Bayesian-Nash equilibrium in a first-price auction
It is actually assumed that $b_i(v_i)$ is of the form $\alpha_i+\beta_i \cdot v_i$. So it is an affine function. Linearity only works if the bottom of the uniform distribution is 0.

A somewhat intuitive reasoning is that if the valuations are uniformly distributed over $[c,d]$ and $b_i(v_i)$ is a symmetric equilibrium, then $b_i(v_i)+k$ should define a symmetric equilibrium when valuations are uniform over $[c+k,d+k]$ as neither the probability of winning nor the gain changes.

Assuming a strategy of the form $\alpha_i+\beta_i \cdot v_i$ you get

$$
=Pr[\alpha_1 + \beta_1 \cdot v_1 &lt; \alpha_2 + \beta_2 \cdot v_2] \cdot ((1- \beta_2) \cdot v_2 - \alpha_2) 
$$

$$
=Pr\left[v_1 &lt; \frac{\alpha_2 - \alpha_1 + \beta_2 \cdot v_2}{\beta_1} \right] \cdot ((1- \beta_2) \cdot v_2 - \alpha_2) 
$$

$$
=\frac{\frac{\alpha_2 - \alpha_1 + \beta_2 \cdot v_2}{\beta_1}-c}{d-c}\cdot ((1- \beta_2) \cdot v_2 - \alpha_2)
$$

Taking derivative w.r.t. $\alpha_2$ and $\beta_2$ you get your first order conditions. Assuming symmetry you get a unique solution. If $c = 0$ then $\alpha_1 = \alpha_2 = 0$.

By the way, I prefer treating $b_2(v_2)$ as the decision variable, then you can
$$
\max_{b_2(v_2)} 
Pr\left[v_1 &lt; \frac{b_2(v_2) - \alpha_1}{\beta_1} \right] \cdot (v_2 - b_2(v_2)) 
$$
and the last fraction is somewhat simpler.