In the practice term test 2 Variant A Problem 1, I had to solve the the following Sturm Liouville problem:
$$X''+\lambda x = 0$$ with boundary conditions $$X'(0) = X'(4\pi) = 0$$ In the answer key, the eigenfunction corresponding to the eigenvalue $\lambda_{0} = 0$ is $X_0 = \frac{1}{2}$. However, if we substituted $\lambda_0 = 0$ into the ODE, we get:
$$X'' = 0$$ which the solution is simply $$X(x) = \alpha + \beta x$$ Plugging into the boundary conditions and we get that $$\beta = 0$$ so The solution is $$X(x) = \alpha$$ where $\alpha \in \mathbb{R}$. How do we get that $ X_0 = \frac{1}{2}$?
