Example 4. Solve the equation

$$
y^{\mathrm{IV}}-y=e^{\alpha x}+e^{-\alpha x}+\cos \beta x
$$
Solution. The characteristic equation

$$
r^{4}-1=0
$$

has roots $r_{1}=1, r_{2}=-1, r_{3}=i, r_{4}=-i$, therefore

$$
y_{0}=C_{1} e^{x}+C_{2} e^{-x}+C_{3} \cos x+C_{4} \sin x
$$

In accordance with formula (11.25), we seek a particular solution:

$$
\bar{y}=\bar{y}_{1}+\bar{y}_{2}+\bar{y}_{3},
$$

where

$$
\bar{y}_{1}=A e^{\alpha x}, \bar{y}_{2}=B e^{-\alpha x}, \bar{y}_{3}=C \sin \beta x+D \cos \beta x
$$

are particular solutions of the equations

$$
y^{\mathrm{IV}}-y=e^{\alpha x}, y^{\mathrm{IV}}-y=e^{-\alpha x}, y^{\mathrm{IV}}-y=\cos \beta x
$$

Finding the derivatives $\bar{y}_{1}^{\mathrm{IV}}, \bar{y}_{2}^{\mathrm{IV}}, \bar{y}_{3}^{\mathrm{IV}}$ and substituting their expressions and $\bar{y}_{1}, \bar{y}_{2}, \bar{y}_{3}$ into the last equations, we get:

$$
A=\frac{1}{\alpha^{4}-1}, B=\frac{1}{\alpha^{4}-1}, C=0, D=\frac{1}{\beta^{4}-1}
$$

The general solution of the original equation is

$$
y=C_{1} e^{x}+C_{2} e^{-x}+C_{3} \sin x+C_{4} \cos x+\frac{e^{\alpha x}+e^{-\alpha x}}{\alpha^{4}-1}+\frac{\cos \beta x}{\beta^{4}-1} .
$$

Remark. In the case where $\alpha=\beta=1$, the general solution will be:

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

In this case, particular solutions should be sought in the form:

$$
\bar{y}_{1}=A x e^{x} ; \bar{y}_{2}=B x e^{-x} ; \bar{y}_{3}=x(C \sin x+D \cos x)
$$