Example 3. Solve the system

$$
\left\{\begin{array}{l}
\frac{d x}{d t}=2 x+y \\
\frac{d y}{d t}=4 y-x
\end{array}\right.
$$
Solution. The characteristic equation

$$
\left|\begin{array}{cc}
2-r & 1 \\
-1 & 4-r
\end{array}\right|=0, \text { or } r^{2}-6 r+9=0
$$

has a multiple root $r_{1}=r_{2}=3$.

The solution should be sought in the form

$$
x=\left(\lambda_{1}+\mu_{1} t\right) e^{3 t}, \quad y=\left(\lambda_{2}+\mu_{2} t\right) e^{3 t}
$$

Substituting (13) into the first equation of the system (12), we get

$$
3\left(\lambda_{1}+\mu_{1} t\right)+\mu_{1}=2\left(\lambda_{1}+\mu_{1} t\right)+\left(\lambda_{2}+\mu_{2} t\right) .
$$

Equating the coefficients of the same powers of $t$ on the left and right sides of (14), we obtain:

$$
\begin{aligned}
3 \lambda_{1}+\mu_{1} & =2 \lambda_{1}+\lambda_{2}, \\
3 \mu_{1} & =2 \mu_{1}+\mu_{2},
\end{aligned}
$$

from which

$$
\lambda_{2}=\lambda_{1}+\mu_{1}, \quad \mu_{2}=\mu_{1}
$$

The quantities $\lambda_{1}$ and $\mu_{1}$ remain arbitrary. Denoting them by $C_{1}$ and $C_{2}$ respectively, we obtain the general solution of the system (12):

$$
x=\left(C_{1}+C_{2} t\right) e^{3 t}, \quad y=\left(C_{1}+C_{2}+C_{2} t\right) e^{3 t}
$$

Remark. It is easy to verify that if (13) is substituted into the second equation of the system (12), we will get the same result (15). Indeed, from the equality

$$
\mu_{2}+3\left(\lambda_{2}+\mu_{2} t\right)=4\left(\lambda_{2}+\mu_{2} t\right)-\left(\lambda_{1}+\mu_{1} t\right)
$$

we get two relations for determining $\lambda_{2}$ and $\mu_{2}$ in terms of $\lambda_{1}$ and $\mu_{1}$ :

$$
\begin{aligned}
\mu_{2}+3 \lambda_{2} & =4 \lambda_{2}-\lambda_{1}, \\
3 \mu_{2} & =4 \mu_{2}-\mu_{1},
\end{aligned}
$$

from which $\lambda_{2}=\lambda_{1}+\mu_{2}, \mu_{2}=\mu_{1}$.