Example 2. Solve the equation $y^{\prime \prime \prime}+y^{\prime}=x^{4}$.
Solution. The right-hand side of the equation is a function of the form (11.18), for which $\alpha=0, m=4$.

The characteristic equation

$$
r^{3}+r=0 \text { or } r\left(r^{2}+1\right)=0
$$

has roots $r_{1}=0, r_{2}=i, r_{3}=-i$, so the general solution of the homogeneous equation $y^{\prime \prime \prime}+y^{\prime}=0$ is given by the formula

$$
y_{0}=C_{1}+C_{2} \cos x+C_{3} \sin x
$$

Since the number $\alpha=0$ is a simple root of the characteristic equation, we seek a particular solution in the form

$$
\bar{y}=x e^{0 \cdot x}\left(A x^{4}+B x^{3}+C x^{2}+D x+E\right)
$$

or

$$
\bar{y}=A x^{5}+B x^{4}+C x^{3}+D x^{2}+E x
$$

We find the derivatives of the function $\bar{y}$:

$$
\begin{gathered}
\bar{y}^{\prime}=5 A x^{4}+4 B x^{3}+3 C x^{2}+2 D x+E \\
\bar{y}^{\prime \prime}=20 A x^{3}+12 B x^{2}+6 C x+2 D \\
\bar{y}^{\prime \prime \prime}=60 A x^{2}+24 B x+6 C
\end{gathered}
$$

Substituting the expressions for $\bar{y}^{\prime}, \bar{y}^{\prime \prime \prime}$ into the given equation, we obtain the identity

$$
60 A x^{2}+24 B x+6 C+5 A x^{4}+4 B x^{3}+3 C x^{2}+2 D x+E=x^{4}
$$

from which

$$
5 A=1,4 B=0,60 A+3 C=0,24 B+2 D=0,6 C+E=0
$$

i.e.

$$
A=\frac{1}{5}, B=0, C=-4, D=0, E=24
$$

Thus

$$
\bar{y}=\frac{1}{5} x^{5}-4 x^{3}+24 x
$$

$\mathbf{4}$

$$
y=C_{1}+C_{2} \cos x+C_{3} \sin x+\frac{x^{5}}{5}-4 x^{3}+24 x
$$