Solve the isoperimentric problem to minimize $$\int_0^1 y(x)\,{\rm d}x$$ subject to the constraint $$\int_0^1 \sqrt{1 + y'(x)^2}\,{\rm d}x = \frac{\pi}{2}$$ with $y(0) = 0$ and $y(1) = 0$.
My Work: $$F_y - \frac{d}{dx}F_y' + \lambda\left(G_y -\frac{d}{dx}G_y'\right) = 0$$
I get $$1 - \lambda \frac{d}{dx} \frac{y'}{\sqrt{1+y'^2}} = 0$$
I'm not sure where to go from here to solve.
As Winther commented, starting with your last equation $$1 - \lambda \frac{d}{dx}\Big(\frac{y'}{\sqrt{1+y'^2}}\Big) = 0$$ you find, after simplification, $$1-\frac{\lambda y''}{\left(1+y'^2\right)^{3/2}}=0$$ Now, set $p=y'$ and you need to solve $$1-\frac{\lambda p'}{\left(1+p^2\right)^{3/2}}=0$$ whih is separable. So $$\int \frac {dp}{\left(1+p^2\right)^{3/2}}=\frac 1 \lambda \int dx\implies \frac{p}{\sqrt{1+p^2}}=\frac x \lambda +c_1$$ Now, you are let with a quadratic in $p$ the solution of which being $$p=\pm\frac{i (x+c_1 \lambda )}{\sqrt{x^2+2 c_1 \lambda x+(c_1^2-1) \lambda ^2}}$$ Integrating once more leads to $$y=\pm i \sqrt{x^2+2 c_1 \lambda x+\left(c_1^2-1\right) \lambda ^2}+c_2$$ Now the given conditions will fix integration constants $c_1$ and $c_2$ as simple function of $\lambda$.