Let $X$ be a smooth projective variety over $\mathbb{C}$. Let $E,F\to X$ be 2 holomorphic vector bundles and $D\hookrightarrow X$ be a smooth divisor. Denote by $\mathcal{O}\_X(D)$ the line bundle associated to the divisor and let $s: \mathcal{O}\_X\to \mathcal{O}\_X(D)$ be a section, such that $s^{-1}(0) = D$. Assume also that we have an isomorphism $E|\_D\cong F|\_D$ of the restrictions to $D$. 


Consider the following perfect complex $\mathcal{E}^\bullet$ on $X$:
\begin{equation\*}
E\oplus F\xrightarrow{
\begin{pmatrix}
\text{id}\otimes s& 0\\
0&\text{id}\otimes s
\end{pmatrix}}
 (E\oplus F)\otimes \mathcal{O}\_X(D)\xrightarrow{
\begin{pmatrix}
\rho\_E&\rho\_F
\end{pmatrix}
} (E\otimes \mathcal{O}\_X(D))|\_D\,,
\end{equation\*}
where $\rho\_{(-)}$ corresponds to restriction of sections to $D$, and we use the isomorphism $F|\_D\cong E|\_D$.


Now it is well known that there exists a locally free resolution of this complex on $X$. However, is it possible to write it down explicitly from what we know?


**Edit:** By a locally free resolution I mean a complex of vector bundles $L^\bullet$ with a quasi-isomorphism $L^\bullet\to \mathcal{E}^\bullet$. It is important to me that I have such a map, as I would like to use it to construct a differential operator. 



Explicit locally free resolution of a perfect complex $E\oplus F\to (E\oplus F)\otimes \mathcal{O}_X(D)\to (E\otimes \mathcal{O}_X(D))|_D$
Projections to the second summands define a morphism from that complex to the complex
$$
F \stackrel{s}\to F(D)\tag{\*}
$$
of locally free sheaves. The cone of this morphism is the complex
$$
0 \to E \stackrel{s}\to E(D) \stackrel{\rho\_E}\to E(D)\vert\_D \to 0
$$
which is acyclic. Therefore, $(\*)$ is a locally free resolution of the original complex.


EDIT. Alternatively, let 
$$
K = \mathrm{Ker}\Big((\rho\_E, \rho\_F) \colon E(D) \oplus F(D) \to E(D)\vert\_D\Big).
$$ 
Then $K$ is locally free, the morphism $(s, s) \colon E \oplus F \to E(D) \oplus F(D)$ factors through $K$, and the complex
$$
E \oplus F \to K
$$
is quasiisomorphic to the original one (via the morphism, which is identical on $E \oplus F$ and is the natural embedding on $K$).


