how to derive the stochastic differential equation of this process

Question

How can I derive the SDE for the vasicek model :

$$r_t = 0.1 + 0.1 e^{-t} + e^{-t}\int_0 ^t e^s dB_s$$

From observation, the SDE vasicek's model is such that:

$$dr_t = b(a-r_t)dt + \sigma dB_t$$

but I do not know how to link the two. The answer is

$$dr_t = (0.1 - r_t)dt + dB_t$$

Answer 1

Let $X_t = \int_0^t e^s dB_s$. Then $$r_t = f(t, X_t) = .1 + .1e^{-t} + e^{-t}X_t$$ By Ito's Lemma: $$dr_t = \frac{\partial f}{\partial x}dX_t + (\frac{\partial f}{\partial t} + \frac{1}{2}\frac{\partial^2f}{dx^2}Var(X_t))dt$$

So: \begin{eqnarray*} \frac{\partial f}{\partial x}dX_t + (\frac{\partial f}{\partial t} + \frac{1}{2}\frac{\partial^2f}{dx^2}Var(X_t))dt &=& e^{-t}dX_t + (-.1e^{-t} + e^{-t}X_t)dt \\ &=& e^{-t}e^tdB_t + (-.1e^{-t} - e^{-t}X_t)dt \\ &=& dB_t + (-.1e^{-t} - e^{-t}e^t(r_t - .1 - .1 e^{-t})dt \\ &=& dB_t + (.1 - r_t)dt \\ \end{eqnarray*}