= Solution
For $y=e^{-x}$, substitution in the unnormalized equation gives $[(x+2)-(x+1)-1]e^{-x}=0$. For $y=ax+b$, the remaining expression is $(x+1)a-(ax+b)=a-b$. Thus an independent second solution is $x+1$. Their <Wronskian> is $-(x+2)e^{-x}$, nonzero on $x\geq0$.
The left boundary solution is $u(x)=x+1$, since $u'(0)=u(0)=1$; the decaying right solution is $v(x)=e^{-x}$. A <Robin half-line Green function from left and right solutions> must be continuous at $x=\xi$, while its derivative jumps by one because the leading coefficient of $L$ is one. These conditions give
$$
G(x,\xi)=\frac{u(\min(x,\xi))v(\max(x,\xi))}{W(\xi)},\qquad W(\xi)=-(\xi+2)e^{-\xi}.
$$
Therefore
$$
\boxed{G(x,\xi)=\begin{cases}-\dfrac{x+1}{\xi+2},&0\leq x<\xi,\\-\dfrac{\xi+1}{\xi+2}e^{\xi-x},&x>\xi.\end{cases}}
$$
The two values agree at $x=\xi$, and $G_x(\xi^+,\xi)-G_x(\xi^-,\xi)=(\xi+1)/ (\xi+2)+1/(\xi+2)=1$. Each branch solves the homogeneous equation and its relevant boundary condition, so these checks prove the distributional identity $LG=\delta$ rather than merely postulate the kernel.
Integrate this <Green function> against the forcing $-(\xi+2)e^{-\xi}$. Splitting at $\xi=x$ gives
$$
y(x)=e^{-x}\int_0^x(\xi+1)d\xi+(x+1)\int_x^\infty e^{-\xi}d\xi=\boxed{\left(\frac{x^2}{2}+2x+1\right)e^{-x}.}
$$
It has $y(0)=y'(0)=1$ and decays at infinity. Direct substitution gives $L[y]=-(x+2)e^{-x}$. Uniqueness follows because a decaying homogeneous solution must be a multiple of $e^{-x}$, and its left boundary condition would require $-c=c$, forcing $c=0$.
Back to article page
