Example 6. Solve the system of differential equations

$$
\left\{\begin{array}{l}
x^{\prime}=3 x+4 y+2 z \\
y^{\prime}=x+4 y+z \\
z^{\prime}=4 x+6 y+5 z
\end{array}\right.
$$
Solution. The characteristic equation of the matrix $A=\left(\begin{array}{lll}3 & 4 & 2 \\ 1 & 4 & 1 \\ 4 & 6 & 5\end{array}\right)$ is: $\left|\begin{array}{ccc}3-\lambda & 4 & 2 \\ 1 & 4-\lambda & 1 \\ 4 & 6 & 5-\lambda\end{array}\right|=0$, or $(3-\lambda)(4-\lambda) \times$ $\times(5-\lambda)+12+16-6(3-\lambda)-4(5-\lambda)-8(4-\lambda)=0 . \quad$ From this, $\lambda^{3}-12 \lambda^{2}+29 \lambda-18=0$. This equation has three simple roots $\lambda_{1}=1, \lambda_{2}=2, \lambda_{3}=9$, which are the three eigenvalues of the matrix $A$.

Now let's find the eigenvectors as solutions to the homogeneous system of linear equations $(A-\lambda E) U=0$, where $E-$ is the identity matrix of the third order, $U=\left(u_{1}, u_{2}, u_{3}\right)^{\top}$ - a column of three unknowns.

1) $\lambda_{1}=1$. The corresponding homogeneous linear system has the form (after the system, its transformation and the eigenvector, whose coordinates we take as integers, are given):

$$
\left\{\begin{array} { r l } 
{ 2 u _ { 1 } + 4 u _ { 2 } + 2 u _ { 3 } } & { = 0 } \\
{ u _ { 1 } + 3 u _ { 2 } + u _ { 3 } } & { = 0 , } \\
{ 4 u _ { 1 } + 6 u _ { 2 } + 4 u _ { 3 } } & { = 0 }
\end{array} \Rightarrow \left\{\begin{array}{r}
u_{1}+u_{2}+u_{3}=0 \\
u_{2}=0
\end{array}\right.\right.
$$

we take $U_{1}=\left(\begin{array}{c}1 \\ 0 \\ -1\end{array}\right)$. The particular solution corresponding to the eigenvalue $\lambda_{1}=1$ and the eigenvector $U_{1}$ is: ( $x_{1}=e^{t}$, $\left.y_{1}=0, z_{1}=-e^{t}\right)$.
2) $\lambda_{2}=2$. We have:

$$
\left\{\begin{array} { r l } 
{ u _ { 1 } + 4 u _ { 2 } + 2 u _ { 3 } } & { = 0 } \\
{ u _ { 1 } + 2 u _ { 2 } + u _ { 3 } } & { = 0 , } \\
{ 4 u _ { 1 } + 6 u _ { 2 } + 3 u _ { 3 } } & { = 0 }
\end{array} \Rightarrow \left\{\begin{array}{rl}
u_{1} & =0 \\
2 u_{2}+u_{3} & =0
\end{array}\right.\right.
$$

we take $U_{2}=\left(\begin{array}{c}0 \\ 1 \\ -2\end{array}\right)$. We write down the triplet of functions - the particular solution of the system: $\left(x_{2}=0, y_{2}=e^{2 t}, z_{1}=-2 e^{2 t}\right)$.
3) $\lambda_{3}=9$. We have:

$$
\left\{\begin{array} { r l } 
{ - 6 u _ { 1 } + 4 u _ { 2 } + 2 u _ { 3 } } & { = 0 } \\
{ u _ { 1 } - 5 u _ { 2 } + u _ { 3 } } & { = 0 , } \\
{ 4 u _ { 1 } + 6 u _ { 2 } - 4 u _ { 3 } } & { = 0 }
\end{array} \Rightarrow \left\{\begin{array}{rl}
-3 u_{1}+2 u_{2}+u_{3} & =0 \\
u_{1}-5 u_{2}+u_{3} & =0
\end{array}\right.\right.
$$

We take $U_{3}=\left(\begin{array}{c}7 \\ 4 \\ 13\end{array}\right)$. The particular solution corresponding to the eigenvalue $\lambda=9$ is: $\left(x_{3}=7 e^{9 t}, y_{3}=4 e^{9 t}, z_{3}=13 e^{9 t}\right)$. We form the general solution of the original system.

$$
\text { Answer. }\left\{\begin{array}{lr}
x=C_{1} e^{t} & +7 C_{3} e^{9 t} \\
y= & C_{2} e^{2 t}+4 C_{3} e^{9 t} \\
z=-C_{1} e^{t}-2 C_{2} e^{2 t}+13 C_{3} e^{9 t}
\end{array}\right.
$$