$y'$ is being eliminated as I solve using Eulers formula what to do?

Question

Investigate the extremals of the functional $\int((y^2 + (x^2)\cdot y'))dx$ with limits $0$ to $1$ under the conditions $y(0) = 0$, $y(1) = A$.

Answer 1

If you solve Euler's equation, you immediately get $y(x)=x,$ without any differential equation. Unfortunately, that solution doesn't satisfy the boundary condition $y(1)=A,$ if $A\neq1.$ So what's going on, here?

By partial integration, using $y(1)=A$, we have $$\int^1_0(y^2+x^2y')\,dx=A+\int^1_0(y^2-2xy)\,dx=A-\frac13+\int^1_0(y-x)^2\,dx,$$ so $A-\frac13$ is the infimum of our functional, because $(y-x)^2\ge0.$ It is attained as a minimum only for $y(x)=x,$ if $A=1.$ Otherwise, the infimum is not attained. But it can be approximated: if $$y(x)=x+(A-1)\,x^n,$$ the boundary condition is satisfied, and we have $$\int^1_0(y-x)^2\,dx=(A-1)^2\int^1_0 x^{2n}\,dx=\frac{(A-1)^2}{2n+1},$$ that's arbitrarily small when $n$ is large.