stuck with how this ordinary differential equation changes in new coordinates? I am studying ordinary differential equations. There are some solved examples in my book to learn the material from. For this one below however, I do not know how this part of the solution is derived.
The book says, $y'''+yy''=0$ , under $r=xy$ and $s=\ln x $, changes to
$$s_r s_{rrr}+r s_r^2 s_{rr}-3s_{rr}^2=0 $$
How this latter equation is derived? I tried chain rule but I had $r_x $ and $s_x$, which clearly can not be correct.
Thank you.
 A: In the question it is assumed that Eq.$(1)$ :
$$y'''+yy''=0 \tag 1$$
is transformed into Eq.$(2)$ :
$$s_r s_{rrr}+r s_r^2 s_{rr}-3s_{rr}^2=0 \tag 2$$
thanks to the change of variables : $\quad\begin{cases} r=xy \\ s=\ln x \end{cases}$
If this assumption is true any solution $y(x)$ of Eq.$(1)$ is transformed into $s(r)$ which is a solution of Eq.$(2)$.
For example try with $y(x)=x$ which is an obvious solution of $(1)$.
The change of variables transforms $y(x)=x$ into $s(r)=\frac12\ln(r)$.
$s_r=\frac{1}{2r} \quad;\quad s_{rr}=-\frac{1}{2r^2} \quad;\quad s_{rrr}=\frac{1}{r^3}$
$s_r s_{rrr}+r s_r^2 s_{rr}-3s_{rr}^2=\frac{1}{2r}(\frac{1}{r^3})+r(\frac{1}{2r})^2(-\frac{1}{2r^2})-3(-\frac{1}{2r^2})^2$
$s_r s_{rrr}+r s_r^2 s_{rr}-3s_{rr}^2=\frac{1}{r^4}(\frac12-\frac{1}{8}-\frac{3}{4})=-\frac{3}{8r^4}\neq 0$
The transformed function $s(r)$ is not solution of Eq.$(2)$.
This proves that the initial assumption is false. Possibly there is a typo in the question or a misunderstanding in the context.
