= Solution
Away from $x=\xi$, the <Dirichlet Green function> solves the homogeneous equation. The endpoint conditions select $Ay_1(x)$ on the left and $By_2(x)$ on the right. Continuity and the unit derivative jump needed for a <Dirac delta function> give
$$
Ay_1(\xi)=By_2(\xi),\qquad
By_2'(\xi)-Ay_1'(\xi)=1.
$$
With <Wronskian> $W=y_1y_2'-y_1'y_2\ne0$, solve to obtain $A=y_2(\xi)/W(\xi)$ and $B=y_1(\xi)/W(\xi)$. Thus
$$
\boxed{G(x,\xi)=
\begin{cases}
y_1(x)y_2(\xi)/W(\xi),&x<\xi,\\
y_2(x)y_1(\xi)/W(\xi),&x>\xi.
\end{cases}}
$$
The first derivative term creates no extra delta because $G$ is continuous. The value at $x=\xi$ is the common limiting value.
For the specified constant-coefficient operator, take
$$
y_1(x)=e^{-x}-e^{-2x},\qquad
y_2(x)=e^{-x}-e^{1-2x},\qquad
W(\xi)=(e-1)e^{-3\xi}.
$$
Therefore its explicit <Green function> is
$$
\boxed{G(x,\xi)=\frac{e^{3\xi}}{e-1}
\begin{cases}
(e^{-x}-e^{-2x})(e^{-\xi}-e^{1-2\xi}),&x<\xi,\\
(e^{-x}-e^{1-2x})(e^{-\xi}-e^{-2\xi}),&x>\xi.
\end{cases}}
$$
For the forcing step and the homogeneous endpoint conditions inherited from this Green-function problem, $y(x)=2\int_{x_0}^1G(x,\xi)\,d\xi$. When $x<x_0$, every integration point lies on the first branch, so
$$
\boxed{y(x)=\frac{2(e^{-x}-e^{-2x})}{e-1}\int_{x_0}^1(e^{2\xi}-e^{1+\xi})\,d\xi
=-\frac{(e-e^{x_0})^2}{e-1}(e^{-x}-e^{-2x}).}
$$
Without those endpoint conditions, an arbitrary homogeneous solution could also be added.
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