= Solution
Use the boundary-adapted <homogeneous solutions>
$$
u(x)=1+x,\qquad v(x)=e^x-\frac e2(1+x),\qquad W(x)=uv'-u'v=xe^x.
$$
For a source point $\xi\ne0$, the <Green function> is
$$
\boxed{G(x,\xi)=\frac1{\xi e^\xi}\begin{cases}u(x)v(\xi),&x\leq\xi,\\u(\xi)v(x),&x\geq\xi.\end{cases}}
$$
It vanishes at both endpoints and is continuous at $x=\xi$. Its first <derivative> has the jump
$$
G_x(\xi^+,\xi)-G_x(\xi^-,\xi)=\frac{u(\xi)v'(\xi)-u'(\xi)v(\xi)}{W(\xi)}=1,
$$
as required for a monic second-order <differential operator> with a <Dirac delta> source. Away from that source it is a <homogeneous solution> and extends smoothly through $x=0$.
\b[The interior singular point requires a qualification.] This is a <Green kernel with a vanishing Wronskian at an interior singular point>, with a $1/\xi$ singularity in its source variable. It is not an ordinary integral inverse for unrestricted forcing. For smooth forcing $f(\xi)=\xi q(\xi)$, however, the product $Gf$ extends continuously at zero and the integral is well defined. This includes part (iii). We use solutions with two continuous derivatives across zero, interpreting the equation there after multiplication by $x$. For such a solution, expansion at zero forces $f(0)=0$; arbitrary nonzero forcing at zero cannot satisfy this regularity. Merely imposing one continuous derivative would leave a piecewise homogeneous ambiguity, so the original singular coefficients alone do not specify an unrestricted boundary-value inverse.
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