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.
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 $b=(\frac{g^2c}{3l^2}-\frac{2c^3}{27})$.