Show that process satisfy given equation 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! 
 A: You seem to be able to prove that $\mathrm dY_t=\mathrm e^{bt}\sigma \mathrm dW_t$. Since $X_t=f(t,Y_t)$ where the function $f:(t,y)\mapsto\mathrm e^{-bt}y$ is smooth, the most basic form of Itô's formula yields 
$$\mathrm dX_t=\partial_yf(t,Y_t)\mathrm dY_t+\partial_tf(t,Y_t)\mathrm dt+\tfrac12\partial^2_{yy}f(t,Y_t)\mathrm d\langle Y,Y\rangle_t,
$$ 
that is, $$\mathrm dX_t=\mathrm e^{-bt}\mathrm dY_t-b\mathrm e^{-bt}Y_t\mathrm dt.$$ 
Replacing $Y_t$ and $\mathrm dY_t$ by their expressions in terms of $X_t$ and $\mathrm dW_t$ respectively, you are done.
A: For any process of the form $$dY_t = \sigma_t \, dW_t + \mu_t \, dt, \tag{1}$$ we have
$$df(t,Y_t) = \left( \frac{\partial f(t,Y_t)}{\partial t} + \mu_t \frac{\partial f(t,Y_t)}{\partial y} + \frac{1}{2} \sigma_t^2 \frac{\partial^2 f(t,Y_t)}{\partial y^2} \right) \, dt + \sigma_t \frac{\partial f(t,Y_t)}{\partial y} \, dW_t. \tag{2} $$
You have already shown that $$dY_t = e^{bt} \sigma \, dW_t,$$ i.e. $\mu_t = 0$, $\sigma_t = \sigma e^{bt}$in $(1)$. Now apply $(2)$ to $f(t,y) := e^{-bt} y$ in order to find an expression for the stochastic differential
$$dX_t = df(t,Y_t) = d(e^{-b t} Y_t)$$
