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$y'' + λy = 0$

BC (Boundary Conditions): $y'(0) = 0, y(π) = 0$

My Work

I set $\lambda = \mu^2$ and $y=e^{rx}$

$$ y^2 + \mu^2y = 0 \\ r^2 + \mu^2 = 0 \\ r = \pm\mu i \\ y=c_1\cos(\mu x) + c_2\sin(\mu x) \\ y' = -\mu c_1\sin(\mu x) + c_2 \mu \cos(\mu x)$$

Plug in BC

$$ y(\pi) = c_1\cos(\mu \pi) + c_2 \sin(\mu \pi) = 0 \\ y'(0) = c_2\mu = 0 \\ c_2 \ne 0, \text{ so } \mu = 0 $$

But this can't be true because the textbook answer is enter image description here

What did I do wrong? How do I solve this equation?

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It's obvious that that two cases when $\lambda =0$ or $\lambda =- \mu^2$ gives trivial solution,but you have to mention that .

$$y=c_1 \cos(\mu x)+c_2 \sin(\mu x) \Rightarrow y'=-c_1\mu \sin(\mu x)+c_2 \mu \cos(\mu x)$$

$$y'(0)=0\Rightarrow c_2=0$$ So $y=c_1 \cos(\mu x)$

Now $$y(\pi)=0\Rightarrow c_1\cos(\mu \pi)=0$$ Which implies that : $$\mu=(2n-1)/2$$

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  • $\begingroup$ Does $2n+1$ also work (instead of $2n-1$)? In the final answer for the eigenfunc, what happens to the arbitrary constant $c_1$? Why is n=1,2,3,..? Shouldn't it be for any integer (i.e n = -3,-2,-1,1,2,3, etc)? $\endgroup$ – 14wml Mar 24 '17 at 4:47
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    $\begingroup$ Yes $2n+1$ also works but n must start from $0$ $c_1$ is not appearing in $y_n$ because we let $c_1 =1$ since it's involved in the constant in the general solution $y=\sum_{n=0}^{\infty} c_n \cos(2n+1)/2 x$ .Now since $cos(x)=cos(-x)$ So for negative value for n doesn't count even for any other problem $y_n=cy_{-n}$ for some c so forget about the negative value for $n$ $\endgroup$ – Mahmoud Hassan Mar 24 '17 at 5:28
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First take the auxiliary equation $m^2 + λ = 0$ then three cases arrives, case $1$ when $λ<0$ , case $2$ when $λ=0$ , and the last case when $λ=0$

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