= Solution
The zero initial value and slope select a forward <Green's function>: set it to zero for $x<\xi$, and choose the right piece to start at zero with slope one. This gives
$$
\boxed{G_2(x,\xi)=\begin{cases}0,&x\leq\xi,\\\tfrac12(x-\xi^2/x),&x\geq\xi.\end{cases}}
$$
For $f=1$ on $(0,1)$ and zero elsewhere,
$$
\boxed{y_2(x)=\begin{cases}x^2/3,&0<x\leq1,\\x/2-1/(6x),&x\geq1.\end{cases}}
$$
Indeed $y_2=\int_0^{\min(x,1)}(x-\xi^2/x)d\xi/2$. It and its first <derivative> match at one, while $y_2,y_2'\to0$ at zero. The only homogeneous solution with these two initial limits is zero, so this solution too is unique. Its growth at infinity is allowed by these boundary conditions.
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