I am trying to determine the symmetries for the solutions to $\ddot{y} = 0$ where $y=y(x)$.
I have determined the 8 vectors and the Lie Algebras.
I have determined the canonical coordinates $(r,s)$ and now need to determine $\frac{ds}{dr}$ for each of my vectors.
In Bluman & Anco, this is defined as: $$ \frac{ds}{dr} = \frac{s_x + s_y y'}{r_x + r_y y'} $$
For one of my vectors I have: \begin{align*} s = & \frac{-1}{x} & & r=\frac{y}{x}\\ \end{align*}
I am calculating: \begin{align*} \frac{ds}{dr} = & \frac{\frac{1}{x^2}+0}{\frac{-y}{x^2}+\frac{1}{x}y'}\\ = & \frac{1}{x^2}\left(\frac{x^2}{-y+xy'}\right)\\ = & \frac{1}{-xr + xy'}\\ = & \frac{1}{x}\left(\frac{1}{y'-r}\right)\\ = & \frac{s}{r-y'}\\ \end{align*}
I can now solve for $s$ by:
\begin{align*} \frac{1}{s}ds = & \frac{1}{\left( r - y' \right) } dr\\ \ln s = & \ln \left( r- y' \right) \\ s = & r - y' \end{align*}
NB I want the simplest solution and so am omitting the constant.
I rearrange and substitute $s(x,y)$ and $r(x,y)$ to find $y$:
\begin{align*} y' = & r - s\\ \frac{dy}{dx} = & \frac{y}{x} + \frac{1}{x}\\ \frac{1}{y+1}dy = & \frac{1}{x}dx \\ \ln \left( y + 1 \right ) = & \ln x\\ y = & x - 1\\ \end{align*}
Is this correct?
Now my second vector is: $ X_2 = y \partial x $
when $\eta = 0$ we set $r = x$ Now the formula to determine $s$ is the same and we get $ s = \ln y $
When I try to find $\frac{ds}{dr}$ as above, I get $$ \frac{ds}{dr} = \frac{y'}{y} $$
In trying to solve for $y$ I have the following:
\begin{align*} \frac{dy}{dx} = \frac{y \ln y}{x}\\ \\ \frac{1}{y \ln y} dy = & \frac{1}{x} dx \\ \\ \ln \left( \ln y \right) = & \ln x \\ \\ \ln y = & x \\ \\ y = e^x \\ \end{align*}
Something is obviously wrong in what I am doing as the solution should be linear.
I think the fact that $ r = x $ and thus $y = y(r) $ comes into play, I just don't know how.
Please help.
Thanks