The Question:
Find an extremal corresponding to $$\int_{-1}^1ydx$$ subject to the conditions $$y(-1)=y(1)=0\qquad \text{and} \qquad \int_{-1}^1(y^2+y'^2)dx=1$$
My Attempt:
Not much of an attempt here, but the point is I keep running into contradictions.
By the Euler-Lagrange equation, the extremal satisfies
$$\frac{d}{dx}\frac{\partial F}{\partial y'}-\frac{\partial F}{\partial y}=0$$
where in this case $F(x,y,y')=y$. Plugging it in, we get $-1=0$???
Similarly, Beltrami's Identity states that if $F$ is not explicitly dependent on $x$, then
$$\frac{d}{dx}\biggl[ y'\frac{\partial F}{\partial y'}-F \biggr ] =0$$
which implies that $y=\text{constant}$.
The conditions $y(-1)=y(1)=0$ imply that $y(x) \equiv 0$, but then this wouldn't satisfy the other condition...
What on earth is going on?
This is a constrained variational problem. The correct Lagrangian should be $$ F(x,y,y')=y-\lambda\left(y^2+y'^2\right), $$ where $\lambda$ a Lagrange multiplier (a constant) to be determined.
Apply Euler-Lagrange equation to the above Lagrangian, and $$ 1-2\lambda y=-2\lambda y'', $$ whose solution, together with the imposed boundary condition $y(-1)=y(1)=0$, reads $$ y(x)=\frac{1}{2\lambda}\frac{1+e^2-2e\cosh x}{1+e^2}. $$ Finally, $\lambda$ shall be determined by using the constraint $$ \int_{-1}^1\left(y^2+y'^2\right){\rm d}x=1, $$ which yields $$ \lambda=\pm\frac{1}{\sqrt{1+e^2}}. $$
To sum up, the very function $y$ you are expecting should be $$ y(x)=\pm\frac{1+e^2-2e\cosh x}{2\sqrt{1+e^2}}. $$