In the study of partial differential equations, it is often considered enough to analyze the principal symbols and their characteristic variety (see for example, <http://www.sciencedirect.com/science/article/pii/S0001870802000993>, and <https://en.wikipedia.org/wiki/Symbol_of_a_differential_operator>). In the first link, only the principal symbol is used to count the number of solutions to a holonomic system. But there are easy examples for which this information does not appear to be enough. Consider the two-dimensional system:
$$
(\partial+1)\phi(z,z')=0
$$
$$
(\partial+\partial')\phi(z,z')=0
$$
and,
$$
(\partial+z)\psi(z,z')=0
$$
$$
(\partial+\partial')\psi(z,z')=0.
$$
Both have the same principal symbols, namely, $p$ and $p+p'$, however, the first has the solution $\phi(z,z')=e^{z'-z}$, while the second has no solution (other than $\psi(z,z')=0$). 


It seems to be suggested in the first link that the second has a solution on an integral curve, perhaps, but it is not clear to me if that is the right context and if so, how does one construct such a curve?



Characteristic Variety of the Principal Symbol solves PDE system?
(*This really should be a comment, but it was too long.*)


Part of the problem is that when you're dealing with a *system* of (linear) PDE's, the principal symbol is not the "right" object. Instead, you want to look at the characteristic ideal. It's defined as follows: 


Let $D$ be the ring of linear partial differential operators in $z\_1,\ldots,z\_n$. A system of PDE's may be viewed as an $r\times s$ matrix $\mathsf P$ with entries in $D$. Let $N$ be the left column space of $\mathsf P$, i.e. the left $D$-module generated by the columns of $\mathsf{P}$. Let $\operatorname{gr}(D)=\{\sigma\_P\mid P\in D\}$ be the (commutative) ring of principal symbols of differential operators, and let $\operatorname{gr}(N)=\{\sigma\_V\mid V\in N\}$ be the $\operatorname{gr}(D)$-module of principal symbols of $N$. The *characteristic ideal* of the system of PDE's is (the radical of) the ideal
$$ \operatorname{Ann}\_{\operatorname{gr}(D)}\bigl(\operatorname{gr}(D)^r/\operatorname{gr}(N)\bigr).$$ 
The reason we need to take the radical is because the un-radicalized ideal depends on choice of coordinates while its radical does not.


There's a coordinate-free definition, but it's a bit more involved to state.


**Edit: In your case:**
System 1: Have $r=1$ and $N=D\{\partial+1, \partial +\partial'\} $. One can show using Gröbner basis theory that $\operatorname{gr}(N)=(\partial, \partial') $.


System 2: Have $r=1$ and $N=D\{\partial+z, \partial +\partial'\} $. You might guess that $\operatorname{gr}(N)$ would be the same, but you'd be wrong! In fact, notice that the commutator $[ \partial +\partial', \partial+z] $ is $1$, so $N=D $, and in particular $\operatorname{gr}(N)= \operatorname{gr}(D) $.


