= Solution
Away from $x=\xi$, the homogeneous equation has solutions $e^x,e^{-x}$. A basis adapted to the two <boundary conditions> is $u(x)=\sinh x$ at the origin and $v(x)=e^{-x}$ at infinity. Their <Wronskian> is $uv'-u'v=-1$. The <Green function> must be continuous at $\xi$, while <integration> of $-G_{xx}+G=\delta(x-\xi)$ across the source gives
$$
G_x(\xi+,\xi)-G_x(\xi-,\xi)=-1.
$$
These conditions give the <Dirichlet half-line Green function for minus d2 plus 1>:
$$
\boxed{G(x,\xi)=\sinh(x_<)e^{-x_>}=\frac12\left(e^{-|x-\xi|}-e^{-(x+\xi)}\right),}
$$
where $x_<=\min(x,\xi)$ and $x_>=\max(x,\xi)$. The second term is the image that enforces $G(0,\xi)=0$; both terms decay as $x\to\infty$.
For the finite forcing interval, $y(x)=\int_0^\mu G(x,\xi)\,d\xi$. When $x\leq\mu$, split the <integral> at $\xi=x$:
$$
y(x)=e^{-x}\int_0^x\sinh\xi\,d\xi+\sinh x\int_x^\mu e^{-\xi}\,d\xi=1-e^{-x}-e^{-\mu}\sinh x.
$$
When $x\geq\mu$, the entire source interval lies to the left, giving the <Dirichlet half-line response to uniform interval forcing>
$$
\boxed{y(x)=\begin{cases}1-e^{-x}-e^{-\mu}\sinh x,&0\leq x\leq\mu,\\(\cosh\mu-1)e^{-x},&x\geq\mu.\end{cases}}
$$
This solution is unique: a homogeneous difference satisfying the origin condition is proportional to $\sinh x$, which cannot decay at infinity unless its coefficient is zero.
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