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Find the extremum of the following isoperimetric problem. \begin{align}\int_0^\pi y'(x)^2dx\end{align} with $y(0) = y(\pi) = 0$ subject to the constraint \begin{align}\int_0^\pi y(x)^2dx = \frac{\pi}{2}.\end{align} Show that there is an infinite set of extrema.

Solving the Euler-Lagrange for $L + \lambda M$, we have \begin{align*} (L + \lambda M)_y = \frac{d}{dx}\bigg[(L+\lambda M)_z\bigg] \iff 2y\lambda - \frac{d}{dx}(2y') = 0 \iff 2y\lambda - 2y'' = 0 \end{align*} which we may write as $y'' - \lambda y = 0$ which is a second-order differential equation. This has a solution ${y} = c_1e^{\lambda t} + c_2e^{-\lambda t}$.

Applying the boundary conditions, we have that $c_1 + c_2 = 0$ and $c_1e^{\lambda \pi} + c_2e^{-\lambda\pi} = 0 \implies c_2(e^{-\lambda\pi} - e^{\lambda\pi}) = 0$ which only holds if $c_2 = 0$ or if $\lambda = 0$.

If $c_2 = 0$, then the solution is trivial, so consider $\lambda = 0$. Then, we have $y'' = 0$. From here, I am stuck. How do I see that there are an infinite amount of extrema?

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    The solutions involve $e^{\sqrt\lambda t}$, which depending on the sign of $\lambda $ can be a trigonometric function. In fact these provide the infinite family of solutions – 2017-02-24

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Everything's fine up to $c_2(e^{\lambda \pi}-e^{-\lambda \pi}) = 0$. You're missing other possibilities that give $e^{2\lambda \pi} = 1$, namely $\lambda = 2ik$ for integer $k$. This gives the infinite family of solutions $$ y_k(x) = A\sin{2\pi k x}. $$