I have to show that process (1) $$X_t=e^{-bt}X_0+\int_0^te^{-b(t-s)}\sigma dW_s$$ satisfies the following equation (2) $$dX_t=-bX_tdt+\sigma dW_t$$ My attempt:
Multiply both sides of (1) by $e^{bt}$ and that yields (3) $$Y_t=X_0+\int_0^te^{bs}\sigma dW_s$$ where $$Y_t:=e^{bt}X_t$$ Now we have two equalities $$dY_t=d(e^{bt}X_t)=e^{bt}\sigma dW_t$$ Now I think I should use Ito formula for $F(x,t)=e^{bt} x$. Ito formula is given by: $$ \begin{align} df(t,X_t) =\left(\frac{\partial f}{\partial t} + \mu_t \frac{\partial f}{\partial x} + \frac{1}{2}\sigma_t^2\frac{\partial^2f}{\partial x^2}\right)dt+ \sigma_t \frac{\partial f}{\partial x}\,dW_t. \end{align} $$ My problem is that I don't know what are $\mu_t$ and $\sigma_t$ in our case or how to find them? I would like to see some explanation. Thanks a lot!