Let $(*)$ be the following boundary value problem $$\begin{align} \begin{cases} -x''(t) + \lambda x(t) = r(t), \, t \in [0,1]\\ x(0) = \alpha\\ x'(1) = \beta \end{cases} \end{align}$$ Find the values of $\lambda \in \mathbb{R}$ for which the problem above has only one solution for all $r \in \mathcal{C}([0,1])$ and arbitrary values of $\alpha, \beta \in \mathbb{R}$.
In order to find all values of $\lambda$, I think I should use Fredholms' Alternative, that is
Or the homogeneous problem has as its only solution the trivial one, which is equivalent to $(*)$ having a unique solution for all $r \in \mathcal{C}([0,1])$ and $\alpha, \beta \in \mathbb{R}$, or the homogeneous problem has a non trivial solution, which is equivalent to $(*)$ having infinite solutions or non at all.
So in my case, I want to solve the homogenous problem for $\lambda$ so that its only solutions is $x(t) \equiv 0$.
The homogeneous problem is (letting $r(t) \equiv 0,\, \alpha, \beta = 0$) $$\begin{cases} -x''(t) + \lambda x(t) = 0\\ x(0) = 0\\ x'(1) = 0 \end{cases}$$ which has a general solution $$x(t) = c_1e^{\sqrt{\lambda} t} + c_2 e^{-\sqrt{\lambda}t}$$ where $c_1$ and $c_2$ satisfy $$\begin{cases} c_1 + c_2 = 0\\ \sqrt{\lambda}\left(c_1e^{\sqrt{\lambda}} - c_2e^{-\sqrt{\lambda}}\right) = 0 \end{cases}$$ Now from the first conditions we get $c_1 = -c_2$, so subtituting in the second equations $$\sqrt{\lambda}\left(c_1e^{\sqrt{\lambda}} - c_2e^{-\sqrt{\lambda}}\right) = \sqrt{\lambda}c_2\cosh(\sqrt{\lambda}) = 0 \qquad (1)$$ Since $c_2 \neq 0$, $\lambda$ must be non zero as well, since we want no solution for $(1)$. Furthermore $$\cosh(x) = 0 \Leftrightarrow x = \left(\frac{\pi}{2} + k\pi\right)i$$ for $k = 0, 1, \dots$,so we want $\sqrt{\lambda}$ not to be of the form $\left(\frac{\pi}{2} + k\pi\right)i$, or in other words, $$\lambda \neq -\left(\frac{\pi}{2} + k\pi\right)^2$$
Is this reasoning correct? Thank you very much!