(\rho_i +  \theta_i (1 - 1 / \delta_i)\widehat{c}_{-i}(t)) f_i(t) + \frac{\epsilon_i^{-\gamma_i}}{\gamma_i} \bar{c}_{-i}(t)^{- \gamma_i \theta_i(1- 1 / \delta_i)} f_i(t)^{1 - \gamma_i}, \label{systeq2}\end{aligned}$$ with $f_i(T) = 1$. Indeed, the first equation gives the best response of agent $i$ in terms of the other agents’ strategies (computed in ) and the function $f_i$. The second equation is exactly the differential equation which determined $f_i$, which we solved explicitly in terms of the other agents’ strategies in . However, now that we have verified the validity of the ansatz for $v_i(x,y,t)$, to resolve the equilibrium it is more convenient to abandon the explicit form for $f_i$ and instead solve the equations and simultaneously. To do this, first plug into the last term of to find $$f_i'(t) + \left(\rho_i +  \theta_i \left(1 - \frac{1}{\delta_i}\right)\widehat{c}_{-i}(t) + \frac{1}{\gamma_i} c_i(t)\right) f_i(t) = 0.$$ Defining the full average $\widehat{c}(t) = \frac{1}{n} \sum_{k = 1}^n c_k(t)$, note that $\widehat{c}_{-i}(t) = \widehat{c}(t) - c_i(t)/n$. Recalling the definition of $\gamma_i$ in , we deduce that $$f_i'(t) + \left(\rho_i +  \theta_i \left(1 - \frac{1}{\delta_i}\right)\widehat{c}(t) + \frac{1}{\delta_i} c_i(t)\right) f_i(t) = 0.$$ Hence, with $f_i(T) = 1$ this leads to $$\begin{aligned}
f_i(t) = \exp \left( \int_t^T \left(\rho_i +  \theta_i \Big(1 - \frac{1}{\delta_i}\Big)\widehat{c}(s) + \frac{1}{\delta_i} c_i(s)\right) 