Here is a geometric picture of how symmetries can aid in the solution of certain ODEs. I'll begin with some technical discussion and conclude with some qualitative remarks.
Let M denote the phase space. This is the manifold where solution curves to the ODE are contained. Let X:M\rightarrow TM denote the vector field that defines the ODE. A geometric way of saying that this ODE possesses symmetry is M admits a (left) Lie group action \Phi:G\times M\rightarrow M that commutes with the flow map of X. Recall that the flow map F_\lambda of X takes a point m\in M and returns the point F_\lambda(m) equal to where the initial condition m ends up after moving \lambda seconds according to the dynamics defined by the ODE. So, restated in symbols, the condition that the ODE possesses a symmetry is
\forall g\in G,\lambda\in\mathbb{R},~~\Phi_g\circ F_\lambda=F_\lambda\circ\Phi_g.
Provided the action \Phi is sufficiently nice (e.g., free and proper), then it is possible to form the quotient manifold M/G, which is just the collection of group orbits (see
http://en.wikipedia.org/wiki/Group_action for more discussion on group actions, particularly the section on continuous group actions). On this smaller manifold M/G, the original ODE induces a new ODE x:M/G\rightarrow T(M/G) via the following construction.
Let \pi:M\rightarrow M/G denote the map that assigns to every point m\in M the group orbit m lives on. For each \lambda\in\mathbb{R}, define the mapping f_\lambda:M/G\rightarrow M/G by f_\lambda(\pi(m))=\pi(F_\lambda(m)).
This is a well-defined mapping precisely because the symmetry condition holds. That is, if m^\prime=\Phi_g(m) lives on the same group orbit as m, then
<br />
f_\lambda(\pi(m^\prime))=\pi(F_\lambda(\Phi_g(m)))=\pi(F_\lambda(m))=f_\lambda(\pi(m)).<br />
The key property this map possesses is that it defines a one-parameter subgroup of M/G, which in symbols means f_{a+b}=f_a\circ f_b for any real numbers a,b. Thus, f_\lambda is the flow map of some vector field x on M/G.
The vector field x:M/G\rightarrow T(M/G) contains all of the "essential" information contained in the original vector field X:M\rightarrow TM. In fact, if it turns out to be possible to solve the ODE defined by x, then the solution to the original ODE can be explicitly reconstructed in terms of integrals of the solution to x (I won't explain how this works here). Note that because x is defined on a space with smaller dimension than M (in fact \text{dim}G dimensions smaller), it is reasonable to expect that x is easier to solve than X.
Finally, I'd like to point out that symmetry is even more useful when your ODE is Hamiltonian. In the Hamiltonian case, it is often possible to reduce the dimensionality of the system by TWICE the dimension of G! This is because Hamiltonian symmetries also give rise to conserved momentum maps. An authoritative source for learning about Hamiltonian symmetries is Abraham and Marsden's Foundations of Mechanics.
