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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!

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    We don't use [SOLVED] in the title. Instead we **accept answers** to indicate that the problem has been solved. You can also post an answer of your own and accept if you wish.2012-10-12
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    Sorry, @AsafKaragila, I have deleted the [SOLVED] now.2012-10-12

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The first differential equation can be integrated directly to give: $$f'' - kf - 3f^2 + a = 0.$$ Multiply this equation by $f'$ $$f'f'' - kff' - 3f^2f' + af' = 0.$$ Now integrate again. $$\frac{1}{2}(f')^2 - \frac{k}{2}(f)^{2} - f^{3} + af +b =0.$$ This is the second differential equation you've provided.