= Solution
The source factor cancels the singular denominator of the <Green function>. Splitting the <integral> at $\xi=x$ gives
$$
y(x)=v(x)\int_{-1}^x(1+\xi)e^{-\xi}d\xi+(1+x)\int_x^1\left[1-\frac e2(1+\xi)e^{-\xi}\right]d\xi.
$$
Using $\int(1+\xi)e^{-\xi}d\xi=-(\xi+2)e^{-\xi}$, these two integrals are $e-(x+2)e^{-x}$ and $5/2-x-e(x+2)e^{-x}/2$, respectively. Simplification gives
$$
\boxed{y(x)=e^{x+1}+\frac{1-e^2}{2}(1+x)-x^2.}
$$
The exponential and linear terms are homogeneous solutions, while direct <differentiation> gives $L(-x^2)=x$ for $x\ne0$. Also $y(-1)=y(1)=0$, and the expression is analytic through zero. Uniqueness in this regular class follows because any difference is a global <linear combination> $Ae^x+B(1+x)$: the left boundary forces $A=0$, and the right then forces $B=0$.
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