Question: In the Calculus of Variations there is a passage from Euler's characteristic equation: $$ \frac {\partial F}{\partial y} - \frac {d}{dx} \left(\frac {\partial F}{\partial y'} \right)=0 $$ in which $$ F=F(x,y,y') $$ If the function F is made explicitly independent of x then: $$ F=F(y,y') $$ Then by integration of Euler's characteristic equation one finds: $$ F - y' \frac{\partial F}{\partial y'}=k $$ I'm having great difficulty trying to do that integration, hence I'd appreciate any help!!!

Answer: 
If you want $A$ to be bounded, then it is necessary and sufficient that there exists $c > 0$ such that $||x|| > c$ implies $|F(x)| > 0$. This follows immediately from the definition of boundedness for sets in $\mathbb{R}^n$.

As a previous post remarked, we could ask $\lim\_{||x||\rightarrow\infty} |F(x)| = \infty$, but certainly the function need not be unbounded for $A$ to be bounded! We could ask the growth condition $\liminf\_{||x||\rightarrow\infty} |F(x)| > 0$. This is sufficient but not necessary. For instance, if $F:\mathbb{R}\rightarrow\mathbb{R}$ is defined to be $F(x)=x$ on $[-1,1]$ and $F(x)=e^{-(|x|-1)^2}$ everywhere else, then the $\liminf = \lim = 0$, while $A=\{0\}$ for this $F$. Outside of $[-1,1]$, the function I defined is just the Gaussian (bell curve).

<https://www.wolframalpha.com/input/?i=F%28x%29%3De%5E%7B-%28%7Cx%7C-1%29%5E2%7D>
(This graph shows the behavior outside $[-1,1]$.)

If there is any question about the definition this notion of $\liminf$, it is covered on Wikipedia:
<http://en.wikipedia.org/wiki/Limit_superior_and_limit_inferior#Functions_from_metric_spaces_to_metric_spaces>
If $x = (x(1), x(2),\ldots, x(n))$ then we need that there is a $C > 0$ such that $|x(k)|<C$ for all $k = 1, 2,\ldots, n$. We can take this to be the condition to guarantee boundedness of $A$.