Let $X = \mathbb{C}^{n}$ and $D_{X}$ be the sheaf of algebraic differential operators. Let $M$ be a coherent $O_{X}$-module. Consider the induced right $D_{X}$-module $M \otimes_{O_{X}} D_{X}$ (the right $D_{X}$-module structure here is the trivial one by multiplying the second factor on the right). If I Hom into $D_{X}$ I obtain a left $D_{X}$-module. That is, consider
$$ L = Hom_{D_{X}}(M \otimes_{O_{X}} D_{X}, D_{X}). $$
I would like to understand what $L$ is in a more useful form and what its left $D_{X}$-module structure is.
By combining tensor-hom adjunction and the coherence of $M$ I believe I get \begin{align*} (\star) Hom_{D_{X}}(M \otimes_{O_{X}} D_{X}, D_{X}) \simeq Hom_{O_{X}}(M, Hom_{D_{X}}(D_{X}, D_{X})) \simeq Hom_{O_{X}}(M, D_{X}) \simeq Hom_{O_{X}}(M, O_{X}) \otimes_{O_{X}} D_{X} \end{align*} where the first $\simeq$ is tensor-hom adjunction and the last $\simeq$ uses the $O_{X}$-coherence of $M$ and the fact that (as $O_{X}$-modules) $D_{X} \simeq \oplus O_{X}$.
Here are my explicit questions: (1) is the above calculation $(\star)$ correct and if not, what would be the correct simplification of $L$?; (2) if the above calculation $(\star)$ is correct, what is the explicit map from LHS to RHS of $(\star)$?; (3) if the above calculation $(\star)$ is correct, what is the left $D_{X}$-module structure on the RHS of $(\star)$?