Finding Green's Functions with Unmixed Boundary Conditions $L= (-\frac{d^2}{dx^2})-k^2$ that satisfies (and this is where I lose it) $u(0)=u(1)$ and $u'(0)=u'(1)$
I'm having trouble figuring out a practice problem I am working on.
I need to find the Green's function. I tried setting the equations equal to each other like $u(0)-u(1)=0$ and $u'(0)-u'(1)=0$ but was unsure if that was a good way to go or even correct for that matter.
 A: The Green's Function is defined as follows:
The Green's function $G(x,x')$ satisfies the ODE
$$\frac{d^2G(x,x')}{dx^2}+k^2G(x,x')=0 \tag 1$$
for $x\ne x'$.  
In addition, $G(x,x')$ is continuous at $x'=x$ and $\frac{dG}{d}$ is discontinuous at $x'=x$ such that
$$\begin{align}\lim_{\epsilon \to 0^+}\left(G(x,x+\epsilon)-G(x,x-\epsilon)\right)&=0 \tag 2\\\\
\lim_{\epsilon \to 0^+}\left(\frac{dG(x,x+\epsilon)}{dx}-\frac{dG(x,x-\epsilon)}{dx}\right)&=-1\tag 3
\end{align}$$
Finally, $G$ satisfies the boundary conditions 
$$G(0,x')=G(1,x') \tag 4$$
$$\left.\frac{dG(x,x')}{dx}\right|_{x=0}=\left.\frac{dG(x,x')}{dx}\right|_{x=1} \tag 5$$
Now, solving $(1)$, we find that 
$$G(x,x')=
\begin{cases}
A\sin(kx)+B\cos(kx)&,x\le x' \tag 6\\\\
C\sin(k(1-x))+D\cos(k(1-x))&,x\ge x'
\end{cases}$$
Finally, to find the constants $A$, $B$, $C$, and $D$, simply apply the two continuity conditions $(2)$ and $(3)$ along with the two boundary conditions $(4)$ and $(5)$ to $(6)$.  
This results in a linear system of $4$ equations with $4$ unknowns, the solution of which is left as an exercise for the reader.
