If $$ \begin{align*} r = \frac{y}{x} & & s= -\frac{1}{y} & & \frac{ds}{dr}=\frac{y'}{r^2\left(xy'-y\right)} \end{align*} $$
determine $$\frac{d^2s}{dr^2}$$
NB: $y' = \frac{dy}{dx}$
NB: this is part of a greater problem regarding solving $y''=0$ using symmetry methods.
Ok so here is my attempt at this very ugly second derivative:
$$ \begin{align*} \frac{d^2s}{dr^2} = &\frac{d}{dr}\left( \frac{y'}{ryy'-r^2y} \right) \\ \\ = & \frac{ y'' \left(ryy'-ry \right) - y'\left(yy' +r(y')^2 + ryy'' - y - ry'\right)}{\left(ryy'-ry\right)^2}\\ \\ \end{align*} $$
Since the initial condition equation I am solving is $y''=0$ I can simplify this (ever so slightly) to get:
$$ \frac{d^2s}{dr^2} = \frac{ -y' \left( yy' +r(y')^2 -y - ry' \right)}{\left(ryy' - ry \right)^2 }\\ \\ $$
The aim is to do the following:
- let $z = \frac{ds}{dr}$
- so $\frac{d^2s}{dr^2}=\frac{dz}{dr}$
- Integrate $\frac{dz}{dr}$ to get $z$ in terms of only $r, s$ and constants
- substitute this form of $z$ into $\frac{ds}{dr}$
- integrate and solve for $s$
- substitute the given values for $s$ and $r$ in terms of $x$ and $y$
- rearrange to get an equation in the form $y=Ax+B$
$$ s\, \, x = \frac{-1}{r}$$
Differentiate w.r.t. $r$ using chain rule, using primes w.r.t $r$ only,obtaining $ s^{'}, s^{''}, r^{'}, r^{''} $ during the differerntiation.
$$ s^{'} x +s x^{'} = \frac{1}{r^2}$$
$$ s^{''} x + 2 s^{'} x^{'} + s x^{''} = \frac{-2}{r^3}.$$