\delta_i)}f_i(t), \label{ansatz-Not}\end{aligned}$$ for a differentiable function $f_i : [0,T] \to \R$ to be determined. Note that the boundary condition requires $f_i(T)=1$. Plugging this into the HJB equation , we find that $v(x,y,t)/f_i(t)$ factors out of each term, and we get $$\label{eqf}
0 = f_i'(t) + \left(\rho_i +  \theta_i \left(1 - \frac{1}{\delta_i}\right)\widehat{c}_{-i}(t)\right) 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},$$ where we define $$\label{rhoi}
\begin{split}
      \rho_i = \left(1 - \frac{1}{\delta_i}\right) & \left( \frac{\gamma_i (1 - \theta_i/n)(\mu_i - \sigma_i \widehat{\sigma \pi}_{-i}\theta_i (1 - 1 / \delta_i))^2}{2\Sigma_i} \right. \\
      &\quad + \left. \frac{1}{2}(\widehat{\sigma \pi}_{-i}^2 + \frac{1}{n}\widehat{(\nu \pi)^2}_{-i})\theta_i^2(1 - 1 / \delta_i) - \theta_i \widehat{\mu \pi}_{-i} + \frac{\theta_i}{2}\widehat{\Sigma \pi^2}_{-i}  \right).
\end{split}$$ Indeed, the last term in comes from the identity $$\begin{split}
v_x(x,y,t)^{1 - \gamma_i}y^{- \gamma_i \theta_i(1- 1 / \delta_i)} = \epsilon_i^{-\gamma_i} \left( 1 - \frac{\theta_i}{n} \right)^{1 - \gamma_i} \left( 1 - \frac{1}{\delta_i}\right) f_i(t)^{- \gamma_i} v(x,y,t).
\end{split}$$ To solve , let us for the moment abbreviate $$\begin{aligned}
a_i(t) &:= \rho_i + \theta_i(1 - 1/\delta_i)\widehat{c}_{-i}(t), \quad\qquad
b_i(t) := \frac{\epsilon_i^{-\gamma_i}}{\gamma_i} \bar{c}_{-i}(t)^{- \gamma_i \theta_i(1- 1 / \delta_i)}. \label{def:a-b-for-f}\end{aligned}$$ Then (\[eqf\]) rewrites as $$f_i' + a_i f_i + b_i f_i^{1 - \gamma_i} = 0.$$ This is an example of a *Bernoulli equation*, and a well 