I am reading through the solution to the Ornstein-Uhlenbeck SDE given here
$$dX_t = a(\mu - X_t)\,d_t + \sigma \,dW_t$$
I understand all of it except for the first simple step.
We want to simplify the SDE by substituting in $Y_t = X_t - \mu$ so that we can proceed using Ito's lemma and the substitution $Z_t = e^{at} Y_t$.
My question is simply how do we arrive at:
$$dY_t = -a Y_t \, d_t + \sigma \, dW_t$$
Intuitively of course this is what I would expect. But I'm not sure of the actual justification/rule used to do this. The step is trivial so that an explanation is not given in the page I linked.
I hope this addresses your problem. I think you only need to use the integral formulation which is associated to the differential formulation.
Suppose that the OU process $X_t$ is given (otherwise it's difficult to do a substitution). By the definition of solution to an SDE this is equivalent to: $$ X_t -X_0 = \int_0^t a(\mu - X_s)ds + \int_0^t \sigma dW_s. $$ Now observing that $$X_t -X_0 = Y_t - Y_0 $$ gives the required result.