Example 2. Solve the Cauchy problem for the system

$$
\begin{gathered}
\left\{\begin{array}{l}
\frac{d x}{d t}=3 x+8 y \\
\frac{d y}{d t}=-x-3 y
\end{array}\right. \\
x(0)=6, \quad y(0)=-2
\end{gathered}
$$
Solution. From the second equation of the system (7), we find

$$
x=-3 y-\frac{d y}{d t}
$$

from which

$$
\frac{d x}{d t}=-3 \frac{d y}{d t}-\frac{d^{2} y}{d t^{2}}
$$

Substituting (9) and (10) into the first equation of the system (7), we obtain the equation $\frac{d^{2} y}{d t^{2}}-y=0$, the general solution of which is

$$
y=C_{1} e^{t}+C_{2} e^{-t}
$$

Substituting (11) into (9), we find

$$
x=-4 C_{1} e^{t}-2 C_{2} e^{-t}
$$

The general solution of the system (7)

$$
x=-4 C_{1} e^{t}-2 C_{2} e^{-t}, \quad y=C_{1} e^{t}+C_{2} e^{-t} .
$$

Given the initial conditions (8), from (12) we obtain the system of equations for determining $C_{1}, C_{2}:$

$$
\left\{\begin{aligned}
-4 C_{1}-2 C_{2} & =6 \\
C_{1}+C_{2} & =-2
\end{aligned}\right.
$$

Solving this system, we find $C_{1}=-1, C_{2}=-1$. Substituting these values of $C_{1}$ and $C_{2}$ into (12), we obtain the solution to the given Cauchy problem:

$$
x=4 e^{t}+2 e^{-t}, \quad y=-e^{t}-e^{-t}
$$