Example 1. Find the general solution of the equation $x y^{\prime \prime}+2 y^{\prime}+x y=0$, if $y_{1}=\frac{\sin x}{x}$ is its particular solution. then
Solution. Let $y=\frac{\sin x}{x} \cdot z$, where $z$ is a new unknown function of $x$;

$$
y^{\prime}=y_{1}^{\prime} z+y_{1} z^{\prime}, \quad y^{\prime \prime}=y_{1}^{\prime \prime} z+2 y_{1}^{\prime} z^{\prime}+y_{1} z^{\prime \prime}
$$

Substituting into the given equation, we get

$$
\left(x y_{1}^{\prime \prime}+2 y_{1}^{\prime}+x y_{1}\right) z+x y_{1} z^{\prime \prime}+2\left(x y_{1}^{\prime}+y_{1}\right) z^{\prime}=0
$$

But since $y_{1}=\frac{\sin x}{x}$ is a particular solution of the given equation, then $x y_{1}^{\prime \prime}+2 y_{1}^{\prime}+ x y_{1}=0$, so we have

$$
x y_{1} z^{\prime \prime}+2\left(x y_{1}^{\prime}+y_{1}\right) z^{\prime}=0
$$

But $y_{1}^{\prime}=\frac{\cos x}{x}-\frac{\sin x}{x^{2}}$, so $x y_{1}^{\prime}+y_{1}=\cos x$, and equation (37) will become

$$
z^{\prime \prime} \sin x+2 z^{\prime} \cos x=0
$$

We can rewrite it as

$$
\frac{z^{\prime \prime}}{z^{\prime}}+2 \frac{\cos x}{\sin x}=0
$$

From this, we have $\left(\ln \left|z^{\prime}\right|+2 \ln |\sin x|\right)^{\prime}=0$, hence

$$
\ln \left|z^{\prime}\right|+2 \ln |\sin x|=\ln \tilde{C}_{1}, \quad \text { or } \quad z^{\prime} \sin ^{2} x=\widetilde{C}_{1} .
$$

Integrating this equation, we find $z=-\tilde{C}_{1} \operatorname{ctg} x+C_{2}$ and, therefore, the general solution of the given equation is

$$
y=-\tilde{C}_{1} \frac{\cos x}{x}+C_{2} \frac{\sin x}{x}, \quad \text { or } \quad y=C_{1} \frac{\cos x}{x}+C_{2} \frac{\sin x}{x} \quad\left(C_{1}=-\tilde{C}_{1}\right)
$$