Nash Equilibrium in 2 bidder auction
Assuming the auction has a pure strategy equilibrium, we can set up the problem as follows:

Each bidder has a value $v_i\sim F$ on support $[\underline{v},\overline{v}]$. The symmetric equilibrium, we posit, is an increasing continuous function $b(v)$.

Now, imagine bidder $i$ bids as if his value were $\widetilde{v}$. Then he wins iff $v_j&lt;\widetilde{v}$ and $i$'s expected payoff is therefore

$$U_i=-[1-F(\widetilde{v})]b(\widetilde{v})+\int_0^{\widetilde{v}}\! \left(v_i-b(v_j)\right)F'(v_j)\,dv_j$$.

The first term is the payoff if $i$ loses (he then pays his own bid, which is $b(\widetilde{v})$ sinc $i$ is using $\widetilde{v}$'s bidding strategy). The second term is the payoff if he wins (he pays his rival's bid, provided his rival's value is less than $\widetilde{v}$).

Now we compute the best $\widetilde{v}$ for $i$ to choose by computing a first-order condition:
$$\frac{\partial U_i}{\partial \widetilde{v}}=-[1-F(\widetilde{v})] b'(\widetilde{v})+v F'(\widetilde{v})=0.$$

We know that, in equilibrium, it must be optimal for $i$ to choose $\widetilde{v}=v_i$ (i.e., it must be a best response for $i$ to use his intended strategy rather than deviate to someone else's strategy):

$$-[1-F(v)] b'(v)+v F'(v)=0.$$

Thus,
$$b(v)=b(\underline{v})+\int_{\underline{v}}^v\!\frac{xF'(x)}{1-F(x)}\,dx.$$

We can check that $$b'(v)=\frac{vF'(v)}{1-F(v)}&gt;0$$ as required. All we need is a boundary condition to pin-down $b(\underline{v})$. For example, we might think it natural that $b(\underline{v})=0$ because a $\underline{v}$ type knows he can never win the auction.