Find the eigenvalues of $$y'' + \lambda y = 0, \; y'(0) = 0, y(1) = 0$$
For $\lambda >0$,
$$y(x) = c_1 \cos(\sqrt{\lambda} x) + c_2 \sin(\sqrt{\lambda}x)$$
We get that $y'(0) = 0 \implies c_2 = 0$, but when I try to solve for $\lambda$ when doing $y(1) = 0$, I run into trouble. Anyone know how to do this sort of problem?
Apply your boundary condition at $x=1$: $$ y(1)=c_1\cos(\sqrt{\lambda})=0\implies \sqrt{\lambda}={(2n-1)\pi\over 2},\ n=1,2,\dots $$ (recall that the zeros of the cosine function occur at odd multiples of $\pi/2$), but then squaring both sides, $$ \lambda=\lambda_n=\left({(2n-1)\pi\over 2}\right)^2,\ n=1,2,\dots $$