Differential invariants of ODEs I have been reading 'Symmetry and Integration Methods for Differential Equations' by Bluman and Anco. I'm trying to make sense of differential invariants and ODEs... It is confusing.
There is an ODE in a general form, for example 
$$F(x,y,y',y'',...,y^{(n)})=0$$
where $x$ is the independent variable, and $y$ is the dependent variable.
Say, the ODE is invariant under the one-parameter Lie group $X_1$.
Say we know the prolongation
$$X_1^{(n)}$$
and the ODE admits the group, i.e. is invariant w/respect to $X_1^{(n)}$
If I do
$$X_1^{(n)}I=0$$
then $I$ is a "differential invariant" of the Lie group, right? 
What does it mean that $I$ is a "differential invariant"? Does it mean it doesn't change under the transformation $X_1^{(n)}$? Or does it mean $I$ doesn't change under the transformations $X_1$ (without the prolongation)? It seems confusing...
What are the properties of $I$?
 A: Chapter 3 of the book  'Symmetry and Integration Methods for Differential Equations' by Bluman and Anco deals with differential invariants and ODEs. There are a number of worked examples given throughout the chapter.
This is the proper use of the term "differential invariant":
Let $F(x,y,y',y''...,y^{(n)})=0$ be an ODE subject to the action of 1-parameter Lie group with generator $X$. $X^{(n)}$ is the n-th extension.  Assume the group is admitted by the ODE:
$$X^{(n)}F=0 \mbox{ when } F=0$$
 Let $u(x,y)$, $v(x,y,y')$ be invariants of the 1-st extension,
$$X^{(1)}u(x,y)=0$$
$$X^{(1)}v(x,y,y')=0$$
Then the derivatives 
$$\frac{du}{dv}, \frac{d^2u}{dv^2}, ..., \frac{d^{n-1}u}{dv^{n-1}}$$
are called "differential invariants" of higher orders. The text calls them 'differential invariants" of the n-th extended group'. In other words, new invariants of higher orders are derived through differentiation.
These differential invariants are used to reduce the order of the ODE. Many examples are found in the book, see e.g. p.127, 138 etc. 
