Find the solution $Y_t$ of the following SDE $$d\ln Y_t=(-\beta \ln Y_i+C)dt+\sigma dW_t,Y_0=y$$
I know how to solve this by letting $X_t:=\ln Y_t$ and apply ito's lemma on $e^{-\beta t}\ln Y_t$. This is very similar to Ornstein-Uhlenbeck process.
I just wondering why we cannot directly differentiate $e^{-\beta t}\ln Y_t$. I mean how do we proceed as follows and why the following approach does not work,$$d(e^{-\beta t}\ln Y_t)=-\beta e^{-\beta t}\ln Y_tdt+\frac{e^{-\beta t}}{Y_t}dY_t-\frac12\frac{e^{-\beta t}}{Y_t^2}dY_tdY_t $$ Another question is the following always true? $$dY_tdY_t=d[Y,Y]_t$$ Is this just two different notation for quadratic variation? Further on that in this case is the following true?$$ d\ln Y_td\ln Y_t=((-\beta \ln Y_i+C)dt+\sigma dW_t)^2=\sigma^2dt$$ My understanding for Stochastic calculus is a bit shaky. Thanks for any clarification.
Note,
$$d(e^{\beta t}\ln Y_t) = e^{\beta t}(Cdt+\sigma dW_t)$$
Integrate to get,
$$e^{\beta t} \ln Y_t - \ln Y_0 = C\int_0^t e^{\beta \tau}d\tau + \sigma \int_0^t e^{\beta \tau}dW\tau$$
or,
$$ \ln Y_t = e^{-\beta t}\ln y + \frac C\beta(1-e^{-\beta t}) +\sigma \int_0^t e^{-\beta (t-\tau)}dW\tau$$
Thus, the solution of the SDE is
$$ Y_t = \exp\left(e^{-\beta t}\ln y + \frac C\beta(1-e^{-\beta t}) +\sigma \int_0^t e^{-\beta (t-\tau)}dW\tau\right)$$