Solutions of $-y''-\lambda y=0$ Consider the problem
$$
\begin{cases}
-y''-\lambda y=0 \\
y(0)=y(L) \tag{1}
\end{cases}
$$
where, $L>0$ is fixed.
Question. It's possible to find a solution $y$ and $\lambda \in \mathbb{R}$  satisfying ($1$)?
I thought about is applying second-order linear ODE resolution methods. But I was not successful, because in that case it would depend on the "constant" $\lambda \in \mathbb{R}$.
Moreover, roughly speaking, can the problem ($ 1 $) be interpreted as a eigenvalue problem?
 A: The ODE $-y''-\lambda y = 0$ corresponds to $y'' + \lambda y = 0$.
If $\lambda < 0$, the general solution of $y'' + \lambda y = 0$ is
$$y(x) = A e^{-\alpha x} + B e^{\alpha x}, $$
where $\alpha = \sqrt{-\lambda}.$
The boundary condition reads as:
$$y(0) = A + B = y(L) = A e^{-\alpha L} + B e^{\alpha L}.$$
A possible way to solve this is the following:
$$A(1 - e^{-\alpha L}) = B(e^{\alpha L} - 1)\Rightarrow$$
$$A = B \frac{e^{\alpha L} - 1}{1 - e^{-\alpha L}}\Rightarrow$$
$$A = B \frac{e^{\alpha L} - 1}{e^{-\alpha L}(e^{\alpha L} - 1)} \Rightarrow$$
$$A = Be^{\alpha L}.$$
If $\lambda > 0$, the general solution of $y'' + \lambda y = 0$ is:
$$y(t) = A \cos (\omega x) + B \sin (\omega x),$$
where $\omega = \sqrt{\lambda}.$
The boundary condition reads as:
$$y(0) = A = y(L) = A \cos (\omega L) + B \sin (\omega L).$$
A  possible way to solve this is the following:
$$A(1 - \cos(\omega L)) = B \sin(\omega L) \Rightarrow$$
$$A = B \frac{\sin(\omega L)}{1 - \cos(\omega L)} \Rightarrow$$
$$A = B \frac{2\sin\left(\frac{\omega L}{2}\right)\cos\left(\frac{\omega L}{2}\right)}{1 - \left(2\cos^2\left(\frac{\omega L}{2}\right)-1\right)} \Rightarrow$$
$$A = B \frac{2\sin\left(\frac{\omega L}{2}\right)\cos\left(\frac{\omega L}{2}\right)}{2\left(1 - \cos^2\left(\frac{\omega L}{2}\right)\right)} \Rightarrow$$
$$A = B \frac{\sin\left(\frac{\omega L}{2}\right)\cos\left(\frac{\omega L}{2}\right)}{\sin^2\left(\frac{\omega L}{2}\right)} \Rightarrow$$
$$A = B \cot\left(\frac{\omega L}{2}\right).$$
