Example 2. Construct the Green's function for the differential equation

$$
x y^{\prime \prime}+y^{\prime}=0
$$

subject to the following conditions:

$$
y(x) \text { is bounded as } x \rightarrow 0, \quad y(1)=\alpha y^{\prime}(1), \quad \alpha \neq 0 .
$$
Solution. First, we find the general solution of equation (23) and verify that conditions (24) are satisfied only when

$$
y(x) \equiv 0
$$

Indeed, denoting $y^{\prime}(x)=z(x)$, we get $x z^{\prime}+z=0$, from which $\ln z= \ln c_{1}-\ln x, z=\frac{c_{1}}{x}$, and thus,

$$
y(x)=c_{1} \ln x+c_{2}
$$

It is clear that $y(x)$, defined by formula (25), satisfies conditions (24) only when $c_{1}=c_{2}=0$, which means that the Green's function for the problem (23)-(24) can be constructed.

Formally, write $G(x, \xi)$ as

$$
G(x, \xi)= \begin{cases}a_{1}+a_{2} \ln x & \text { for } \quad 0<x \leqslant \xi \\ b_{1}+b_{2} \ln x & \text { for } \quad \xi \leqslant x \leqslant 1\end{cases}
$$

From the continuity of $G(x, \xi)$ at $x=\xi$, we get

$$
b_{1}+b_{2} \ln \xi-a_{1}-a_{2} \ln \xi=0
$$

and the jump of $G_{x}^{\prime}(x, \xi)$ at the point $x=\xi$ is $\frac{1}{\xi}$, so

$$
b_{2} \cdot \frac{1}{\xi}-a_{2} \cdot \frac{1}{\xi}=\frac{1}{\xi}
$$

Setting

$$
c_{1}=b_{1}-a_{1}, \quad c_{2}=b_{2}-a_{2}
$$

we have

$$
\left\{\begin{aligned}
c_{1}+c_{2} \ln \xi & =0 \\
c_{2} & =1
\end{aligned}\right.
$$

from which

$$
c_{1}=-\ln \xi, \quad c_{2}=1
$$

Now, use conditions (24). The boundedness of $G(x, \xi)$ as $x \rightarrow 0$ gives us $a_{2}=0$, and from the condition $G(1, \xi)=\alpha G_{x}^{\prime}(1, \xi)$ we get $b_{1}=\alpha b_{2}$. Considering (27) and (28), we obtain the values of all coefficients in (26):

$$
a_{1}=\alpha+\ln \xi, \quad a_{2}=0, \quad b_{1}=\alpha, \quad b_{2}=1
$$

Thus,

$$
G(x, \xi)= \begin{cases}\alpha+\ln \xi, & 0<x \leqslant \xi \\ \alpha+\ln x, & \xi \leqslant x \leqslant 1\end{cases}
$$