Here is a probably quite basic question on holonomic D-modules, but I am only a physicist so please bear with me.
If I have the weyl algebra $D$ in $n$ variables, and an exact sequence of holonomic $D$-modules \begin{equation} 0\rightarrow M_1 \rightarrow M_2 \rightarrow M_3\rightarrow 0 \end{equation} I know that the holonomic ranks must satisfy $rank(M_2)=rank(M_1)+rank(M_3)$. To me this seems to imply that there is an exact sequence \begin{equation} 0\rightarrow Sol(M_3)\rightarrow Sol(M_2) \rightarrow Sol(M_1)\rightarrow 0\;, \end{equation} where $Sol(\underline\;)$ the solution space of the module, since these are all just vector spaces.
However, I can also consider the solution space as $Hom(\underline\;,\mathcal{O}_X)$ for an appropriate function space $\mathcal{O}_X$. Applying this functor to the exact sequence above then leads to the exact sequence \begin{equation} 0\rightarrow Hom(M_3,\mathcal{O}_X)\rightarrow Hom(M_2,\mathcal{O}_X)\rightarrow Hom(M_1,\mathcal{O}_X)\rightarrow Ext_1(M_3,\mathcal{O}_X)\;. \end{equation} It seems to me that these facts combined somehow seem to imply that $Ext_1(\underline\;,\mathcal{O}_X)=0$ for all holonomic $D$-modules but this seems wrong somehow since I don't see why $\mathcal{O}_X$ should be an injective module.
Could someone spot the mistake I made or explain why it is actually correct?
I am mostly interested in if the exact sequence of solution spaces hold so if someone knows this I'd already be very thankful.
Edit: Could it be related to $Ext^i(M,D)=0$ for $i\neq n$ and $M$ a holonomic $D$-module?