Example 4. Solve the equation $y^{\text {IV }}-4 y^{\prime \prime \prime}+8 y^{\prime \prime}-8 y^{\prime}+4 y=0$.
Solution. Let's form the characteristic equation

$$
r^{4}-4 r^{3}+8 r^{2}-8 r+4=0
$$

Transforming the left side of this equation, we get

$$
\begin{gathered}
r^{4}-4 r^{3}+8 r^{2}-8 r+4=r^{4}-4 r^{3}+4 r^{2}+4 r^{2}-8 r+4= \\
=\left(r^{2}-2 r+2\right)^{2}=\left(r^{2}-2 r+2\right)\left(r^{2}-2 r+2\right)=0
\end{gathered}
$$

from which $r^{2}-2 r+2=0, r^{2}-2 r+2=0$.

Therefore, the complex conjugate roots of the characteristic equation $r_{1,2}=1 \pm i$ have multiplicity $k=2$. Since in this case $\alpha=1, \quad \beta=1$, the general solution, according to formulas (11.14) and (11.8), is

$$
y=C_{1} e^{x_{1}} \cos x+C_{2} e^{x} \sin x+C_{3} x e^{x} \cos x+C_{4} x e^{x} \sin x
$$

or

$$
y=e^{x}\left[\left(C_{1}+C_{3} x\right) \cos x+\left(C_{2}+C_{4} x\right) \sin x\right]
$$