Ordinary differential operators satisfying braid relation? Let $W$ be the algebra of linear ordinary differential operators with analytic coefficients $C^{\omega}(\mathbb{R})[\partial_x]$ (with multiplication given by composition). Do there exist two elements $L_1$, $L_2\in W$ of degree (in $\partial_x$) at least 1 such that 
$$
L_1L_2L_1=L_2L_1L_2,
$$
i.e. $L_1$ and $L_2$ satisfy the braid relation?
The reason I am asking this is that the study of commuting ordinary differential operators (aka Krichever theory) exhibits a lot of interesting algebraic geometry. 
For example, if two differential operators $L_1$, $L_2$ commute then there necessarily exists a polynomial $R(x, y)$ such that $R(L_1, L_2)=0$, i.e. one can associate an affine curve $C$ to the pair $(L_1, L_2)$. The common eigenfunctions define a holomorphic vector bundle $B$ on $C$. It can be extended to a vector bundle $\hat{B}$ on the projective completion $\hat{C}$ of $C$; one can reconstruct the commutative algebra generated by $L_1$ and $L_2$ from $\hat{C}$, a marked point, $\hat{B}$ and its trivialization at the neighbourhood of marked point. 
I was wondering whether the study of 'braided differential operators' as explained above also exhibits some rich algebraic geometry. 
P.S.: googling 'braided differential operators' does give a lot of results, but if I understand correctly they are not related to this question. 
Remark: we assume $L_2 \neq L_1$, of course.
 A: I don't know the answer to this question, but the algebra of differential operators is almost commutative. So by looking at the principal symbols (with respect to the standard filtration by the degree of $D=\partial_x$), you can conclude that $\sigma(L_1)=\sigma(L_2)$. You can push it further a bit by considering other filtrations if, say, $L_i$ are polynomial coefficient operators, and by looking at the quasi-classical approximation given by the Poisson bracket. 


NB: We assume that the coefficients of the differential operators belong to a commutative domain (polynomial, rational, algebraic, exp-algebraic, etc functions). The situation is quite different for various matrix-valued operators, where it is more reasonable to expect braid relations.


As an illustration, it is straightforward to show that in degree one, $L_1=L_2$. Indeed, it is well-known (and easy to prove) that any first order linear differential operator can be conjugated to $a\partial$ in a suitable differential extension (this requires adding exponentials of antiderivatives, as in solving $(\partial+p)f=0$). So without loss of generality, we may assume that $L_1=a\partial$ and $L_2=a\partial+b$, where $a,b$ are functions. Looking at the subprincipal terms, i.e. equating the coefficients of $\partial^2$ in the braid relation, we conclude that $a^2b=0$ and so $b=0$, i.e. $L_1=L_2$.  
