= Solution
The factor satisfying the left condition is $x$, while $x^{-1}$ satisfies decay at infinity. Matching the value and unit derivative jump gives
$$
\boxed{G_1(x,\xi)=\begin{cases}-x/2,&x\leq\xi,\\-\xi^2/(2x),&x\geq\xi.\end{cases}}
$$
For the specified compactly supported forcing, integrating over $0<\xi<1$ gives
$$
\boxed{y_1(x)=\begin{cases}x^2/3-x/2,&0<x\leq1,\\-1/(6x),&x\geq1.\end{cases}}
$$
For example, on $0<x<1$, the two pieces contribute $-x^2/6$ and $-x(1-x)/2$. Both $y_1$ and $y_1'$ match at one, and the function tends to zero at both ends. A homogeneous difference $ax+b/x$ with those two limits must vanish, establishing uniqueness.
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