How to find Sturm-Liouville problem eigenvalue and function? 
So I have the following Sturm-Liouville problem:
  $$
 y'' + \lambda y = 0 
$$
  Such that $ \lambda > 0 $ and the initial conditions are as follows:
  $$ y (0) + y'(0) = 0 $$
$$ y(1) + y'(1) = 0 $$

So my attempt at this goes something like this:
I know that the $\lambda$ is positive so the solution must be:
$$ y(t) = A\cos(\sqrt(\lambda)t) + B\sin(\sqrt(\lambda)t)$$
and:
$$ y'(t) = -A\sqrt{\lambda}\sin(\sqrt{\lambda}t) + B\sqrt{\lambda}\cos(\sqrt{\lambda})t)$$
Such that $A$ and $B$ are constants.
So I can evaluate the solution at the first initial condition:
\begin{align}
&=A\cos(\sqrt{\lambda}0) + B\sin(\sqrt{\lambda}0) + -A\sqrt{\lambda})\sin(\sqrt{\lambda}0) + B\sqrt{\lambda}\cos(\sqrt{\lambda}0)\\
&=A  + B\sqrt{\lambda}
\end{align}
Thus I know:
$$ A = -B\sqrt{\lambda}$$
Evaluating at the second initial condition:
\begin{align}
&=A\cos(\sqrt{\lambda}1) + B\sin(\sqrt{\lambda}1)  -A\sqrt{\lambda}\sin(\sqrt{\lambda}1) + B\sqrt{\lambda}\cos(\sqrt{\lambda}1)\\
&=A\cos(\sqrt{\lambda})  + B\sin(\sqrt{\lambda}) -A\sqrt{\lambda}\sin(\sqrt{\lambda}) + B\sqrt{\lambda}\cos(\sqrt{\lambda})
\end{align}
I'm not sure where to go from here, I factored the second initial condition by cos and sin but that led me to a trivial solution for A and B like this:
$$(A + B\sqrt{\lambda})\cos(\sqrt{\lambda}) + (B - A\sqrt{\lambda})\sin(\sqrt{\lambda}) = 0$$
Plugging in $$ A = -B\sqrt{\lambda}$$.
$$(-B\sqrt{\lambda} + B\sqrt{\lambda})\cos(\sqrt{\lambda}) + (B + B\sqrt{\lambda}\sqrt{\lambda})\sin(\sqrt{\lambda}) = 0$$
$$ (B + B\lambda)\sin(\sqrt{\lambda}) = 0$$
So assuming $B$ isn't $0$, then I know $\lambda = (n\pi)^2$ but how do I find the value of B?
Any guidance would be greatly appreciated! 
Thank you.
 A: Start by solving $y''+\lambda y = 0$ subject to
$$
                y'(0)+y(0)=0 \\
                y(0)=1.
$$
The second condition is arbitrary; you could take $y(0)-y'(0)=1$ for example. The point is that no solution can satisfy $y'(0)+y(0)=0$ and $y(0)=0$ unless it is identically the $0$ solution, which means you can always scale $y$ so that $y'(0)+y(0)=0$ and $y(0)=1$. The solution of this problem is
$$
                y_{\lambda}(x)=-\frac{\sin(\sqrt{\lambda}x)}{\sqrt{\lambda}}+\cos(\sqrt{\lambda}x).
$$
Notice that the above is the correct solution at $\lambda=0$ when interpreted as a limit as $\lambda\rightarrow 0$, where it gives $-x+1$. In fact, $y_{\lambda}$ is a power series in $\lambda$. This will always be the case.
The added condition that $y'(1)+y(1)=0$ gives an equation in $\lambda$:
$$
 -\cos(\sqrt{\lambda})-\sqrt{\lambda}\sin(\sqrt{\lambda})-\frac{\sin(\sqrt{\lambda})}{\sqrt{\lambda}}+\cos(\sqrt{\lambda}) = 0.
$$
The solutions $\lambda$ are the zeros of a power series in $\lambda$, and $\lambda=0$ is not a solution. The solutions $\lambda$ must satisfy
$$
       \sqrt{\lambda}\sin(\sqrt{\lambda})+\frac{\sin(\sqrt{\lambda})}{\sqrt{\lambda}}=0.
$$
Looks like $\lambda=0$ is not a solution, but $\sqrt{\lambda}=n\pi$ for $n=1,2,3,\cdots$ are solutions. And $\lambda=-1$ looks like a solution. $y_{-1}$ is an exponential solution $e^{-x}$.
A: $$y'' + \lambda y = 0 \qquad \lambda>0$$
Since $\lambda>0\quad $ let $\quad \lambda=\omega^2$ .
$$y'' + \omega^2 y = 0$$
$$y(x)=c_1\cos(\omega x)+c_2\sin(\omega x)$$
$$y'=-c_1\omega \sin(\omega x)+\omega c_2\cos(\omega x)$$
Conditions :
$y(0)=c_1$
$y(1)=c_1\cos(\omega)+c_2\sin(\omega)$
$y'(0)=\omega c_2$
$y'(1)= -c_1\omega \sin(\omega)+\omega c_2\cos(\omega)$
$$\begin{cases}
y(0)+y'(0)=c_1+\omega c_2=0 \\
y(1)+y'(1)=c_1\cos(\omega)+c_2\sin(\omega)-c_1\omega \sin(\omega)+\omega c_2\cos(\omega)=0
\end{cases}$$
$c_1= -\omega c_2$
$(-\omega c_2)\cos(\omega)+c_2\sin(\omega)-(-\omega c_2)\omega \sin(\omega)+\omega c_2\cos(\omega)=0$
A first solution is :
$$c_2=0 \quad\implies\quad c_1=0\quad\implies\quad y(x)=0$$
This trivial solution was visible at first sight.
Non trivial case  $\quad c_2\neq 0 $ :
$-\omega\cos(\omega)+\sin(\omega)+\omega^2\sin(\omega)+\omega\cos(\omega)=0$
$(1+\omega^2)\sin(\omega)=0$
$\omega=\pm n\pi\quad$ any integer $n$ .
If $\lambda\neq (n\pi)^2\quad$the case $c_2\neq 0$ is impossible. The only solution of the problem is $y(x)=0$.
If $\lambda= (n\pi)^2\quad$ They are an infinity of solutions :
$\omega=\pm n\pi \quad;\quad c_1= \mp n\pi c_2\quad$ any $c_2$ .
$y(x)=\mp n\pi c_2\cos(\pm n\pi x)+c_2\sin(\pm n\pi x)$
$y(x)=\mp n\pi c_2\cos(n\pi x)\pm c_2\sin(n\pi x)$
$y(x)=\pm c_2\left(-n\pi\cos(n\pi x)+\sin(n\pi x)\right)$
Since $c_2$ is any positive or negative constant, without loss of generality :
$$y(x)=c_2\left(-n\pi\cos(n\pi x)+\sin(n\pi x)\right)\quad\text{if}\quad \lambda= (n\pi)^2$$
