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Consider the boundary value problem $$ -y'' + 2 y' = \lambda y $$ with boundary conditions $y(0) = y(L) = 0$. I need to determine the eigenvalues and eigenfunctions $y_n$.

So we have the second order linear ODE $$ y'' - 2y' + \lambda y = 0. $$ The characteristic equation of this is $r^2 - 2r + \lambda = 0. $ The roots of this are $r_{1,2} = 1 \pm \sqrt{1 - \lambda}$. I could write down a general solution now as $y(x) = c_1 e^{r_1 x} + c_2 e^{r_2 x}$, but I'm sure how to determine the eigenvalues and eigenfunctions from that?

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You just have to check under which condition on $\lambda$ you can find a solution with $y(0)=y(L)=0$. This is equivalent to the fact that the determinant of the matrix with $1,1$ on the first line $e^{r_1L}, e^{r_2L}$ on the second line is $0$. Thus $e^{r_1L-r_2L}=1$, or $r_1L-r_2L=2ik\pi$, $\sqrt {1-\lambda} L=ik\pi$, $1-\lambda= {k^2\over L^2}\pi^2$, for $k\in \bf Z$

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  • $\begingroup$ how do you deduce that $r_1 L - r_2 L = 2i k \pi $? If I compute the determinant with the first row all 1, and second row $e^{r_1 L}$ and $e^{r_2 L}$, I get the condition that $e^{r_2 L} - e^{r_1 L} = 0$ ? $\endgroup$ – Kamil Jan 13 '16 at 20:46
  • $\begingroup$ $e^{(r_2-r_1)L}=1$ or $(r_2-r_1)L=0$ modulo $i.2\pi$ $\endgroup$ – Thomas Jan 14 '16 at 6:13

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