Finding Green's Function for BVP Find Green's function for the boundary value problem 
$$\begin{cases}
y''+ \lambda y =0 \quad\text{ in } [0,1]\\
y(0)=0, y(1)=0
\end{cases}$$
is given for $\lambda > 0$, $\lambda \neq n^{2} \pi^{2}$ where $(n=1,2,3,...).$
 A: As there is no forcing (RHS $= 0$), the Green's function is just the zero function. I presume that the correct RHS should be $\delta(x-\xi)$, the point forcing at $x=\xi$.
First you need to a test for the existence of Green's function: check the completely homogeneous problem and see if it only admits the trivial zero solution. 
The homogeneous problem has solution of the form
\begin{equation}
y\left(x\right)=c_{1}\cos\left(\sqrt{\lambda}x\right)+c_{2}\sin\left(\sqrt{\lambda}x\right)
\end{equation}
Using the boundary data, 
\begin{equation}
0=y\left(0\right)\implies c_{1}=0 \\
0=y\left(1\right)\implies c_{2}\sin\left(\sqrt{\lambda}\right)=0\implies c_{2}=0
\end{equation}
because $\lambda\neq n^{2}\pi^{2}$. Hence the solution to the homogeneous problem is indeed the trivial solution. There exists a Green's function.
Consider the Green's function
\begin{equation}
G\left(x,\xi\right)=\begin{cases}
G_{-}\left(x,\xi\right), & 0\leq x<\xi\leq1\\
G_{+}\left(x,\xi\right) & 0\leq\xi<x\leq1
\end{cases}
\end{equation}
where $G_{-}\left(x,\xi\right)$ and $G_{+}\left(x,\xi\right)$ solve the homogeneous ODE and also satisfy the following four conditions,
\begin{equation}
G_{-}\left(0\right)=0 \\
G_{+}\left(1\right)=0 \\
\text{Continuity of } G \text{ at some point } x=\xi \\
\text{Jump discontinuity of } \frac{dG}{dx} \text{ at some point } x=\xi \text{ such that } \left[\frac{dG}{dx}\right]_{\xi^{-}}^{\xi^{+}} = 1 \\
\end{equation}
We already have a general form of the solution to the homogeneous problem, i.e.
\begin{equation}
G_{-}\left(x\right)=A\cos\left(\sqrt{\lambda}x\right)+B\sin\left(\sqrt{\lambda}x\right) \\
G_{+}\left(x\right)=A\cos\left(\sqrt{\lambda}x\right)+D\sin\left(\sqrt{\lambda}x\right)
\end{equation}
where $A,B,C,D$ are constants to be determined by the above four conditions.
The boundary conditions force $A=C=0$. Continuity and jump condition yield two equations with two unknowns, 
\begin{equation}
B\sin\sqrt{\lambda}\xi=D\sin\sqrt{\lambda}\left(\xi-1\right)\\
\sqrt{\lambda}D\cos\left(\sqrt{\lambda}\left(\xi-1\right)\right)-\sqrt{\lambda}B\cos\left(\sqrt{\lambda}\xi\right)=1
\end{equation}
After using some trigonometric identities in solving this system(the angle difference formula for $\sin\left(X-Y\right)=\sin X\cos Y-\cos X\sin Y$), we have 
\begin{equation}
B=\frac{\sin\left(\sqrt{\lambda}\left(\xi-1\right)\right)}{\sqrt{\lambda}\sin\sqrt{\lambda}}\\
D=\frac{\sin\left(\sqrt{\lambda}\xi\right)}{\sqrt{\lambda}\sin\sqrt{\lambda}}
\end{equation}
Plugging these two constants back into the definition of $G$, we finally have constructed the Green's function, 
\begin{equation}
G\left(x,\xi\right)=\begin{cases}
\frac{\sin\left(\sqrt{\lambda}\left(\xi-1\right)\right)\sin\left(\sqrt{\lambda}x\right)}{\sqrt{\lambda}\sin\sqrt{\lambda}}, & 0\leq x<\xi\leq1\\
\frac{\sin\left(\sqrt{\lambda}\left(x-1\right)\right)\sin\left(\sqrt{\lambda}\xi\right)}{\sqrt{\lambda}\sin\sqrt{\lambda}} & 0\leq\xi<x\leq1
\end{cases}
\end{equation}
