TITLE: $x_t := a_t -b_t c_t $ , with $dx_t = \theta (\mu-x_t) dt+ \sigma dW_t$
QUESTION [5 upvotes]: I would like to solve the following equation explicitly using Ito's lemma:
$$
x_t := a_t -b_t c_t  ,
$$
where $x_t$ is an Ornstein-Uhlenbeck process (see here)
$$
dx_t = \theta (\mu-x_t) dt+ \sigma dW_t
$$
Can I simply use the solution provided here? So:
$$ f(x_t, t) = x_t e^{\theta t}$$
and then:
\begin{align}
df(x_t,t) & =  \theta x_t e^{\theta t}\, dt + e^{\theta t}\, dx_t \\[6pt]
& = e^{\theta t}\theta \mu \, dt + \sigma e^{\theta t}\, dW_t.
\end{align}
$$
x_t  = x_0 e^{-\theta t} + \mu(1-e^{-\theta t}) + e^{-\theta t}\int_0^t \sigma e^{\theta s}\, dW_s.
$$

REPLY [3 votes]: You have correctly found $x_t$ as 
$$x_t=x_0e^{-\theta t}+\mu (1-e^{-\theta t})+\sigma e^{-\theta t}\int_0^te^{\theta s}dW_s$$
We can rewrite this as 
$$x_t=a_t-b_tc_t$$
where
$$\begin{align}
&a_t=x_0e^{-\theta t}+\mu (1-e^{-\theta t})\\\\
&b_t=-\sigma e^{-\theta t}\\\\
&c_t=\int_0^te^{\theta s}dW_s
\end{align}$$

Note, 
$$dx_t=da_t-c_tdb_t-b_tdc_t$$
with 
$$\begin{align}
&da_t=\theta(\mu- a_t)dt\\\\
&db_t=-\theta b_tdt\\\\
&dc_t=e^{\theta t}dW_t
\end{align}$$
Then, 
$$\begin{align}
dx_t&=da_t-c_tdb_t-b_tdc_t\\\\
&=\theta(\mu-a_t+b_tc_t)dt-(-\sigma e^{-\theta t})(e^{\theta t}dW_t)\\\\
&=\theta(\mu-x_t)dt+\sigma dW_t
\end{align}$$
as expected!!