Consider the following SDE: $dX(t)=(A(t) X(t) + a(t)) dt + (B(t)X(t) + b(t))dW(t)$.
The solution is given by $$X(t)= Z(t) \cdot \left(x + \frac{a(t) -B(t)b(t)}{Z(t)}dt+ \frac{b(t)}{Z(t)} dW(t)\right)$$ where $Z(t)$ is the solution of $$d Z(t) = Z(t) A(t) dt + Z(t)B(t) dW(t)$$ and $X(0)=x$.
To prove this, denote the right hand side of the solution by $Y(t)$, such that $X(t)=Z(t)Y(t)$.
Applying Ito's product rule yields: \begin{align} Z(t)Y(t)&= x+ Z(t)dY(t) + Y(t)dZ(t) + dZ(t)dY(t)\\ &= x +Z(t)\left[(A(t) Y(t) + a(t)) dt + (B(t)Y(t) + b(t))dW(t)\right] \\ &+ Y(t)[Z(t) A(t) dt + Z(t)B(t) dW(t)] \\ &+ (B(t)Y(t) + b(t))Z(t)B(t) dW(t). \end{align}
How can I simplify that further to get to the solution?
We have the SDE $dX=(AX+a)\,dt+(BX+b)\,dW$ where we omit $\cdot(t)$ for convenience.
With the substitution $X=ZY$ where $dY=F\,dt+G\,dW$ for some $F,G$, we have \begin{align}d(ZY)=(AZY+a)\,dt+(BZY+b)\,dW.\end{align} Simultaneously, from Itô's product rule we obtain \begin{align}d(ZY)&=Z\,dY+Y\,dZ+dZ\,dY\\&=Z(F\,dt+G\,dW)+Y(AZ\,dt+BZ\,dW)+BZG\,dt\\&=(ZF+AZY+BZG)\,dt+(ZG+BZY)\,dW\end{align} using the definition that $dZ=AZ\,dt+BZ\,dW$. Equating the two expressions yields $$(ZF+BZG-a)\,dt+(ZG-b)\,dW=0$$ so that $G=b/Z$ and $F=(a-Bb)/Z$. Thus $$X(t)=Z(t)\left(x+\int_0^t\frac{a(s)-B(s)b(s)}{Z(s)}\,ds+\int_0^t\frac{b(s)}{Z(s)}\,dW_s\right).$$