Maximizing $\int_0^\infty (1+xy')^2e^y dx$ subject to $\int_0^\infty e^ydx = 1$

Solution 1:

The derivative $x\frac{d}{dx}$ suggests that we should change coordinates $$t~=~\ln x, \qquad q~=~y+t.\tag{A}$$ Then OP's Lagrangian becomes $$ \tilde{L}~=~(\dot{q}^2 +\lambda)e^q, \qquad t~\in~\mathbb{R}. \tag{B}$$ The EL equation becomes $$ (2\ddot{q}+\dot{q}^2-\lambda)e^q~=~0,\tag{C}$$ with solution $$ e^{q(t)}~=~\left\{ \begin{array}{lcr} A\cosh^2\frac{\sqrt{\lambda}(t-t_0)}{2}&{\rm for}& \lambda~>~ 0,\cr A\cos^2\frac{\sqrt{-\lambda}(t-t_0)}{2}&{\rm for}& \lambda~<~ 0,\cr A(t-t_0)^2&{\rm for}& \lambda~=~ 0.\end{array}\right.\tag{D}$$ To keep the constraint integral finite, we must choose a trivial solution $$A~=~0\quad\Leftrightarrow\quad q(t)~=~-\infty,\tag{E}$$ but even then the constraint is not satisfied.