Can anyone spot why this equivalence is true?
$f=f(x)$
$f'''-(k+6f)f'=0$ and $c_1+c_2f+{k\over 2}f^2+f^3-{1\over 2} (f')^2=0$ where $c_1,c_2$ can be any pair of constants.
I think the ${1\over 2}(f')^2$ is very suggestive, but I just can't see it.
EDIT: Ah, I have just spotted the link! Thanks for reading anyway!