Sheaves and Differential Equations How do sheaves arise in studying solutions to ordinary differential equations?
EDIT: Is it possible to construct non-isomorphic sheaves on a domain $D \subset \mathbb{R}^n$ using solution sets to differential equations?
EDIT: Is the sheaf of vector spaces arising from the solution set of a linear ODE necessarily a vector bundle?
 A: One way is through $D$-modules, perverse sheaves, and the Riemann-Hilbert correspondence. A good reference is: "D-Modules, Perverse Sheaves, and Representation Theory", by Hotta, 
Takeuchi & Tanisaki. 
A: Let $U$ be an open subset of $\mathbb R^n$, and let $X$ be a vector field on $U$. You can construct a sheaf $\mathcal F$ of solutions of the ODE $Xf=0$ by letting $\mathcal F(U)$, for each open subset $V\subseteq U$, be the vector space of all $C^\infty$ functions $f$ on $V$ such that $Xf=0$.
By changing the field $X$ you can certainly change the isomorphism clas of $\mathcal F$.
Let $U=\mathbb R^2\setminus\{(0,0)\}$, define fields $X_1(x,y)=\Bigl((\frac1r-1)\frac xr-y,(\frac1r-1)\frac yr+x\Bigr)$ and $X_2(x,y)=(y,-x)$ and consider the corresponding sheaves $\mathcal F_1$ and $\mathcal F_2$. It is not difficult show show that $\mathcal F_1(U)$ is one-dimensional as a real vector space, while $\mathcal F_2(U)$ is infinite dimensional. It follows that $\mathcal F_1\not\cong\mathcal F_2$.

Notice that $\mathcal F_1$ and $\mathcal F_2$ are locally isomorphic. This follows easily from the fact that the fields $X_1$ and $X_2$ are non-zero on their domain.
A: Jet bundles.
A: I will start commenting on Mariano's answer. I believe it is a perfect answer for the question 

How do sheaves arise in studying
  solutions of differential equations ?

but not for the question 

How do sheaves arise in studying
  solutions to ordinary differential
  equations ? 

According to the current terminology a function $f$ satisfying $X(f)=0$ is not a solution of the vector field $X$ but a first integral. Moreover, if $X = a(x,y) \partial_x + b(x,y) \partial_y$ then 
$$
X(f) = a \partial_x f + b \partial_y f  .
$$
Thus $X(f)=0$ is a PDE and not an ODE. Indeed t3suji made the same point at a  comment on Mariano's answer.  I understand the solutions of (the ODE determined by)  $X$ as  functions $\gamma : V \subset \mathbb R \to U$ satisfying $X(\gamma(t))=\gamma'(t)$ for every $t \in V$. Notice that here indeed we have a system of ODEs. 
A vector field can be thought as autonomous differential equation and I do not see clearly how to consider the sheaf of its solutions. 
On the other hand when we have a non-autonomous ordinary differential equation  then there is its sheaf of solutions. This sheaf is a sheaf over the time variable 
only and not the whole space. ( At this point it is natural  to talk about connections and/or jet bundles but I will try to keep things as elementary as possible. ) 
Note that in general the sheaf of solutions will not be a sheaf of vector spaces: the sum of two solutions, or the multiplication of a solution by a constant need not to be a solution. This will  occur only when the differential equation is linear.
The differential equations $y'(t) = y$ and $y'(t) = y^2$, both defined over the whole real line, are examples of differential equations with non-isomorphic sheaves of solutions. The solutions of the first ODE are the multiples of $\exp t $ and define a sheaf of $\mathbb R$-modules. The solutions of the second ODE are zero and $\frac{1}{\lambda - t}$ with $ \lambda \in \mathbb R$. They do define a sheaf of sets, but  not a sheaf of $\mathbb R$-modules.
To obtain examples of linear differential equations with non-isomorphic sheaves, one has to have nontrivial  fundamental group on the time-variable of the differential equation. Thus it is natural to consider complex differential equations over $\mathbb C^{\ast}$. 
The equations  $y'(z) = \frac{ \lambda y(z)}{z}$ parametrized by $\lambda \in \mathbb C$ have non-isomorphic sheaves of solutions. More precisely,


*

*if $\lambda \in \mathbb Z$ then the solution sheaf is the free $\mathbb C$-sheaf of rank one (solutions of the ODE are complex multiples of $z^{ \lambda }$); 

*if $\lambda \in \mathbb Q - \mathbb Z $ then the solution sheaf has no global sections but some tensor power of it does;

*if $\lambda \in \mathbb C - \mathbb Q$ then the solution sheaf has no global sections nor any of its powers does.

A: Being a solution to a differential equation is a local condition, so solutions to a differential equation are naturally a sheaf.
