Let $W=\{W_t\}_{t\in[0;T]}$ be a real-valued Brownian motion, $\{F_t\}_{t\in [0;T]}$ the filtration generated by $W$, augmented with the nullsets, let $C\in (0;\infty)$ and $\{a_t\}_{t\in[0;T]}$ be a bounded and adapted process. Let $A_t := \int^t_0 a_s \mathrm ds$.
Question: Is the BSDE $$ \mathrm dX_t = C X_t \big( a_t + X_t \big) \mathrm dt + Z_t \mathrm dW_t, \quad \lim_{t\to T} X_t = \infty $$ explicitly solvable? (I only require that $Z \in L^2(\Omega\times[0;t])$ for every $t < T$.)
If $a$ was deterministic, then $$ X_t := \frac {\exp(C A_t)} {C \int^T_t \exp(C A_s) \mathrm ds} $$ would be a solution.
What I've tried: As in the case of linear BSDEs, I set $X_t = B_t M_t$ where $\mathrm dB_t = b_t \mathrm dt$ and $\mathrm dM_t = m_t \mathrm dW_t$. Then $$ M_t b_t \mathrm dt + B_t m_t \mathrm dW_t = M_t \mathrm dB_t + B_t \mathrm dM_t = \mathrm dX_t = C X_t \big( a_t + X_t \big) \mathrm dt + Z_t \mathrm dW_t = C B_t M_t \big( a_t + B_t M_t \big) \mathrm dt + Z_t \mathrm dW_t $$ Dividing both absolutely continuous parts by $M_t$ leads to the ODE $$ \mathrm dB_t = C B_t \big( a_t + B_t M_t \big) \mathrm dt $$ This leads to the general solution $$ B_t = \frac {\exp(C A_t)} {K - C \int^t_0 M_s \exp(A_s) \mathrm ds} $$ Now it seems that I would have a solution if I found a process $M$ being a martingale on every $[0;t]$ with $t < T$ such that $$ C \int^T_0 M_s \exp(A_s) \mathrm ds = K. $$ Since we can multiply $B$ by a constant and and divide $M$ by the same constant, we can just set $K=C$, this leads to the condition $$ \int^T_0 M_s \exp(A_s) \mathrm ds = 1. $$