 Solution

Choose a left-end solution $u(x)=e^{-x}\sin x$ and a right-end solution $v(x)=e^{-x}(\sin x+\cos x)$. They satisfy $u(0)=0$ and $v^{\prime }(2\pi )=0$. Their Wronskian, with the order used here, isThe Green's function is $u(x_{<})v(x_{>})/W(\xi )$, where $x_{<}=\min (x,\xi )$ and $x_{>}=\max (x,\xi )$. Explicitly,The two expressions agree at the source. Their derivative difference is $[u(\xi )v^{\prime }(\xi )-u^{\prime }(\xi )v(\xi )]/W(\xi )=1$, checking both normalization and sign. The first-order derivative term means the kernel need not be symmetric under exchange of $x$ and $\xi$.