I have tried solving this equation from several manners but no luck. Can it be solved? $$\frac{d f}{d t} = A f^2 +g(t)$$
The solution for the homogeneous is (I think; somebody should confirm) $$f_h(t) = -\frac{1}{At + C},$$ where $C$ is a constant. So substituting back would be $$\frac{\frac{dC(t)}{dt}+A}{(C(t)+A t )^2} = A \left(\frac{-1}{At + C}\right)^2 + g(t)$$
but that's where I stopped... Thank you in advance.
$$\frac{d f}{d t} = A f^2 +g(t)$$ Let $f(t)=-\frac{1}{A\:h(t)}\frac{d h}{dt}$ which leads to : $$\frac{d^2h}{dt^2}+A\,g(t)h(t)=0$$ It is well-known that this kind of second order linear ODE can be analytically solved for some forms of functions $g(t)$ and cannot for many others.
So, it is impossible to answer to the question if the function $g(t)$ is not explicitly defined.