Sturm Liouville with periodic boundary conditions Background and motivation: I'm given the boundary value problem:
$$y''(x)+2y(x)=-f(x)$$ 
subject $y(0)=y(2\pi)$ and $y \, '(0)=y \, '(2\pi)$. 
EDIT: These were not given to be zero !! Maybe this helps... 
The text (Nagle Saff and Snider, end of Chapter 11 technical writing exercise) asks us to construct the Green's function for the problem. At the moment, I'm a bit stumped because there is no $\lambda$ in the given problem. Let me elaborate, if we were given:
$$ (py')'+qy+\lambda r y= 0 $$
where $p,p',q$ and $r$ were continuous, real-valued, periodic functions with period $2\pi$ then I think I'd be able to get started. I know the usual solutions then only fit the given boundary conditions for particular choices of $\lambda$. So, my initial observation is that $p=1$ is certainly continuous and periodic so we can set $p=1$.


*

*Question: what should I see as $q$ and $r$ for the problem stated at the start of this post? How can we massage the given problem into the standard form of Sturm Liouville?
I suppose it is important to note we must choose $r>0$ as it serves as the weight function in the inner product which is paired with the eigenspace of solutions for this problem.
Added: here is a picture of the problem from the text:


I will write $a$ for $\sqrt{2}$ for simplicity.
The general solution of the homogenous equation $y''+2y=0$ has the form
$$y(x)=C\cos(ax)+D\sin(ax)$$
Using the variation of parameters method to find a particular solution of the non homogenous problem, we have to determine $C$ and $D$ with
$$
\eqalign{C'\cos(ax)+D'\sin(ax)&=0\cr
-C'\sin(ax)+D'\cos(ax)&=-\frac{1}{a}f(x)
}
$$
This yields
$$
C'=\frac{1}{a}\sin(ax)f(x),\quad D'=-\frac{1}{a}\cos(ax)f(x)
$$
Thus, the general solution of $y''+2y+f=0$ is given by
$$\eqalign{
y(x)&=c\cos(ax)+d\sin(ax)+\frac{\cos(ax)}{a}\int_0^x\sin(at)f(t)dt-\frac{\sin(ax)}{a}\int_0^x\cos(at)f(t)dt\\
&=c\cos(ax)+d\sin(ax)-\frac{1}{a}\int_0^x\sin(ax-at)f(t)dt}
$$
and
$$
y'(x)=-ac\sin(ax)+ad\cos(ax)- \int_0^x\cos(ax-at)f(t)dt
$$
Now, the conditions $y(0)=y(2\pi)$ and $y'(0)=y'(2\pi)$ give us two equations:
$$\eqalign{c
&=c\cos(2a\pi)+d\sin(2a\pi )-\frac{1}{a}\int_0^{2\pi}\sin(2a\pi-at)f(t)dt\cr
ad&=
-ac\sin(2a\pi)+ad\cos(2a\pi)- \int_0^{2\pi}\cos(2a\pi-at)f(t)dt}
$$
This can be arranged as follows
$$\eqalign{c\sin(a\pi)-d\cos(a\pi)
&=-\frac{1}{2a\sin(a\pi)}\int_0^{2\pi}\sin(2a\pi-at)f(t)dt\cr
c\cos(a\pi)+d\sin(a\pi)&=
- \frac{1}{2a\sin(a\pi)}\int_0^{2\pi}\cos(2a\pi-at)f(t)dt}
$$
Solving for $c$ and $d$ we obtain
$$\eqalign{c&=-\frac{1}{2a\sin(a\pi)}\int_0^{2\pi}(\cos(2a\pi-at)\cos(a\pi)+\sin(a\pi)\sin(2a\pi-at))f(t)dt\cr
&=-\frac{1}{2a\sin(a\pi)}\int_0^{2\pi} \cos(a\pi-at)f(t)dt\cr
d&=-\frac{1}{2a\sin(a\pi)}\int_0^{2\pi}(\cos(2a\pi-at)\sin(a\pi)-\cos(a\pi)\sin(2a\pi-at))f(t)dt\cr
 &=\frac{1}{2a\sin(a\pi)}\int_0^{2\pi} \sin(a\pi-at)f(t)dt\cr
}
$$
Finally
$$\eqalign{
y(x)&
=\frac{-1}{2a\sin(a\pi)}\int_0^{2\pi} \big(\cos(ax)\cos(a\pi-at)-\sin(ax)\sin(a\pi-at)\big)f(t)dt\\
&\phantom{=}-\frac{1}{a}\int_0^x\sin(ax-at)f(t)dt\\
&=\frac{-1}{2a\sin(a\pi)}\int_0^{2\pi}\cos(a\pi+ax-at)f(t)dt-\frac{1}{a}\int_0^x\sin(ax-at)f(t)dt
}
$$
This expression of $y$ can be simplified as follows:
$$\eqalign{
y(x)&
=\frac{-\cot(a\pi)}{2a }\int_0^{2\pi} \cos(ax-at)f(t)dt\cr
&\phantom{=}+\frac{1}{2a }\int_0^{2\pi} \sin(ax-at)f(t)dt 
-\frac{1}{a}\int_0^x\sin(ax-at)f(t)dt\cr
&=\frac{-\cot(a\pi)}{2a }\int_0^{2\pi} \cos(ax-at)f(t)dt+\frac{1}{2a }\int_x^{2\pi} \sin(ax-at)f(t)dt\cr
&\phantom{=} 
-\frac{1}{2a }\int_0^{x} \sin(ax-at)f(t)dt \cr
&=\frac{-\cot(a\pi)}{2a }\int_0^{2\pi} \cos(ax-at)f(t)dt-\frac{1}{2a }\int_0^{2\pi} \sin(a|x-t|)f(t)dt\cr
&=\frac{-1}{2a\sin(a\pi)}\int_0^{2\pi} \cos(a\pi-a|x-t|)f(t)dt
}
$$
Therefore,
$$
y(x)=\int_0^{2\pi}G(x,t)f(t)dt
$$
with
$$
G(x,t)=\frac{-\cos(\sqrt{2}(\pi-|x-t|))}{2\sqrt{2}\sin(\sqrt{2}\pi)}.
$$
which is the desired conclusion.$\qquad\square$.
