Example 2. Find the general solution of the equation $y^{\prime \prime}+\frac{2}{x} y^{\prime}+y=\frac{1}{x},(x \neq 0)$.
Solution. The general solution of the corresponding homogeneous equation has the form (see Example 1)

$$
y_{0.0}=C_{1} \frac{\sin x}{x}+C_{2} \frac{\cos x}{x}
$$

and therefore, its fundamental system of solutions will be

$$
y_{1}=\frac{\sin x}{x}, \quad y_{2}=\frac{\cos x}{x}
$$

We will seek the general solution of the given equation by the method of variation of arbitrary constants:

$$
y=C_{1}(x) \frac{\sin x}{x}+C_{2}(x) \frac{\cos x}{x}
$$

where $C_{1}(x), C_{2}(x)$ are unknown functions of $x$ to be determined. To find them, we will set up the following system:

$$
\left\{\begin{array}{l}
C_{1}^{\prime}(x) \frac{\sin x}{x}+C_{2}^{\prime}(x) \frac{\cos x}{x}=0 \\
C_{1}^{\prime}(x) \frac{x \cos x-\sin x}{x^{2}}+C_{2}^{\prime}(x) \frac{-x \sin x-\cos x}{x^{2}}=\frac{1}{x}
\end{array}\right.
$$

From here we find: $C_{1}^{\prime}(x)=\cos x, C_{2}^{\prime}(x)=-\sin x$. Integrating, we get

$$
C_{1}(x)=\sin x+\widetilde{C}_{1}, \quad C_{2}(x)=\cos x+\widetilde{C}_{2}
$$

Substituting these values of $C_{1}(x)$ and $C_{2}(x)$ into the expression for $y$, we find the general solution of the given equation

$$
y=\widetilde{C}_{1} \frac{\sin x}{x}+\widetilde{C}_{2} \frac{\cos x}{x}+\frac{1}{x}
$$