= Solution
Choose the <causal Green function> convention $L_xG(x,\xi)=\delta(x-\xi)$ for each source point $\xi>0$, with $G(x,\xi)=0$ when $x<\xi$. This automatically gives the two zero-origin limits. The kernel is continuous at $x=\xi$, and integrating its equation across the source requires the unit <derivative> jump $\partial_xG(\xi^+,\xi)-\partial_xG(\xi^-,\xi)=1$.
For $x>\xi$, the homogeneous basis from part (a) gives a continuous kernel proportional to $x^r\log(x/\xi)$. Its derivative at $x=\xi$ is $\xi^{r-1}$, so the normalization is $\xi^{1-r}$. Hence the <causal Green function for a repeated-root Euler operator> is
$$
\boxed{G(x,\xi)=\begin{cases}x^r\xi^{1-r}\log(x/\xi),&x\geq\xi,\\0,&0\leq x<\xi,\end{cases}\qquad r=\frac{1-\alpha}{2},\quad\xi>0.}
$$
The same formula follows from $[y_1(\xi)y_2(x)-y_2(\xi)y_1(x)]/W(\xi)$ on the right of the source, with $W(\xi)=\xi^{-\alpha}$.
The singular endpoint matters for uniqueness. If $\alpha<-1$, then $r>1$, and both $x^r$ and $x^r\log x$, together with their first <derivatives>, vanish at zero. Adding any source-dependent linear combination of these homogeneous solutions to $G$ preserves the two stated origin conditions. Thus those conditions alone do not determine a unique <Green's function> in that parameter range. The causal requirement used here selects the displayed kernel. For $\alpha\geq-1$ no nonzero homogeneous solution satisfies both zero-origin limits, so this ambiguity is absent.
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