Solve the given initial value problem.

$$\begin{gathered} \mathit{x}\mathbf{\text{'}}\mathbf{=}\mathit{z}\mathbf{-}\mathit{y}\mathbf{;}\mathbf{}\mathbf{}\mathit{x}\left(0\right)\mathbf{=}\mathbf{0} \\ \mathit{y}\mathbf{\text{'}}\mathbf{=}\mathit{z}\mathbf{;}\mathbf{ }\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathit{y}\left(0\right)\mathbf{=}\mathbf{0} \\ \mathit{z}\mathbf{\text{'}}\mathbf{=}\mathit{z}\mathbf{-}\mathit{x}\mathbf{;}\mathbf{}\mathbf{}\mathbf{}\mathit{z}\left(0\right)\mathbf{=}\mathbf{2} \end{gathered}$$

Substituting $\mathbf{z}\mathbf{=}\mathbf{y}\mathbf{\text{'}}$into the first and the third equation one will get:

$$\begin{gathered} \mathbf{x}\mathbf{\text{'}}\mathbf{-}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{y}\mathbf{=}\mathbf{0} \\ \mathbf{x}\mathbf{+}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{y}\mathbf{\text{'}}\mathbf{=}\mathbf{0} \end{gathered}$$$$\begin{gathered} \mathbf{x}\mathbf{\text{'}}\mathbf{-}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{y}\mathbf{=}\mathbf{0} \\ \mathbf{x}\mathbf{+}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{y}\mathbf{\text{'}}\mathbf{=}\mathbf{0} \end{gathered}$$

One will solve this system using the elimination method. First, one will rewrite this system in operator form:

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

Now, one can eliminate y from the system by "multiplying" the first equation by $D$ and then adding those two equations together:

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

A corresponding auxiliary equation is ${\mathbf{r}}^{\mathbf{2}}\mathbf{+}\mathbf{1}\mathbf{=}\mathbf{0}$, and its roots are ${\mathbf{r}}^{\mathbf{2}}\mathbf{=}\mathbf{-}\mathbf{1}\Rightarrow {\mathbf{r}}_{\mathbf{1}\mathbf{,}\mathbf{2}}\mathbf{=}\mathbf{\pm }\mathbf{i}$ and the general solution to $x$ is $\mathbf{x}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{c}}_{\mathbf{1}}\mathbf{cost}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}\mathbf{sint}$

To derive a general solution to $y$one will multiply the second equation of the system (1) by $-\mathrm{D}$ and then add those two equations together:

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

The auxiliary equation is ${\mathbf{r}}^{\mathbf{3}}\mathbf{-}{\mathbf{r}}^{\mathbf{3}}\mathbf{+}\mathbf{r}\mathbf{-}\mathbf{1}\mathbf{=}\mathbf{0}$and its roots are

$$\begin{gathered} {\mathbf{r}}^{\mathbf{3}}\mathbf{-}{\mathbf{r}}^{\mathbf{3}}\mathbf{+}\mathbf{r}\mathbf{-}\mathbf{1}\mathbf{=}\mathbf{(}\mathbf{r}\mathbf{-}\mathbf{1}\mathbf{)}\left({\mathbf{r}}^{\mathbf{2}}\mathbf{+}\mathbf{1}\right)\mathbf{=}\mathbf{0} \\ {\mathbf{r}}_{\mathbf{1}}\mathbf{=}\mathbf{1}\mathbf{,}{\mathbf{r}}_{\mathbf{2}\mathbf{,}\mathbf{3}}\mathbf{=}\mathbf{\pm }\mathbf{i} \end{gathered}$$

So, the general solution to $y$be$\mathbf{y}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{d}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{t}}\mathbf{+}{\mathbf{d}}_{\mathbf{2}}\mathbf{cost}\mathbf{+}{\mathbf{d}}_{\mathbf{3}}\mathbf{sint}$

Butone has that$\mathbf{x}\mathbf{+}\mathbf{D}\left(\mathbf{D}\mathbf{-}\mathbf{1}\right)\left[\mathbf{y}\right]\mathbf{=}\mathbf{0}$.So, one will find the first and the second derivative of $y$ and substitute it into the previous equation to obtain the relation between constants ${\mathbf{c}}_{\mathbf{i}}\mathbf{,}\mathbf{i}\mathbf{=}\overline{\mathbf{1}\mathbf{,}\mathbf{2}}$and${\mathbf{d}}_{\mathbf{i}}\mathbf{,}\mathbf{i}\mathbf{=}\overline{\mathbf{1}\mathbf{,}\mathbf{3}}$.

So, one has that:

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

So, the general solution to x is $\mathbf{x}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\left({\mathbf{d}}_{\mathbf{2}}\mathbf{+}{\mathbf{d}}_{\mathbf{3}}\right)\mathbf{cost}\mathbf{+}\left({\mathbf{d}}_{\mathbf{3}}\mathbf{-}{\mathbf{d}}_{\mathbf{2}}\right)\mathbf{sint}$

Now one can find the general solution for $z$. One will find it from $\mathbf{z}\mathbf{=}\mathbf{y}\mathbf{\text{'}}$. One has already found , so $\mathbf{z}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{d}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{t}}\mathbf{+}{\mathbf{d}}_{\mathbf{3}}\mathbf{cost}\mathbf{-}{\mathbf{d}}_{\mathbf{2}}\mathbf{sint}$

It remains to find the constants ${\mathbf{d}}_{\mathbf{1}}\mathbf{,}{\mathbf{d}}_{\mathbf{2}}$and ${\mathbf{d}}_{\mathbf{3}}$. One will find them from initial conditions which are $\mathbf{x}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{z}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{2}$, so one has,

$$\begin{gathered} x\left(0\right)=d_2+d_3=0 \\ y\left(0\right)=d_1+d_2=0 \\ z\left(0\right)=d_1+d_3=2 \end{gathered}$$

The first two equations give us that ${\mathbf{d}}_{\mathbf{1}}\mathbf{=}\mathbf{-}{\mathbf{d}}_{\mathbf{2}}\mathbf{=}{\mathbf{d}}_{\mathbf{3}}$, so substituting ${\mathbf{d}}_{\mathbf{1}}\mathbf{=}{\mathbf{d}}_{\mathbf{3}}$ into the third equation one gets that $\mathbf{2}{\mathbf{d}}_{\mathbf{2}}\mathbf{=}\mathbf{2}\mathbf{,}$so ${\mathbf{d}}_{\mathbf{3}}\mathbf{=}\mathbf{1}$. No one has that ${\mathbf{d}}_{\mathbf{1}}\mathbf{=}\mathbf{1}\mathbf{,}{\mathbf{d}}_{\mathbf{2}}\mathbf{=}\mathbf{-}\mathbf{1}\mathbf{,}{\mathbf{d}}_{\mathbf{3}}\mathbf{=}\mathbf{1}$so, substituting the values for ${\mathbf{d}}_1\mathbf{,}{\mathbf{d}}_2$ and ${\mathbf{d}}_{\mathbf{3}}$one hasthose solutions for $\mathbf{x}\left(\mathbf{t}\right)\mathbf{,}\mathbf{y}\left(\mathbf{t}\right)$and $\mathbf{z}\left(\mathbf{t}\right)$are;

$$\begin{gathered} \mathbf{x}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{2}\mathbf{sint} \\ \mathbf{y}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{e}}^{\mathbf{t}}\mathbf{-}\mathbf{cost}\mathbf{+}\mathbf{sint} \\ \mathbf{z}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{e}}^{\mathbf{t}}\mathbf{+}\mathbf{cost}\mathbf{+}\mathbf{sint} \end{gathered}$$