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Find the eigenvalues of $$y'' + \lambda y = 0, \; y'(0) = 0, \; y'(1) = 0$$


In my work, I have:

For $\lambda > 0$, $y(x) = c_1\cos(\sqrt{\lambda}x) + c_2\sin(\sqrt{\lambda}x)$,

$$y'(x) = -c_1\sin(\sqrt{\lambda}x) + c_2\cos(\sqrt{\lambda}x).$$

$$0 = y'(0) \implies c_2 = 0$$

$$0 = y'(1)= -c_1\sin(\sqrt{\lambda}) + c_2\cos(\sqrt{\lambda})$$

In the second equation, how do I solve for $\lambda$? In Eigenvalue problem $y'' + \lambda y = 0,$ $y'(0) = 0$, $y(1) = 0$ this was more easily done, but now I do not see how.

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  • $\begingroup$ $y'$ lacks a factor $\sqrt{\lambda}$. $\endgroup$ – mvw Dec 9 '14 at 20:57
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You have already concluded that $c_2 = 0$, so when using the second condition, you have simply $-c_1\sin(\sqrt\lambda) = 0.$

You need $c_1 \neq 0$ (otherwise all solutions are trivial), so you must have $\sin(\sqrt\lambda) = 0$. That is, $\sqrt\lambda = n\pi : n \in \mathbb Z$.

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    $\begingroup$ Oh, that makes total sense. Wow. Thank you. $\endgroup$ – diffeBlows Dec 9 '14 at 19:16

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