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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 $x\ge 0$.

The left boundary solution is $u(x)=x+1$, since $u^{\prime }(0)=u(0)=1$; 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 giveThereforeThe 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 $\xi =x$ givesIt has $y(0)=y^{\prime }(0)=1$ 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 $c=0$.