Solve the given initial value problem.

$$\begin{gathered} \mathbf{x}\mathbf{\text{'}}\mathbf{=}\mathbf{y}\mathbf{+}\mathbf{z}\mathbf{;}\mathbf{x}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{2} \\ \mathbf{y}\mathbf{\text{'}}\mathbf{=}\mathbf{x}\mathbf{+}\mathbf{z}\mathbf{;}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{2} \\ \mathbf{z}\mathbf{\text{'}}\mathbf{=}\mathbf{x}\mathbf{+}\mathbf{y}\mathbf{;}\mathbf{z}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{1} \end{gathered}$$

Let's rewrite the given system in operator form:

$$\begin{gathered} \mathbf{D}\left[\mathbf{x}\right]\mathbf{-}\mathbf{y}\mathbf{-}\mathbf{z}\mathbf{=}\mathbf{0} \\ \mathbf{x}\mathbf{-}\mathbf{D}\left[\mathbf{y}\right]\mathbf{+}\mathbf{z}\mathbf{=}\mathbf{0} \\ \mathbf{x}\mathbf{+}\mathbf{y}\mathbf{-}\mathbf{D}\left[\mathbf{z}\right]\mathbf{=}\mathbf{0} \end{gathered}$$

One will solve this system using the method of elimination. Let eliminate$\mathbf{z}$. We can obtain the first equation without $\mathbf{z}$ by adding the first and the second equation together, so one will get:

$\mathbf{D}\left[\mathbf{x}\right]\mathbf{+}\mathbf{x}\mathbf{-}\mathbf{D}\left[\mathbf{y}\right]\mathbf{-}\mathbf{y}\mathbf{=}\mathbf{0}$

To obtain the second equation one will add the second equation of the initial system "multiplied" by to the third equation:

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

So now we are solving a new system:

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

By subtracting the second equation from the first, one will get ${\mathbf{D}}^{\mathbf{2}}\mathbf{\left[}\mathbf{y}\mathbf{\right]}\mathbf{-}\mathbf{D}\mathbf{\left[}\mathbf{y}\mathbf{\right]}\mathbf{-}\mathbf{2}\mathbf{y}\mathbf{=}\mathbf{0}$

The corresponding auxiliary equation is ${\mathbf{r}}^{\mathbf{2}}\mathbf{-}\mathbf{r}\mathbf{-}\mathbf{2}\mathbf{=}\mathbf{0}$ and its roots are:

${\mathbf{r}}_{\mathbf{1}\mathbf{,}\mathbf{2}}\mathbf{=}\frac{\mathbf{1}\mathbf{\pm }\sqrt{\mathbf{1}\mathbf{+}\mathbf{8}}}{\mathbf{2}}\Rightarrow {\mathbf{r}}_{\mathbf{1}}\mathbf{=}\mathbf{-}\mathbf{1}\mathbf{,}{\mathbf{r}}_{\mathbf{2}}\mathbf{=}\mathbf{2}$

So, the general solution for$y$is $\mathbf{y}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{-}\mathbf{t}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\mathbf{2}\mathbf{t}}$

One will now find the general solution for$x$. The equation $\mathbf{D}\left[\mathbf{x}\right]\mathbf{+}\mathbf{x}\mathbf{-}\mathbf{D}\left[\mathbf{y}\right]\mathbf{-}\mathbf{y}\mathbf{=}\mathbf{0}$ gives us that$\mathbf{D}\left[\mathbf{x}\right]\mathbf{+}\mathbf{x}\mathbf{=}\mathbf{D}\left[\mathbf{y}\right]\mathbf{+}\mathbf{y}$.

So, one will find the first derivative of y and substitute it into the previous equation to obtain$\mathbf{D}\left[\mathbf{x}\right]\mathbf{-}\mathbf{x}$.

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

One can find a homogeneous solution. The auxiliary equation is$\mathbf{r}\mathbf{+}\mathbf{1}\mathbf{=}\mathbf{0}$and its solution is $\mathbf{r}\mathbf{=}\mathbf{-}\mathbf{1}$ so the homogeneous solution for $x$ is ${\mathbf{x}}_{\mathbf{h}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{d}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{-}\mathbf{t}}$

Now, one can find a particular solution using the method of undetermined coefficients.

