I have a problem with solving the following problem:
I.e. I want to show that $X_t$ is a solution to the SDE by employing the Itō formula. Now the problem is I don't get how I should set the function to show this with the Itō formula/lemma. I've tried setting $Z_t:=\ \int_0^{t}... ds$ and then $X_t:=f(t,X_t), f(t,z)=x_0e^{\sigma B_t-\sigma^2t/2}+\alpha z$ but I ended up with another expression for $dX_t$. Could anyone help me how to show that $X_t$ is the solution to the SDE with the help of the Itô formula?
Suppose that $(Y_t)_{t \geq 0}$ and $(Z_t)_{t \geq 0}$ are one-dimensional Itô processes and $f \in C_b^2$. Then Itô's formula states
$$\begin{align*} f(t,Y_t,Z_t)-f(0,Y_0,Z_0) &= \int_0^t f_y(s,Y_s,Z_s) \, dY_s + \int_0^t f_z(s,Y_s,Z_s) \, dZ_s + \frac{1}{2} \int_0^t f_{yy}(s,Y_s,Z_s) \, d\langle Y \rangle_s \\ &\quad + \int_0^t f_{xz}(s,Y_s,Z_s) \, d\langle Y,Z \rangle_s + \frac{1}{2} \int_0^t f_{zz}(s,Y_s,Z_s) \, d\langle Z \rangle_s \\ &\quad + \int_0^t f_t(s,Y_s,Z_s) \, ds. \end{align*}$$
Apply this to
$$Y_t := B_t \qquad \quad Z_t := \int_0^t e^{-\sigma B_s + \frac{\sigma^2}{2}s} \, ds \qquad f(t,y,z) := e^{\sigma y - \sigma^2/2 t} (x_0+\alpha z).$$ (Note that $X_t = f(t,Y_t,Z_t)$.)