Green fuctions and sturm liouville problem I have the following problem. Find the Green function in the problem of Sturm-Liouville:
$$Ly=-y''.$$
$$y'(0)=y(0),\quad y'(1)+y(1)=0$$
I can not find the eigenvalues ​​and the eigenfunctions but how do I find the Green's function?
The solution in the book is 
$\frac{1}{3}\left\{\begin{matrix}
(x+1)(2-\xi ), 0\leq x \leq \xi & \\ (\xi +1)(2-x), \xi\leq x\leq 1
 &
\end{matrix}\right.$ 
Thanks for all and any help
 A: Recall that I did find your eigenvalues and eigenfunctions, as well as the Green's function in terms of an expansion in those eigenfunctions.
To find the Green's function directly in the form you want, note that you are solving
$$\frac{d^2}{d x^2} G(x,x') = -\delta(x-x')$$
The general solution to this equation may be immediately written down as
$$G(x,x') = \begin{cases}\\A x + B & 0 \le x\le x'\\ C x+D & x' < x \le 1 \end{cases}$$
The constants are determined through boundary conditions, as well as conditions imposed on the Green's function itself through the differential equation.  The boundary conditions at $x=0$ imply that $A=B$.  The boundary conditions at $x=1$ imply that
$$C+C+D = 0 \implies D=-2 C$$
We need two other conditions to completely determine the Green's function.  First, we require that $G$ be continuous at $x=x'$:
$$A x'+B = C x'+D$$
The last condition is a little tricky.  We integrate the diff. equation with respect to $x$ in a small neighborhood about $x=-x'$; note that the integral of a delta function that goes through $x=x'$ is $1$.  Thus, we have
$$\lim_{\epsilon \to 0}\left(\left[\frac{d}{dx} G(x,x')\right]_{x=x'+\epsilon}-\left[\frac{d}{dx} G(x,x')\right]_{x=x'-\epsilon}\right) = -1$$
This is saying that the derivative of $G$ is discontinuous at $x=x'$. What this means for our solution is that 
$$C-A=-1$$
You then solve, four equations for four unknowns.  Actually, these equations are simple enough, and you will find that
$$A = B = \frac13(2-x')$$
$$C=-\frac13(1+x')$$
$$D=\frac{2}{3}(1+x') $$
The sought result then follows:
$$G(x,x') = \begin{cases}\\\frac13 (2-x')(1+x) & 0 \le x\le x'\\ \frac13 (1+x')(2-x) & x' < x \le 1 \end{cases}$$ 
