Example 3. Solve the system of equations

$$
\left\{\begin{array}{l}
x^{\prime}=3 x-y+z \\
y^{\prime}=-x+5 y-z \\
z^{\prime}=x-y+3 z
\end{array}\right.
$$
Solution. We will reduce the given system to a single third-order equation using the method described above in point $2^{\circ}$. The order of using one or another equation can be arbitrary.

1) Differentiate (with respect to $t$) the first equation of the system: $x^{\prime \prime}=3 x^{\prime}-$ $-y^{\prime}+z^{\prime}$. In this equality, replace the difference $z^{\prime}-y^{\prime}$, found from the other two equations: $z^{\prime}-y^{\prime}=2 x-6 y+4 z$.

We obtain a second-order equation in terms of $x$, but with extraneous unknown functions $y$ and $z: x^{\prime \prime}=3 x^{\prime}+2 x-6 y+4 z$.

2) Differentiate this equality again with respect to $t: x^{\prime \prime \prime}=3 x^{\prime \prime}+$ $+2 x^{\prime}-6 y^{\prime}+4 z^{\prime}$ and express $-6 y^{\prime}+4 z^{\prime}$ in terms of the variables $x, y, z$. For this, we need to perform some transformations of the original system: multiply the second equation by -6, the third by 4, and add the results (we get the desired expression): $-6 y^{\prime}+4 z^{\prime}=10 x-34 y+18 z$.

Substituting this expression into the previous one, we get: $x^{\prime \prime \prime}=3 x^{\prime \prime}+$ $+2 x^{\prime}+10 x-34 y+18 z$.

3) Now let's form a new system of three equations: the first equation of the original system; the equation obtained at the end of step 1); the last equation we obtained before this paragraph. We will write the equations as follows:

$$
\left\{\begin{aligned}
x^{\prime}-3 x & =-y+z, \\
x^{\prime \prime}-3 x^{\prime}-2 x & =-6 y+4 z, \\
x^{\prime \prime \prime}-3 x^{\prime \prime}-2 x^{\prime}-10 x & =-34 x+18 z
\end{aligned}\right.
$$

In this system, we also have extraneous unknowns $y$ and $z$, which we will eliminate as follows. Transform the first two equations of this system: first, multiply the first equation by -6 and add it to the second equation; then multiply this same first equation by -4 and add it to the second equation. We will get two auxiliary equations, which we will combine into a system:

$$
\left\{\begin{array}{rr}
x^{\prime \prime}-9 x^{\prime}+16 x=-2 z, & 9 \\
x^{\prime \prime}-7 x^{\prime}+10 x=-2 y ; & -17
\end{array}\right.
$$

in one of them $y$ is absent, and in the other $z$ is absent.
4) Now let's turn to this system: multiply the first equation by 9, the second by -17, and add the results to the last equation of the system formed at the beginning of step 3). By doing this, we eliminate the extraneous functions from the third-order equation in terms of $x: x^{\prime \prime \prime}-11 x^{\prime \prime}+36 x^{\prime}-36 x=0$. We obtain the desired third-order differential equation with one unknown function $x$, which we solve in the usual way.

The characteristic equation $k^{3}-11 k^{2}+36 k-36=0$ has three simple roots: $k_{\mathrm{I}}=2, k_{2}=3, k_{3}=6$, and the general solution of the differential equation is $x=C_{1} e^{2 t}+C_{2} e^{3 t}+C_{3} e^{6 t}$.

5) The other unknown functions $y$ and $z$ will be found from the intermediate equations (relations) obtained above in step 3). From this, in particular, we get $y=\frac{1}{2}\left(-x^{\prime \prime}+7 x^{\prime}-10 x\right), z=\frac{1}{2}\left(-x^{\prime \prime}+9 x^{\prime}-16 x\right)$. It remains to substitute these expressions for $x$ and its derivatives (we suggest doing this independently).

$$
\text { Answer. }\left\{\begin{array}{lr}
x=C_{1} e^{2 t} & +C_{2} e^{3 t}+C_{3} e^{6 t} \\
y= & C_{2} e^{3 t}-2 C_{3} e^{6 t} \\
z=-C_{1} e^{2 t}+C_{2} e^{3 t}+C_{3} e^{6 t}
\end{array}\right.
$$

Note. We suggest finding a more economical way to obtain the differential equation from step 4).