I am trying to understand how we come up with the PDE to be solved to find $u(x,t) = E[f(X(t)]$ in the context where $X(t)$ is a solution of the SDE $$dX(t) = \mu(X(t))dt + \sigma(X(t))dB(t)$$
For example I understand for a Brownian motion with drift: $$dX(t) = cdt + dB(t)$$
The generator is $\frac{1}{2}\frac{\partial^2}{\partial x^2} + c \frac{\partial}{\partial x}$ and so the PDE to be solved is: $$u_{t} = \frac{1}{2} u_{xx} + cu_{x}$$
However for $dX(t) = X(t)dt + \sqrt{2}dB(t)$, I obtain the generator of $\frac{\partial^2}{\partial x^2} + X(t) \frac{\partial}{\partial x}$. However it seems the PDE to solve is: $$u_{t} = u_{xx} + xu_{x} + u$$
May I ask where the $u$ comes from?
