I am trying to understand calculus of variations. I have $I(y)=\int_1^2(x^2y'^2+2y^2)dx$ and boundary conditions $y(1)=4$ and $y(2)=1$. I have to find extremal $y_0$ od the functional $I(y)$. I started that with EL equation for $\mathcal{L}=x^2y'^2+2y^2$: $$ \frac{d}{dx}\left(\frac{\partial \mathcal{L}}{\partial y'}\right) = \frac{\mathcal{L}}{dy}. $$ I get the differential equation $4y=4xy'+2x^2y''$, which I than simplified to $x^2y''+2xy'-2y=0$ and solved for $y=x^{\lambda x}$. The solution is $y=C_1x+C_2x^{-1}$, and after evaluating for boundary conditions I get $y_0=\frac{4}{x^2}$.
Now I would like to know if $y_0$ is min or max of functional $I$. I know from the textbook, that $y_0$ is min, but I do not understand how can I determine that. I would also like to know, how to prove, that functional $I(y)$ doesn't have supremum, eg. $sup(I(y))$ doesn't exist. How do one show that?
Without boundary conditions considerations we have that $y = c_1x+\frac{c_2}{x^2}$ is the solution for the Euler-Lagrange equation $x^2 y''+2x y'-2y=0$. Now substituting the solution into the integral we have
$$ I(y) = \frac{14 c_1^2}{3}+c_1 c_2 \log (16)+\frac{7 c_1}{3}+\frac{7 c_2^2}{12}-\text{c2} \ln(4) $$
which is a quadratic form with minimum but not maximum (depending on $c_1,c_2$)