You can directly apply separation of variables, but in fact this PDE can get the more simplified form when applying separation of variables by applying the following change of variables:
Let $\begin{cases}x_1=a+bx\\t_1=t\end{cases}$ ,
Then $\dfrac{\partial u}{\partial x}=\dfrac{\partial u}{\partial x_1}\dfrac{\partial x_1}{\partial x}+\dfrac{\partial u}{\partial t_1}\dfrac{\partial t_1}{\partial x}=b\dfrac{\partial u}{\partial x_1}$
$\dfrac{\partial^2u}{\partial x^2}=\dfrac{\partial}{\partial x}\left(b\dfrac{\partial u}{\partial x_1}\right)=\dfrac{\partial u}{\partial x_1}\left(b\dfrac{\partial u}{\partial x_1}\right)\dfrac{\partial x_1}{\partial x}+\dfrac{\partial u}{\partial t_1}\left(b\dfrac{\partial u}{\partial x_1}\right)\dfrac{\partial t_1}{\partial x}=b^2\dfrac{\partial^2u}{\partial x_1^2}$
$\dfrac{\partial u}{\partial t}=\dfrac{\partial u}{\partial x_1}\dfrac{\partial x_1}{\partial t}+\dfrac{\partial u}{\partial t_1}\dfrac{\partial t_1}{\partial t}=\dfrac{\partial u}{\partial t_1}$
$\therefore\dfrac{\partial u}{\partial t_1}=\dfrac{b^2c^2}{2}\dfrac{\partial^2u}{\partial x_1^2}+bx_1\dfrac{\partial u}{\partial x_1}+fu$
With reference to Change variables into Fokker-Planck PDE,
Let $\begin{cases}x_2=x_1e^{bt_1}\\t_2=t_1\end{cases}$ ,
Then $\dfrac{\partial u}{\partial x_1}=\dfrac{\partial u}{\partial x_2}\dfrac{\partial x_2}{\partial x_1}+\dfrac{\partial u}{\partial t_2}\dfrac{\partial t_2}{\partial x_1}=e^{bt_1}\dfrac{\partial u}{\partial x_2}=e^{bt_2}\dfrac{\partial u}{\partial x_2}$
$\dfrac{\partial^2u}{\partial x_1^2}=\dfrac{\partial}{\partial x_1}\left(e^{bt_2}\dfrac{\partial u}{\partial x_2}\right)=\dfrac{\partial u}{\partial x_2}\left(e^{bt_2}\dfrac{\partial u}{\partial x_2}\right)\dfrac{\partial x_2}{\partial x_1}+\dfrac{\partial u}{\partial t_2}\left(e^{bt_2}\dfrac{\partial u}{\partial x_2}\right)\dfrac{\partial t_2}{\partial x_1}=e^{2bt_2}\dfrac{\partial^2u}{\partial x_2^2}$
$\dfrac{\partial u}{\partial t_1}=\dfrac{\partial u}{\partial x_2}\dfrac{\partial x_2}{\partial t_1}+\dfrac{\partial u}{\partial t_2}\dfrac{\partial t_2}{\partial t_1}=bx_1e^{bt_1}\dfrac{\partial u}{\partial x_2}+\dfrac{\partial u}{\partial t_2}$
$\therefore bx_1e^{bt_1}\dfrac{\partial u}{\partial x_2}+\dfrac{\partial u}{\partial t_2}=\dfrac{b^2c^2e^{2bt_2}}{2}\dfrac{\partial^2u}{\partial x_2^2}+bx_1e^{bt_1}\dfrac{\partial u}{\partial x_2}+fu$
$\dfrac{\partial u}{\partial t_2}-fu=\dfrac{b^2c^2e^{2bt_2}}{2}\dfrac{\partial^2u}{\partial x_2^2}$