How to find Sturm-Liouville problem eigenvalue and function?

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 $e^{-x}$.

$$y'' + \lambda y = 0 \qquad \lambda>0$$ 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 :

$$y(x)=c_2\left(-n\pi\cos(n\pi x)+\sin(n\pi x)\right)\quad\text{if}\quad \lambda= (n\pi)^2$$