Given $$y"+p(x,\lambda)~y'+q(x,\lambda)~y,~~~~~0<x<1$$ with boundary condition $$a~y(0)+by'(0)=0$$ $$c~y(1)+dy'(1)=0$$ where $a,b,c,d,\lambda$ are constant
By getting the auxiliary equation,take $y=e^\alpha x$,we will get something like: $$\alpha~=~f\pm~i~\omega~~~~~~~~~~~~~(Take~the~complex~one?)$$
I was told the general solution of this kind of BVP is: $$y(x)=e^fx(Asin~\omega~t+Bcos~\omega~t)$$
As from my experience of doing homogeneous BVP, we consider different cases of $\lambda$, however this is always true to take the complex one ! I am wondering if there is a proof, to figure it out. thanks