Differential equation for first-price auction

Question

I want to solve a differential equation of the first-price auction. In particular, from Jonathan Levin's October 2004 lecture notes, we have the following differential equation:

$$b'(s) = (s-b(s))(n-1)\frac{f(s)}{F(s)}$$

Solving this, we should get

$$b(s) = s - \frac{1}{F^{N-1}} \int_\bar{s}^{s_i} F^{N-1}(\tilde{s}) \,{\rm d}\tilde{s}$$

I would like to understand how to get this equation.

I found a November 2011 paper by Timothy P. Hubbard & Harry J. Paarsch discussing this part. On page 6, although it is a regular expression of the differential equation, I should solve the following equation:

$$y=\frac{1}{\mu(x)}\int_{x_0}^x \mu(u)q(u) {\,\rm d}u + k$$

Could you help how to solve this equation to get $b(s)$?

Answer 1

For notational simplicity, let me define the distribution $G(s) = F^{N-1}(s)$ with density $g(s)$. Let $\underline{s} = 0$ (for simplicity).

We have $$ b'(s)G(s)+b(s)g(s)=s g(s) $$ Integrating to $x$ gives us $$ \int_0^x b'(s)G(s)+b(s)g(s) ds = \int_0^x s g(s)ds $$ Notice that $\frac{\partial }{\partial s}(b(s)G(s)) = b'(s)G(s)+b(s)g(s)$, so $$ b(x)G(x) = \int_0^x s g(s) ds $$ where we use the boundary condition that $b(0) = 0$.

Now, $$ b(x) = \frac{1}{G(x)}\int_0^x s g(s) ds $$

Finally, apply integration by parts to $\int_0^x s g(s)$ to get $$ b(x) = \frac{1}{G(x)}\left[ G(x) x - \int_0^xG(y) dy\right]= x - \int_0^x\frac{G(y)}{G(x)} dy $$