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Given the equation $y''+\lambda y=0$ and boundary conditions $y(1)=0$ and $y(0)+y'(0)=0$.

Let $r=\pm\sqrt{-\lambda}.$

If $\lambda >0$ we have $y=C_1\cos(\sqrt{\lambda}x)+C_2\sin(\sqrt{\lambda}x).$ Applying our boundary conditions I get to this point: $-\sqrt{\lambda}=\tan(\sqrt{\lambda}).$ At this point I'm not sure what to do .

Similarly for the case of $\lambda <0$. This means that $r=\pm\sqrt{\lambda}$. I believe this means I have to apply the conditions to $y=C_1e^{\sqrt{\lambda}x}+C_2e^{-\sqrt{\lambda}x}$.

I want to make sure my approach for the second case is correct and I'm not sure how to finish up the first case.

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Application of your boundary condition should give you \begin{equation} 0 = C_{1} \cos(\sqrt{\lambda})+C_{2}\sin(\sqrt{\lambda}) \qquad (y(1)=0) \end{equation} and \begin{equation} 0 = C_{1} + C_{2}\sqrt{\lambda} \qquad (y(0) + y'(0)=0) \end{equation} Take it from here.

EDIT: Can't add comment so I'll re-edit me partial solution.

Your roots are of the form \begin{equation} \pm i \sqrt{\lambda} \end{equation} Remembering that \begin{eqnarray} e^{z} &=& e^{a+ib} \\ &=& e^{a}(\cos(b)+i\sin(b)) \end{eqnarray} You can then look at substituing in these forms and plug in your boundary conditions.

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  • $\begingroup$ I think the derivative was messed up. $y'(0)=C_1+C_2\sqrt{\lambda}$. Also, $y(1)=0$. $\endgroup$ – emka Nov 13 '14 at 15:48
  • $\begingroup$ I have those equations. I'm not sure how to handle the cases that arise from $\lambda$ being positive and negative. $\endgroup$ – emka Nov 13 '14 at 15:59
  • $\begingroup$ I see that I made a sign error with. I'm still yielding $\sqrt{\lambda}=tan(\sqrt{\lambda})$. $\endgroup$ – emka Nov 13 '14 at 16:16

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