# Differential of integrating factor $d(e^{at}r_t)$ in Vasicek model

I am attempting to solve the Vasicek model SDE (using Wikipedia parametrisation):

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

Every solution is proceeding to multiply both sides of the equation by the integrating factor $$e^{at}$$ (akin to solving linear ODEs). After multiplication and rearrangement we get the following equation:

$$e^{at}dr_t + e^{at}ar_tdt= e^{at}(abdt + \sigma dW_t)$$ Now the left hand side is apparently equal to $$d(e^{at}r_t)$$. How is that exactly the case?

Is it by Ito product rule? If so what is $$X(t)$$ and $$Y(t)$$?

Is it by Ito's lemma but then what is the $$f(x,t)$$

\begin{align} d\left(e^{at}r_t\right)&=e^{at} dr_t+r_t de^{at} \\[6pt] &=e^{at} dr_t+r_t e^{at} d(at) \\[6pt] &=e^{at} dr_t+r_t e^{at} a dt \end{align}
• By the "cov of a deterministic and stochastic term" do you mean $dr_tde^{at}$ from Ito's Product rule which when we multiply out and substitute for $dr_t$ we get $[a(b-r_t)dt + \sigma dW_t]de^{at}$ and hence we only get the $dtdt$ and $dtdW_t$ terms which are all 0? – user1568263 Oct 24 '18 at 18:39