Okay i've been at it for far too long now. It comes from a bigger question from working with a PDE. I did seperation of variables and now I am stuck near the end of the problem. Here is the ODE in question \Phi''(y)= \lambda^2\cdot\Phi(y) with the following initial condition \Phi'(H)=0 where $H$ is a positive number. Also I know $\displaystyle \lambda = \frac{n\pi}{L} >0$. There is another condition but I dont think it can help $ \Phi(0) = \begin{cases} 0 & x > L/2 \\ 1 & x < L/2 \end{cases}$ sorry, i dont know how to do cases in latex and yes, that is an $x$ in the initial condition. Like i said this is a bigger problem that has both x and y.
The solution should be $\Phi(y) = B \cdot \cosh{(\lambda(H-y))}$
I've tried going through the following general solutions $\Phi(y) = Ae^{\lambda y} + B e^{-\lambda y}$ and $\Phi(y) = A \sinh{\lambda y} + B \cosh{\lambda y}$ but no luck that way :(