Laurent series expansion for ODE. OK, then I read Frobenius method in mathworld (I learned when I took ODE 2):
http://mathworld.wolfram.com/FrobeniusMethod.html
My question is:
Are there any ODEs where the solution is given by full Laurent series?, i.e its negative indexed coeffecients are not zero starting from some negative integer.
Thanks in advance.
 A: Take any two linearly independent  functions analytic in an annulus and you can find a second-order linear DE that has these as a fundamental set of solutions.
A: Every analytic function in a ring has a Laurent expansion. Thus if your differential equation
has a solution analytic in a ring it has a Laurent expansion. See the examples given in other answers. 
The difference between a full Laurent
series (I understand that "full" means infinitely many coefficients in both directions) and
a Frobenius series (which is infinite in only one direction) is that you cannot manipulate
with a full Laurent series formally. In particular, you cannot multiply two such series:
the coefficient of the product is a series, rather than a finite sum.
For this reason, the use of full Laurent series is limited.
The advantage of the one-suded series is that you can substitute a formal one-sided series to the differential equation and effectively determine its coefficients (like in Frobenius method). You cannot do this
with a full Laurent series.
Thus a full Laurent series solution (when exists) is usually impossible to find explicitly.
A: There are two related notions for linear ODEs with meromorphic coefficients that have a pole at say $z=0$. The notion of regular singular point (which means that any solution grows no faster at $0$ than $|z|^{-d}$ for some $d$) and the notion of the Fuchs condition (which means that the orders of the poles at $0$ of the meromorphic coefficients satisfy some inequalities). The growth condition implies that the Laurent expansion of any solution necessarily starts at some finite negative order (determined by $d$). According to the second link, these two conditions are equivalent. Unfortunately, I don't have a handy reference for where this is discussed in more detail.
For a non-linear equation, it's probably more difficult to figure out when a singular point is regular in the above sense.
Here's an example of an ODE with an irregular singular point at $z=0$, where the solution has a non-terminating Laurent series: $z^2 \frac{du}{dz}=u$, $u(z) = C \exp(-1/z)$.
