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

1. Only one eigenfunction
2. Two eigenfunctions
3. Two linearly independent eigenfunctions
4. Two orthogonal eigenfunctions

I have tried to solve the problem but could not get my calculations right. Can somebody help?

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.