Extremum of $ \int_{1}^e x^2y''^2dx $

Question

The way to find an extremum of functional is to use Euler(-Poisson) equation. $$ \frac {\partial F} {\partial y} - \frac {d}{dx} \frac {\partial} {\partial y'} + \frac {d^2}{dx^2} \frac {\partial} {\partial y'} - ... = 0 $$ For the functional $ J[y] = \int F(x, y, y', y'') dx = \int_{1}^e x^2y''^2dx $ it will be $$ \frac {d^2}{dx^2} (x^2y'')=0 $$ and I have no clue how to solve this kind of differential equation. The fact that confuses me is that it contains full and partial derivatives. Also if it was just $ \frac {d^2}{dx^2} y''=0 $ it would have resolved into $ y''''=0 $, but this $ x^2 $ multiplier should likely be differentiated by $ \frac {d^2}{dx^2} $ and give additional summand (or shouldn't ?). So how this equation should be solved ? I have already browsed through my books on differential equations and haven't found the answer yet. Any help will be appreciated.

Answer 1

Let be $h(x) = x^2y''(x)$. The general solution of the equation $h'' = 0$ is $h(x) = Ax + B$, i.e., your equation is equivalent to $$x^2y''(x) = Ax + B,$$ $$y''(x) = \frac{A}x + \frac{B}{x^2},$$ $$\cdots$$ Can you continue?