Recall that in the theory of ring and modules, we have the concept of modules generated by one element \(M = Rm\). Such modules are isomorphic to \(R/I\) where \(I\) is the kernel of the map \(R\to Rm=M\), or the annihilator of \(m\). i.e.
\[ Rm \cong R/\mathrm{ann}(m). \]
We say such \(m\) is a cyclic vector for the cyclic module \(M\).
Let \(K\) be a differential field, for a differential \(K\)-module \(M\), or a \(K[\partial]\)-module, we say \(M\) is a differential cyclic module if there is a cyclic vector \(m\) that is actually a cyclic vector over \(K[\partial]/K\), i.e. such that \(M\) is generated by the \(K\)-basis \(m, \partial m, \dots, \partial^{n-1}m\) as an \(n\)-dimensional \(K\)-vector space. In this case the matrix of \(\partial\) under this basis is
\[ \begin{pmatrix} & & & -a_0 \\ 1 & & & -a_1 \\ & 1 & & \dots \\ & & 1 & -a_{n-1} \\ \end{pmatrix} , \quad \partial^n m + a_{n-1} \partial^{n-1} m + \dots + a_0 m = 0. \]
i.e. let \(e_i = \partial^i m\), we have
\[ \partial e_i = e_{i+1}, \quad i=0,\dots,n-2; \quad \partial e_{n-1} = - \sum a_i e_i. \]
If we start with an \(n\)-dimensional \(K[\partial]\)-module \(M\), i.e. \(\dim_K M=n\), if \(M\) is cyclic as a \(K[\partial]\)-module, then by the fact that ideals in \(K[\partial]\) are principal (warning: \(K[\partial]\) is not commutative), we can show actually \(M \cong K[\partial]/(P)\) for some
\[ P = \partial^n + a_{n-1}\partial^{n-1} + \dots + a_1 \partial + a_0. \]
Then take the preimage \(m\) of \(1\) of that isomorphism, we see that \(m, \partial m,\dots \partial^{n-1} m\) actually generates \(M\) as \(K\)-vector space.
That is to say, a cyclic differential module over differential field \(K\) is just a finite dimensional cyclic \(K[\partial]\)-module.
Let \(K\) be a differential field of characteristic \(0\) and with non-zero derivation \(\partial\). Then any finite dimensional (over \(K\)) differential module \(M\) is cyclic, i.e. \(M\) has a cyclic vector.
The idea of the proof is similar to the primitive element theorem, which considers a generic linear combination \(a+tb\).
For a differential module \(M\), \(m\) is the cyclic vector in \((M,\partial)\) as \(K[\partial]\)-module iff \(m\) is a cyclic vector in \((M,u\partial)\) as a \(K[u\partial]\)-module, because \((u\partial)^n = u (\partial^n + (\dots) \partial^{n-1} + \dots)\) which is monic. If \(\partial^i m\) generates \(M\), so is \((u\partial)^i m\) and vice versa.
This means by assuming \(\partial x = y\neq 0\), and replacing the derivation with \(\frac{x}{y}\partial\), we can assume there exists \(x\) such that \(\partial x = x\).
Assume \(\dim M = n\) and let \(r\) be the dimension of a maximal cyclic submodule. If \(r=n\) there is nothing to prove. Assume for now \(r < n\), so any \(r+1\) elements of the form
\[ a, \partial a, \dots, \partial^r a \]
must be linearly dependent, the wedge form \(a \wedge \partial a \wedge \dots \wedge \partial^r a = 0\) vanishes. Take a one-parameter family of elements \(a + tb\) where \(t\in \mathbb{Q}\), into the wedge form
\[ (a + tb) \wedge \partial (a + tb) \wedge \dots \wedge \partial^r (a + tb) = 0, \]
by the characteristic \(0\) assumption the field is infinite so each coefficients of \(t^i\) vanishes as well, taking the coefficient of \(t\) we have
\[ \sum_{i=0}^r a\wedge \partial a \wedge \dots \wedge \partial^{i-1} a \wedge \partial^i b \wedge \partial^{i+1} a \wedge \dots \wedge \partial^r a = 0. \]
Now let \(b = x^s c\), compute and plug in
\[\partial^i b = \sum_{j=0}^i \binom{i}{j} s^j x^s \partial^{i-j} c, \]
to get a polynomial in \(s\) which vanishes on integers. It has to be zero by the assumption that \(K\) has characteristic \(0\). Now consider the \(s^r\) coefficient
\[ a \wedge \partial a \wedge \dots \wedge \partial^{r-1} a \wedge (s^r x^s c) = 0 \Rightarrow a \wedge \partial a \wedge \dots \wedge \partial^{r-1} a\wedge c = 0.\]
Since \(c\) is arbitrary, we have in fact for any \(a\),
\[ a \wedge \partial a \wedge \dots \wedge \partial^{r-1} a = 0. \]
This means the maximal cyclic submodule have dimension \(<r\), which is a contradiction. So we can only have \(r=n\) and \(M\) is cyclic.
The above proof actually tells us that a generic linear combination \(a + tb\) is a cyclic vector, so in some sense the vectors which does not generate the module form a subvariety of the projective space of \(M\).