understanding a bribing function in a second bid auction with one-round bribe In this paper about bribing in second price auctions, they describe a model of variable bribe, where the bribing player may decide what amount to offer to the accepting player as a function of his value
On page $16$, proposition $6$ they come to a conclusion if the bribing function $b()$ is continuous then it is the unique solution to the following ode:
$$ b'(\theta_j) = \begin{cases} \frac{F'(\theta_j + b(\theta_j))(\theta_j - b(\theta_j))}{F(\theta_j + b(\theta_j)) - F'(\theta_j +b(\theta_j))(\theta_j-b(\theta_j))} & \text{if $\theta_j+b(\theta_j) \lt 1$} \\
0 & \text{else}\end{cases} $$
They include a short explanation on why this is true, but I didn't fully understand it.
I understand each part of the equation on its own, but not sure why $b$ must satisfy this ODE
 A: Since in the continuous equilibrium the bribing player gives away their type when bribing, the bribed player will accept exactly if $\theta_i-\theta_j\le b(\theta_j)$, or equivalently if $\theta_i\le\theta_j+b(\theta_j)$. Thus the gain that the bribing player expects from the bribe being accepted is
$$
P\left(\theta_i\le\theta_j+b(\theta_j)\right)\left(\theta_j-b(\theta_j)\right)=F\left(\theta_j+b(\theta_j)\right)\left(\theta_j-b(\theta_j)\right)\;.
$$
If we weren't dealing with a continuous equilibrium, we'd simply set the derivative of this expression with respect to $b(\theta_j)$ to zero to find the equilibrium condition
$$
F'\left(\theta_j+b(\theta_j)\right)\left(\theta_j-b(\theta_j)\right)-F\left(\theta_j+b(\theta_j)\right)=0\;.
$$
However, in the continuous equilibrium, this doesn't take into account that the bribed player is inferring the bribing player's type from the bribe. Whereas the $\theta_j$ in the gain per accepting opponent, $\theta_j-b(\theta_j)$, is the actual type of the bribing player, the $\theta_j$ in the probability of acceptance, $F\left(\theta_j+b(\theta_j)\right)$, is the inferred type of the bribing player, the type that the bribed player infers based on the bribe. The bribed player still has the same beliefs and strategy, so we need to treat this inferred $\theta_j$ as $\theta(b)$, the inverse function of $b(\theta)$ that the bribed player uses to infer the bribing player's type from the bribe. Thus, the bribing player's expected gain is
$$
F\left(\theta(b(\theta_j))+b(\theta_j)\right)\left(\theta_j-b(\theta_j)\right)\;.
$$
This leads to an additional term in the derivative with respect to $b(\theta_j)$, and the equilibrium condition becomes
$$
F'\left(\theta_j+b(\theta_j)\right)\left(\theta'(b(\theta_j))+1\right)\left(\theta_j-b(\theta_j)\right)-F\left(\theta_j+b(\theta_j)\right)=0\;.
$$
Then using $\theta'(b(\theta_j))=b'(\theta_j)^{-1}$ yields a differential equation for $b(\theta_j)$, which you can bring into the form given in the paper by solving for $b'(\theta_j)$.
Edit:
I'll rewrite this with more rigorous notation, since the way I wrote it above apparently left room for confusion.
So let $b(\theta_j)$ denote the function that the bribing player uses to determine the bribe in the original strategy, and let $\theta(b)$ denote its inverse, the function that the bribed player uses to infer the bribing player's type from the bribe.
To derive the equilibrium condition, we consider some fixed $\theta_0$, and we want to see whether the bribing player can gain by offering a different bribe $b_0$ at $\theta_0$ than the strategy $b(\theta_j)$ would prescribe. By the definition of an equilibrium, the bribed player's strategy remains the same, so the bribed player continues to use the function $b(\theta)$ to infer the bribing player's type from the bribe.
The bribing player's expected gain from the bribe $b_0$ being accepted is
$$
P\left(\theta_i\le\theta(b_0)+b_0\right)\left(\theta_0-b_0\right)=F\left(\theta(b_0)+b_0\right)\left(\theta_0-b_0\right)\;.
$$
Here $\theta_0-b_0$ is the bribing player's gain per accepting player; we have to use the actual type $\theta_0$ since the bribing player calculates knowing her own type. By contrast, $\theta_i\le\theta(b_0)+b_0$ is the acceptance condition for the bribed player, and here we have to use the inferred type $\theta(b_0)$, since the bribed player calculates with this inferred type and doesn't know the actual type of the bribing player.
Setting the derivative with respect to $b_0$ to zero yields
$$
F'\left(\theta(b_0)+b_0\right)\left(\theta'(b_0)+1\right)\left(\theta_0-b_0\right)-F\left(\theta(b_0)+b_0\right)=0\;.
$$
This condition must hold at $b_0=b(\theta_0)$, so we substitute that into the equation,
$$
F'\left(\theta(b(\theta_0))+b(\theta_0)\right)\left(\theta'(b(\theta_0))+1\right)\left(\theta_0-b(\theta_0)\right)-F\left(\theta(b(\theta_0))+b(\theta_0)\right)=0\;,
$$
and then we use the fact that $\theta$ is the inverse of $b$ and thus $\theta(b(\theta_0))=\theta_0$:
$$
F'\left(\theta_0+b(\theta_0)\right)\left(\theta'(b(\theta_0))+1\right)\left(\theta_0-b(\theta_0)\right)-F\left(\theta_0+b(\theta_0)\right)=0\;.
$$
Since the fixed $\theta_0$ that we picked was arbitrary, we can now replace it with the general $\theta_j$, yielding the equilibrium condition as a differential equation:
$$
F'\left(\theta_j+b(\theta_j)\right)\left(\theta'(b(\theta_j))+1\right)\left(\theta_j-b(\theta_j)\right)-F\left(\theta_j+b(\theta_j)\right)=0\;.
$$