Assume that${\mathbf{x}}_{\mathbf{p}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{A}{\mathbf{e}}^{\mathbf{2}\mathbf{t}}$. One has that${\mathbf{x}}_{\mathbf{p}}^{}\mathbf{\text{'}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{2}\mathbf{A}{\mathbf{e}}^{\mathbf{2}\mathbf{t}}$, so

$$\begin{gathered} \mathbf{D}\left[{\mathbf{x}}_{\mathbf{p}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\right]\mathbf{+}{\mathbf{x}}_{\mathbf{p}}{\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{x}}_{\mathbf{p}}^{\mathbf{\backslash }\mathbf{cent}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{+}{\mathbf{x}}_{\mathbf{p}}\mathbf{\left(}\mathbf{t}\mathbf{\right)} \\ \mathbf{=}\mathbf{2}{\mathbf{Ae}}^{\mathbf{2}\mathbf{t}}\mathbf{+}{\mathbf{Ae}}^{\mathbf{2}\mathbf{t}} \\ \mathbf{=}\mathbf{3}{\mathbf{Ae}}^{\mathbf{2}\mathbf{t}} \\ \mathbf{=}\mathbf{3}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\mathbf{2}\mathbf{t}} \\ \mathbf{A}\mathbf{=}{\mathbf{c}}_{\mathbf{2}} \\ {\mathbf{x}}_{\mathbf{p}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\mathbf{2}\mathbf{t}} \end{gathered}$$

So, the general solution for $x$is $\mathbf{x}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{x}}_{\mathbf{h}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{+}{\mathbf{x}}_{\mathbf{p}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{d}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{-}\mathbf{t}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\mathbf{2}\mathbf{t}}$.

Let find$z$. One has from$\mathbf{D}\left[\mathbf{x}\right]\mathbf{-}\mathbf{y}\mathbf{-}\mathbf{z}\mathbf{=}\mathbf{0}$, one has that

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

It remains to find the constants$c_1$ , c₂ and d₁. One will find them from the initial conditions which are $\mathbf{x}\left(\mathbf{0}\right)\mathbf{=}\mathbf{y}\left(\mathbf{0}\right)\mathbf{=}\mathbf{2}$ and$\mathbf{z}\left(\mathbf{0}\right)\mathbf{=}\mathbf{-}\mathbf{1}$, so one has to solve the following system:

$$\begin{gathered} \mathbf{x}\left(\mathbf{0}\right)\mathbf{=}{\mathbf{d}}_{\mathbf{1}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}\mathbf{=}\mathbf{2} \\ \mathbf{y}\left(\mathbf{0}\right)\mathbf{=}{\mathbf{c}}_{\mathbf{1}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}\mathbf{=}\mathbf{2} \\ \mathbf{z}\left(\mathbf{0}\right)\mathbf{=}\mathbf{-}{\mathbf{c}}_{\mathbf{1}}\mathbf{-}{\mathbf{d}}_{\mathbf{1}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}\mathbf{=}\mathbf{-}\mathbf{1} \end{gathered}$$

From the first and the second equation one has that${\mathbf{c}}_{\mathbf{1}}\mathbf{=}{\mathbf{d}}_{\mathbf{1}}$. So, substituting it into the third equation one will get that$\mathbf{-}\mathbf{2}{\mathbf{c}}_{\mathbf{1}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}\mathbf{=}\mathbf{-}\mathbf{1}$. Also, from the second equation one has that${\mathbf{c}}_{\mathbf{2}}\mathbf{=}\mathbf{2}\mathbf{-}{\mathbf{c}}_{\mathbf{1}}$, so one has that:

$$\begin{gathered} \mathbf{-}{\mathbf{c}}_{\mathbf{1}}\mathbf{-}{\mathbf{c}}_{\mathbf{1}}\mathbf{+}\mathbf{(}\mathbf{2}\mathbf{-}{\mathbf{c}}_{\mathbf{1}}\mathbf{)}\mathbf{=}\mathbf{-}\mathbf{1} \\ \mathbf{-}\mathbf{3}{\mathbf{c}}_{\mathbf{1}}\mathbf{+}\mathbf{2}\mathbf{=}\mathbf{-}\mathbf{1} \\ \mathbf{-}\mathbf{3}{\mathbf{c}}_{\mathbf{1}}\mathbf{=}\mathbf{-}\mathbf{1}\mathbf{-}\mathbf{2} \\ \mathbf{-}\mathbf{3}{\mathbf{c}}_{\mathbf{1}}\mathbf{=}\mathbf{-}\mathbf{3} \\ \Rightarrow {\mathbf{c}}_{\mathbf{1}}\mathbf{=}\mathbf{1} \\ {\mathbf{d}}_{\mathbf{1}}\mathbf{=}{\mathbf{c}}_{\mathbf{1}}\mathbf{=}\mathbf{1} \\ {\mathbf{c}}_{\mathbf{2}}\mathbf{=}\mathbf{2}\mathbf{-}{\mathbf{c}}_{\mathbf{1}} \\ \mathbf{=}\mathbf{2}\mathbf{-}\mathbf{1} \\ {\mathbf{c}}_{\mathbf{2}}\mathbf{=}\mathbf{1} \end{gathered}$$

So, one has that the solution for the given initial value problem is:

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