Example 2. Solve the system

$$
\left\{\begin{array}{l}
\frac{d x}{d t}=x-5 y \\
\frac{d y}{d t}=2 x-y
\end{array}\right.
$$
Solution. Let's write down the system for determining $\lambda$ and $\mu$:

$$
\left\{\begin{array}{l}
(1-r) \lambda-5 \mu=0 \\
2 \lambda-(1+r) \mu=0
\end{array}\right.
$$

The characteristic equation

$$
\left|\begin{array}{cc}
1-r & -5 \\
2 & -1-r
\end{array}\right|=0
$$

has roots $r_{1}=3 i, r_{2}=-3 i$. Substituting $r_{1}=3 i$ into (8), we get two equations for determining $\lambda_{1}$ and $\mu_{1}$:

$$
(1-3 i) \lambda_{1}-5 \mu_{1}=0, \quad 2 \lambda_{1}-(1+3 i) \mu_{1}=0
$$

one of which is a consequence of the other (due to the fact that the determinant of system (8) is zero).

Let's take $\lambda_{1}=5, \mu_{1}=1-3 i$, then the first particular solution is written as:

$$
x_{1}=5 e^{3 i t}, \quad y_{1}=(1-3 i) e^{3 i t} .
$$

Similarly, substituting the root $r_{2}=-3 i$ into (8), we find the second particular solution:

$$
x_{2}=5 e^{-3 i t}, \quad y_{2}=(1+3 i) e^{-3 i t}
$$

Let's transition to a new fundamental system of solutions:

$$
\begin{array}{ll}
\tilde{x}_{1}=\frac{x_{1}+x_{2}}{2}, & \tilde{x}_{2}=\frac{x_{1}-x_{2}}{2 i} \\
\tilde{y}_{1}=\frac{y_{1}+y_{2}}{2}, & \tilde{y}_{2}=\frac{y_{1}-y_{2}}{2 i}
\end{array}
$$

Using the known Euler's formula $e^{ \pm \text {ait }}=\cos \alpha t \pm i \sin \alpha t$, from (9), (10), and (11) we get

$$
\begin{array}{ll}
\widetilde{x}_{1}=5 \cos 3 t, & \widetilde{x}_{2}=5 \sin 3 t, \\
\bar{y}_{1}=\cos 3 t+3 \sin 3 t, & \bar{y}_{2}=\sin 3 t-3 \cos 3 t .
\end{array}
$$

The general solution of system (7) will be

$$
\begin{aligned}
& x=C_{1} \tilde{x}_{1}+C_{2} \tilde{x}_{2}=5 C_{1} \cos 3 t+5 C_{2} \sin 3 t \\
& y=C_{1} \tilde{y}_{1}+C_{2} \widetilde{y}_{2}=C_{1}(\cos 3 t+3 \sin 3 t)+C_{2}(\sin 3 t-3 \cos 3 t)
\end{aligned}
$$

Remark. Having found the first particular solution (9), we could have immediately written the general solution of system (7) using the formulas

$$
x=C_{1} \operatorname{Re} x_{1}+C_{2} \operatorname{Im} x_{1}, \quad y=C_{1} \operatorname{Re} y_{1}+C_{2} \operatorname{Im} y_{1}
$$

where $\operatorname{Re} z$ and $\operatorname{Im} z$ denote the real and imaginary parts of the complex number $z$, i.e., if $z=a+b i$, then $\operatorname{Re} z=a, \operatorname{Im} z=b$.

V. The case of multiple roots.