Let $L$ be some differential operator of the form $$ Ly = y^{(n)}(x) + a_{n-1}(x) y^{(n-1)}(x) + \dots + a_0(x) y(x) $$ with all $a_k(x)$ being smooth.
Let also $M$ be the frozen at $x=0$ operator $L$ (note that $M$ has constant coefficients): $$ My = y^{(n)}(x) + a_{n-1}(0) y^{(n-1)}(x) + \dots + a_0(0) y(x). $$ Consider also fundamental solution $u$ of the operator $M$, i.e. $$ Mu = \delta(x). $$ Is that true that $$ Lu - \delta(x) $$ is a smooth function? In other words, does $Lu$ have singularity only at $x = 0$?
My approach: Let $y$ be the fundamental solution of $L$. Assume that it has a form $$ y(x) = P(x) \theta(x). $$ Then $$ Ly = L[P\theta]\\ L[P\theta] = \sum_{k=0}^n a_k(x) \frac{d^k}{dx^k}\left[P(x)\theta(x)\right] = \sum_{k=0}^n a_k(x) \sum_{m=0}^k {k \choose m} P^{(k-m)}(x) \theta^{(m)}(x) = \\ = \sum_{m=0}^n \left(\sum_{k=m}^n{k \choose m} a_k(x) P^{(k-m)}(x) \right) \theta^{(m)}(x) = \\ = \left(\sum_{k=0}^n a_k(x) P^{(k)}(x) \right) \theta(x) + \sum_{m=0}^{n-1} \left(\sum_{k=m+1}^n {k \choose m+1} a_k(x) P^{(k-m-1)}(x) \right) \delta^{(m)}(x). $$ Using the fact $$ a(x) \delta^{(m)}(x) = \sum_{k=0}^m {m \choose k} (-1)^k a^{(k)}(0) \delta^{(m-k)}(x) $$ the last term in $L[P\theta]$ can be brought to form $$ L[P\theta] = L[P]\theta + \sum_{m=0}^{n-1}\sum_{k=0}^{n-1} \alpha_{m,k} P^{(k)}(0) \delta^{(m)}(x) $$ so we've got a Cauchy problem for $P(x)$: $$ LP = 0\\ \sum_{k=0}^{n-1} \alpha_{m,k} P^{(k)}(0) = \begin{cases} 1, &m = 0\\ 0, &m \neq 0\end{cases}. $$ Freezing $L$ keeps $L[P]$ value at $x = 0$, so $M[P]\theta(x)$ remains smooth, but the singluar part seems to be altered completely, since $\alpha_{m,k}$ depend on $a_k(x)$ derivatives at $x = 0$ (and I don't see a reason why should they not). But it appears that freezing derivatives of $a_k(x)$ up to $n - k - 1$ orders leaves the $\alpha_{m,k}$ the same, so the $Mu-\delta = M[P]\theta$ is a smooth function.
This proof is complicated and I might be missing something. Also I have a feeling that the Cauchy problem for $P$ can be expressed much simpler, without nesting four sumations.