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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

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  • $\begingroup$ I am not quite sure I understand your problem. Should the initial differential equation be equal to $0$? What is $f$? I think your question lacks clarity. $\endgroup$ – Fimpellizieri Apr 21 '16 at 23:13
  • $\begingroup$ sorry for misleading you! f is a constant. and the boundary conditions were given exactly like the above one. Im considering the one with equal to zero. $\endgroup$ – Alana Apr 21 '16 at 23:15

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