Determine canonical coordinates I am trying to learn how to solve an ODE using symmetry.
I have got to the point where I have the generators, now I need to use them to get canonical coordinates, $r$ and $s$ where $r(x,y)$ and $s(x,y)$.
If $\eta(x,y)=xy$ and $\xi(x,y)=x^2$
Then the equations to solve in order to determine $r$ and $s$ are :
$x^2r_x + xyr_y =0$
and 
$x^2s_x + xys_y=1$
however, I don't know how to move forward from here.
It has been eons since I have solved differential equations.
Please can someone walk me through this?
UPDATE: I have tried to refresh my memory wrt to solving PDEs.  Here is my attempt:
For the first equation I have the following:
divide by $x^2$ to get  
$\frac{\partial r}{\partial x} + \frac{y}{x}\frac{\partial r}{\partial y}=0$
so I need to solve $\frac{dy}{dx}=\frac{y}{x}$
if I separate variables and integrate I get 
$\ln y =\ln x+C$  
$\ln(y/x)=C$
$\frac{y}{x}=C$
Does this mean which means $r(x,y) = \frac{y}{x}$ ?
I need to know if this is correct and I need help solving the second equation to find $s(x,y)$.
Thanks
 A: Actually,
your system is solvable virtually by inspection.
The equation for r is instantly rewritable as 
$$
\frac{\partial r }{\partial (\ln x)} + \frac{\partial r }{\partial (\ln y)}=0,  
$$
So, since r is invariant upon differentiation w.r.t. to the sum of the gradients of the two effective variables, it must be a function of the difference of these two variables, $\ln x -\ln y=\ln (x/y)$, so then,  any function of x/y will do. 
So check that 
$$
r=f(x/y)~,
$$ 
for arbitrary "reasonable" f solves the equation. 
Now consider the equation in s,
$$
\frac{\partial s }{\partial (\ln x)} + \frac{\partial s }{\partial (\ln y)}=1/x ~,  
$$
where, any solution for s can be augmented by an additional arbitrary function g(x/y) which solves the homogeneous equation, as above! 
Since the r.h.s. is a function of x, take $s=z(x)$ as an Ansatz for the inhomogeneous equation, so the derivative w.r.t. ln y drops out. 
The solution for z(x) dictated by $ dz/dx=1/x^2$ is then $z=-1/x$,
so that
$$
s=g(x/y) -1/x ~,
$$
for an arbitrary "reasonable" function g, completely unrelated to f.
To sum up, your vector 
$$
\begin{pmatrix} \xi \\ \eta \end{pmatrix}=  \begin{pmatrix} x^2 \\ xy \end{pmatrix},
$$
is orthogonal to $\nabla r$ and with unit projection to $\nabla s$. 
