Consider $f \in H^{2}(-1,1) \cap H_{0}^{1}(-1,1)$. The first eigenvalue of the problem $$ -f^{\prime \prime} = \lambda f \ \ \text{with} \ \ f(-1) = f(1) = 0 $$ As it shows that the first eigenvalue is $\frac{\pi^{2}}{4}$ with eigenfunction $\cos(\frac{\pi x}{2})$??
I know that ($\lambda > 0$)
If $F''(x)+\lambda F=0$, then $$F(x)=c_1\cos \frac{\sqrt{\lambda}\pi}{2}x + c_2\sin \frac{\sqrt{\lambda}\pi}{2}x.$$ How $ 0= F(1) = F(-1) = c_2 \sin \frac{\sqrt{\lambda}\pi}{2}$. Soon $\lambda = 4n^{2}$, $n \in \mathbb{N}$ or $c_2 = 0 $.
And now? How to continue?
$$F''+\lambda F=0$$ For $\lambda > 0$: $$\implies F(x)=c_1\cos \sqrt{\lambda}x + c_2\sin \sqrt{\lambda}x.$$ Then apply initial conditions: $$F(1)=c_1\cos \sqrt{\lambda} +c_2\sin \sqrt{\lambda}=0$$ $$F(-1)=c_1\cos \sqrt{\lambda} - c_2\sin \sqrt{\lambda}=0$$ Add both equations: $$2c_1\cos \sqrt{\lambda}=0$$ $$\cos \sqrt{\lambda}=0$$ $$\sqrt {\lambda}=\dfrac {\pi}{2}+k\pi$$ First value of $\lambda$ is for $k=0$ $$ \implies \lambda =\dfrac {\pi^2}{4}$$ Now we have $$F(1)=0 \implies c_2=0$$ $$F(x)=c_1 \cos \left( \dfrac {\pi}{2} x \right)$$