$y'' + λy = 0$
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

What did I do wrong? How do I solve this equation?
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 : $$\mu=(2n-1)/2$$