Differential Invariants I am trying to solve a simple second order differential equation.
I have been going through Hydon's text.
In Chapter 5, he introduces differential invariants.
Page 76:
The fundamental differential invariants of the group generated by $\mathcal{L}$
are solutions of the system:
$$\begin{bmatrix}
\xi_1 & \eta_1 & \eta^{(1)}_1 & \cdots & \eta^{(R)}_1 \\
\xi_2 & \eta_1 & \eta^{(1)}_2 & \cdots & \eta^{(R)}_2 \\
\vdots & \vdots & \vdots & & \vdots \\
\xi_R & \eta_R & \eta^{(1)}_R & \cdots & \eta^{(R)}_R \\
\end{bmatrix} 
\begin{bmatrix}
I_x \\ I_y \\ I_{y'} \\ \vdots \\ I_{y^(R)} \\
\end{bmatrix} 
= \begin{bmatrix}0 \\ 0 \\ \vdots \\ 0 \end{bmatrix} $$
Example 5.2 on page 77 gives an ODE with the 3 parameter Lie group of point symmetries generated by:
$ X_1 = \partial x, \hspace{10mm} X_2 = x\partial x + y\partial y, \hspace{10mm} X_3=x^2 \partial x + 2xy \partial y. $ 
The fundamental differential invariants are found by solving:
$$
\begin{bmatrix}
1 & 0 & 0 & 0 & 0 \\
x & y & 0 & -y^{''} & -2y^{'''}\\
x^2 & 2xy & 2y & 2(y'-xy^{''}) & -4xy^{'''}
\end{bmatrix}
\begin{bmatrix}
I_x \\ I_y \\ I_{y'} \\ I_{y^{''}} \\ I_{y^{''''}}
\end{bmatrix} = 0$$
When I try to find the first matrix, I am getting the following:
$$
\begin{bmatrix}
1 & 0 & 0 & 0 & 0 \\
x & y & 0 & 0 & 0 \\
x^2 & 2xy & 2y & 0 & 0\\
\end{bmatrix}
$$
what am I doing wrong?
I thought each term in the row after the second column was a derivative of the entry before it.
Is $\eta^{(1)}_1 $ not the derivative of $\eta_1$ ?
 A: No, it is not the derivative. 
You are given a Lie algebra of vector fields (or a transformation group) on $\mathbb R^2$, and would like to understand how this transformation group acts on graphs of functions of the form $y=f(x)$. The transformation group naturally transforms the graphs of functions. For example, $X_1$ corresponds to a translation $x \mapsto x+t$ which just moves the graph along the $x$-axis. Notice that the derivative of a function is not changed by this transformation. The transformation $X_2$ corresponds to a scaling transformation $(x,y) \mapsto (e^s x, e^s y)$. The first derivative of a function does not change under this transformation, but the second derivative does. On the second-order jet space $J^2 \simeq \mathbb R^4(x,y,y',y'')$, the transformation will be $(x,y,y',y'') \mapsto (e^s x, e^s y,y',y'' e^{-s})$, and the corresponding vector field is $X_2^{(2)}= x \partial_x+y \partial_y-y'' \partial_{y''}$ (this corresponds to the first four terms in your second row). 
I am not familiar with Hydon's text, but the formulas for $\eta_i^{(j)}$ are probably given there. For instance, if $X=a(x,y) \partial_x+b(x,y) \partial_y$, then $\eta^{(1)}=- (y')^2 a_y+y' (b_y-a_x)+b_x$.
