If $y(t)$ is a stationary function of $$J[y]=\int_{-1}^1(1-x^2)y'^2\,dx$$ with B.C. $y(-1)=1, y(1)=1$ subject to $\int_{-1}^1y^2=1$. Then which is correct ?
(A) $y$ is unique.
(B) $y$ is always a polynomal of even degree.
(C) $y$ is always a polinomial of odd degree.
(D) No such $y$ exists.
Let, $F=(1-x^2)y'^2+\lambda y^2$, where $\lambda$ is Lagranges multiplier. Then Euler-Lagrange equation gives,
$$(1-x^2)y''-2xy'-\lambda y=0.$$ which is very complicated to solve. At this stage, how can we approach to solve this question ?
I do not know the depth with which you are supposed to solve this problem, but this differential is a very famous one called the Legendre Differential Equation. It is well known that this equation may only lead to bounded solutions when the eigenvalue here $\lambda$ is nonpositive, and specifically unless $\lambda=-\ell(\ell+1)$ we cannot have convergence at both endpoints $-1$ and $1$. The fact that your boundary conditions impose convergence at the endpoints means that we require this restriction for $\lambda$.
The solutions to this class of equations are all polynomials of degree $\ell$. They are often denoted as $P_\ell(x)$ with the first few as
$$ P_0(x)=1, \\ P_1(x)=x, \\ P_2(x)=\frac{3x^2-1}{2}, \\ P_3(x)=\frac{5x^3-3x}{2}, \\ ... $$
Many formulae exist for these polynomials, but that is not quite the point here. A major point to note about these functions is that they all satisfy
$$ P_\ell(1)=1,\;\;\;\;P_\ell(-1)=(-1)^\ell $$
By this property we know that only the cases where $\ell$ is even could possibly solve this problem is when $\ell$ is even, since we need $P_\ell(-1)=1$. This would imply that we are on track for the second solution, but we have the issue of normalization. The Legendre polynomials are not normalized over this interval, with
$$ \int_{-1}^1P_\ell(x)^2\,dx=\frac{2\ell+1}{2} $$
Hence, the solutions to the differential equation when normalized will not satisfy the boundary conditions. The meaning here is that through the combination of convergent boundary conditions, normalization, and specifically assigned values at the boundary, we have no solutions.
Hence the answer is $(D)$; no such solution exists.
Side Note: If you are looking for how to solve the differential equation by hand, consider the Frobenius method and solve term by term, noticing that unless $\lambda$ satisfies a certain criteria then the convergence at the endpoints is not guaranteed.