf{t}}\mathbf{+}{\mathbf{c}}_{\mathbf{3}}\mathbf{sint}\mathbf{-}{\mathbf{c}}_{\mathbf{4}}\mathbf{cost} \\ \mathbf{y}\mathbf{\text{'}}\left(\mathbf{0}\right)\mathbf{=}\mathbf{-}{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{-}\left(\mathbf{0}\right)}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\left(\mathbf{0}\right)}\mathbf{+}{\mathbf{c}}_{\mathbf{3}}\mathbf{sin}\left(\mathbf{0}\right)\mathbf{-}{\mathbf{c}}_{\mathbf{4}}\mathbf{cos}\left(\mathbf{0}\right) \\ \mathbf{-}\mathbf{1}\mathbf{=}\mathbf{-}{\mathbf{c}}_{\mathbf{1}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}\mathbf{-}{\mathbf{c}}_{\mathbf{4}} \\ \mathbf{-}{\mathbf{c}}_{\mathbf{1}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}\mathbf{-}{\mathbf{c}}_{\mathbf{4}}\mathbf{=}\mathbf{-}\mathbf{1} \dots \left(10\right) \end{gathered}$$

First, solve the equations (7) and (9). Then, equations (8) and (10).

$$\begin{gathered} {\mathbf{c}}_{\mathbf{1}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}\mathbf{+}{\mathbf{c}}_{\mathbf{3}}\mathbf{+}{\mathbf{c}}_{\mathbf{1}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}\mathbf{-}{\mathbf{c}}_{\mathbf{3}}\mathbf{=}\mathbf{3}\mathbf{+}\mathbf{1} \\ \mathbf{2}{\mathbf{c}}_{\mathbf{1}}\mathbf{+}\mathbf{2}{\mathbf{c}}_{\mathbf{2}}\mathbf{=}\mathbf{4} \dots \left(11\right) \end{gathered}$$

$$\begin{gathered} \mathbf{-}{\mathbf{c}}_{\mathbf{1}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}\mathbf{+}{\mathbf{c}}_{\mathbf{4}}\mathbf{-}{\mathbf{c}}_{\mathbf{1}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}\mathbf{-}{\mathbf{c}}_{\mathbf{4}}\mathbf{=}\mathbf{1}\mathbf{-}\mathbf{1} \\ \mathbf{-}\mathbf{2}{\mathbf{c}}_{\mathbf{1}}\mathbf{+}\mathbf{2}{\mathbf{c}}_{\mathbf{2}}\mathbf{=}\mathbf{0} \dots \left(12\right) \end{gathered}$$

Then,

Case (1):

$$\begin{gathered} 4{\mathbf{c}}_{\mathbf{2}}\mathbf{=}4 \\ {\mathbf{c}}_2\mathbf{=}\mathbf{1} \end{gathered}$$

Case (2):

$$\begin{gathered} \mathbf{2}{\mathbf{c}}_{\mathbf{1}}\mathbf{+}\mathbf{2}\mathbf{=}\mathbf{4} \\ \mathbf{2}{\mathbf{c}}_{\mathbf{1}}\mathbf{=}\mathbf{2} \\ {\mathbf{c}}_{\mathbf{1}}\mathbf{=}\mathbf{1} \end{gathered}$$

Case (3):

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

Case (4):

$$\begin{gathered} \mathbf{-}\mathbf{1}\mathbf{+}\mathbf{1}\mathbf{+}{\mathbf{c}}_{\mathbf{4}}\mathbf{=}\mathbf{1} \\ {\mathbf{c}}_{\mathbf{4}}\mathbf{=}\mathbf{1} \end{gathered}$$

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

$\mathbf{x}\left(\mathbf{t}\right)\mathbf{=}{\mathbf{e}}^{\mathbf{-}\mathbf{t}}\mathbf{+}{\mathbf{e}}^{\mathbf{t}}\mathbf{+}\mathbf{cost}\mathbf{+}\mathbf{sint}$

$\mathbf{y}\left(\mathbf{t}\right)\mathbf{=}{\mathbf{e}}^{\mathbf{-}\mathbf{t}}\mathbf{+}{\mathbf{e}}^{\mathbf{t}}\mathbf{-}\mathbf{cost}\mathbf{-}\mathbf{sint}$

So, the solution is founded.