Given $\displaystyle F_y - \frac{d}{dx} F_{y'} + \frac{d^2}{dx^2}F_{y''}=0$ (where $y\in C_4$) , if $F(x,y,y',y'')$ does not depend on $x$ explicitly, then the system has the first integral $\displaystyle y''F_{y''} -y'\left(\frac{d}{dx}F_{y''}-F_y\right)=C.$
I've tried expanding out $\displaystyle F_y - \frac{d}{dx} F_{y'} + \frac{d^2}{dx^2}F_{y''}$ by $\displaystyle \frac{d}{dx}F_{y'}= \frac{dF}{dx}+\frac{dF}{dy}y' + \frac{dF}{dy'}y'' + \frac{dF}{dy''}y'''$ (and a similar way for $\displaystyle \frac{d^2}{dx^2}F_{y''}$, and using the fact that there is no explicit x-dependence, then $\displaystyle \frac{dF}{dx} = 0$. But with this method, things don't seem to cancel out nicely in such a way that after I integrate with respect to $x$, I obtain the result I seek.
Another thing i've noted is that I can express what I start with as $\displaystyle F_{y'} = \frac{d}{dx}\left(F_{y'}-\frac{d}{dx}F_{y''}\right)$ and what I need to show as $\displaystyle F-y''F_{y''}+C = y'\left(F_{y'}-\frac{d}{dx}F_{y''}\right)$, so i've tried integrating $\displaystyle F_{y'} = \frac{d}{dx}\left(F_{y'}-\frac{d}{dx}F_{y''}\right)$ from the start, so that $\displaystyle \int F_{y'}dx = F_{y'}+\frac{d}{dx}F_{y''} + C$.
Am I approaching this right? Any idea how to proceed?