A problem For the boundary value problem, $y''+\lambda y=0$, $y(-π)=y(π)$ , $y’(-π)=y’(π)$ For the boundary value problem,
$y''+\lambda y=0$
$y(-π)=y(π)$   ,   $y’(-π)=y’(π)$
to each eigenvalue $\lambda$, there corresponds   


*

*Only one eigenfunction   

*Two eigenfunctions   

*Two linearly independent eigenfunctions   

*Two orthogonal eigenfunctions   


I have tried to solve the problem but could not get my calculations right. Can somebody help?
 A: This is elementary. Find for each $\lambda$ general form of a solution of the equation without boundary conditions, and try to apply the boundry condition. 
You won't get stuck in the calcualtions again, as there is almost nothing to calculate.
A: Here is a related problem. First solve the differential equation
$$ y(x)=c_1\,\sin \left( \sqrt {\lambda}x \right) + c_2\,\cos\left( \sqrt {\lambda}x \right) .$$
Applying the boundary conditions to the solution results in the two equations
$$ y(\pi)=y(-\pi) \implies 2{ c_1}\,\sin \left( \sqrt {\lambda}\pi  \right)= 0 \rightarrow (1) $$
$$y'(\pi)=y'(-\pi)\implies 2{c_2}\sqrt{\lambda}\sin \left( \sqrt {\lambda}\pi  \right) = 0 \rightarrow (2), $$
where $c_1$ and $c_2$ are arbitrary constants. From $(1)$, since $c_1$ is an arbitrary constant, then we have 
$$ \sin\left( \sqrt {\lambda}\pi \right) = 0 \implies \sqrt {\lambda}\pi = n\pi \implies \lambda = n^2. $$
Eq. $(2)$ gives the same eigenvalues with the other eigenvalue $\lambda=0$, since $c_2$ is an arbitrary constant. So, can you find the eigenvectors now and answer the question? 
Note: This problem is known as "The Sturm-Liouville Eigenvalue Problems". One of the properties of this kind of problems is that "for each eigenvalue $\lambda_n$ there exists an eigenfunction $\phi_n$.
