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I'm stuck on transforming this equation and am not sure where to begin. I know I need to define $x$ as some multiple of $u$ and somehow cancel the coefficient of the $x^2$ term but am not sure how to do it! Any help appreciated :)

Write the following in the form of $(u')^2=u^3 + au + b$, with $a$, $b$ constants?

$$x'^2+\frac{g}{l}x^3+cx^2+\frac{g}{l} x=0$$

$c$ is a constant of integration from a previous part of the question.

Start wearing purple
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GGracie
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    If you know you're supposed to write $x$ as a multiple of $u$, why not try it? Let $x=ku$, substitute that in, and see what $k$ has to be in order to get the coefficient of $u^3$ to be 1. Then you have to do $u$ to $u-r$ for some $r$ to get rid of the square term, so, try it and see what $r$ has to be. It's just algebra, you can do it! – Gerry Myerson Jul 19 '13 at 08:59
  • Thanks, I'll work through what you've suggested and see where it gets me! @O.L. , at the moment I just need to transform into the form without the x^2 which I have stated in the question. The next step is to solve it and I will take a look at what you have suggested. – GGracie Jul 19 '13 at 09:05
  • Please discard my previous comment - I have not understood the question correctly at the first reading. – Start wearing purple Jul 19 '13 at 09:09
  • Thank you for your help @GerryMyerson, I have figured it out! Though it has to be said, the constants aren't pretty :P – GGracie Jul 19 '13 at 09:20
  • Good! Now you can write it up and post it as an answer; then later you can accept it. This helps clear up the Unanswered Questions queue. – Gerry Myerson Jul 19 '13 at 09:24
  • Related: Equation of motion for the pendulum – ccorn Jul 19 '13 at 10:17

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Let $x=kv$. Then: $$k^2v'^2+\frac{g}{l}k^3v^3+ck^2v^2+\frac{g}{l}kv=0$$ $$v'^2=-\frac{g}{l}kv^3-cv^2-\frac{g}{lk}v$$ So $k=-\frac{l}{g}$, giving: $$v'^2=v^3-cv^2+\frac{g^2}{l^2}v$$ Now let $v=u+r$, giving: $$u'^2=u^3+3u^2r+3ur^2+r^3-cu^2-cr^2-2cur+\frac{g^2}{l^2}(u+r)$$ We want the coefficient of $u^2$ to be $0$ (i.e $3u^2r=cu^2$).
Thus $r=\frac{c}{3}$.
The cubic then has the form $u'^2=u^3+au+b$: $$u^3+(\frac{g^2}{l^2}-\frac{c^2}{3})u+(\frac{g^2c}{3l^2}-\frac{2c^3}{27}),$$ where $a=(\frac{g^2}{l^2}-\frac{c^2}{3})$ and $b=(\frac{g^2c}{3l^2}-\frac{2c^3}{27})$.

GGracie
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