Example 2. Solve the equation $y^{\prime \prime \prime}-7 y^{\prime \prime}+15 y^{\prime}-9 y=0$.
Solution. Let's form the characteristic equation

$$
r^{3}-7 r^{2}+15 r-9=0
$$

one of the roots of which $r_{1}=1$ can be obtained by trial and error.

Since

$$
\left(r^{3}-7 r^{2}+15 r-9\right):(r-1)=r^{2}-6 r+9
$$

the equation becomes

$$
(r-1)\left(r^{2}-6 r+9\right)=0
$$

from which $r_{2}=r_{3}=3$. Thus, the characteristic equation has one simple (single) root $r_{1}=1$ and a double root $r_{2}=3(k=2)$.

According to formulas (11.12) and (11.8), we obtain the general solution

$$
y=C_{1} e^{x}+C_{2} e^{3 x}+C_{3} x e^{3 x}
$$

or

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