Solving the system of Determining Equations I am practicing solving differential equations using symmetry.  I was working through an example from P.Hydon's paper "An introduction to symmetry methods in the solution of differential equations that occur chemistry and chemical biology".
One of the equations he solves is :$$ y''=\frac{1}{y^2}y' $$
The system of determining equations for $\eta(x,y)$ and $\xi(x,y)$ are as follows:
$$ y^3\eta_{xx} - y\eta_{x}=0$$
$$2\eta - y\xi_x +2y^3\eta_{xy}-y^3\xi_{xx}=0$$
$$y^3\xi_{yy}-2y^3\xi_{xy}-2y\xi_y =0$$
$$y^3\xi_{yy}=0$$
I need to solve these.  Hydon uses a command in Maple to do this.  I need to figure out how to solve them without computer software.
Here is what I have:
consider $y^3\xi_{yy}=0$
$$ \xi_{yy}=0 \\ \xi_{y}=A(x) \\ \xi = A(x)y + B(x) $$
**consider $y^3 \eta_{yy} - 2y^3\xi_{xy}-2y\xi_{y}=0$ **
$$y^3\eta_{yy} -2y^3A'(x) - 2yA(x) = 0 \\ y^3\eta_{yy} = 2y^3A'(x) +2yA(x) \\ \eta_{yy} = 2A'(x) + \frac{2}{y^2}A(x) \\ \eta_{y} = 2A'(x)y - \frac{1}{y}A(x) + C(x) \\ \eta = A'(x)y^2 - \ln\left|y\right|A(x) + C(x)y + D(x) $$
call this eqn i
**consider $2y^3\eta_{xy} - y^3\xi_{xx} - y\xi_{x} +2\eta = 0$ **
$$ 2y^3\left( 2A''(x)y - \frac{1}{y}A'(x) + C'(x) \right) - y^3\left(A''(x)y + B''(x) \right) - y\left(A'(x)y + B'(x) \right) + 2\eta = 0 \\
3y^4A''(x) - 3y^2A'(x) + 2y^3C'(x) - y^3B''(x) -yB'(x) +2\eta = 0 \\
\eta = \frac{-3}{2}y^4A''(x) +\frac{3}{2}y^2A'(x) - y^3C'(x) +\frac{1}{2}y^3B''(x) + \frac{1}{2}yB'(x) $$
call this eqn ii
now compare eqn i and eqn ii
$$ \frac{3}{2}y^2A'(x) = y^2A'(x) \\ A'(x) = 0 \\ A(x) = c_1 \\C(x) = \frac{1}{2}B'(x) \\ 2C(x) = B'(x) \\ B''(x) = 2C'(x) \\ D(x) = 0 $$
What I end up getting is:
$$\eta = -y^3C'(x) +y^3C'(x) + yC(x) \\
\eta = yC(x) \\ \xi = yc_1 + B'(x) $$
Now, Hydon has $$\eta = c_1y \\ \xi = 2c_1x + c_2 $$
If I rewrite in terms of B'(x) I get:
$$\eta = y\frac{1}{2}B'(x) \\ \xi = yc_1 + B'(x)$$
Where am I going wrong ???
PLease help.

Equating the expressions for $\eta$ we get that $2C(x) + B'(x)=0$, $D(x)=0$ and $A(x)=0$.
Thus $\xi_y=0$ and so $\xi = B(x)$, a function of only $x$.
Furthermore we have $\eta = C(x)y = \frac{1}{2}B'(x)y$.
Using the only equation in the system we haven't yet touched, we find,
$$y^3\eta_{xx} -y\eta_x=0\implies y^2C''(x) - C'(x)=0\\
\implies C'(x)=0\implies C(x) = c_1\implies B(x) = 2c_1x+c_2$$
As far as where you went wrong, when equating the two expressions for $\eta$ you didn't conclude that $A(x)=0$ but this is the only way to kill the log term. After that, you should go back to the system you started with and ask yourself what you haven't used yet to get more information.
