Stochastic differential equation substitution reasoning? I need help to explain reasoning behind why we choose certain substitutions to solve SDE. After choosing the correct substitution the solution of the SDE are often quite easy. However I have trouble understanding why we choose certain substitutions. 
For example 
1. Solve the following differential equation:
$dX_t = (\beta - \alpha X_t)dt + \sigma dB_t, \quad X_0 = x_0$
Question 1: Here we set $Y_t = e^{\alpha t}X_t$, find $dY_t$ and then solving the rest of the SDE is easy. Why do we set $Y_t = e^{\alpha t}X_t$? and how do we know how to choose it?
2. and the following:
$dX_t = [1-\ln(X_t)]X_tdt + \sigma X_t dB_t$
Here we first set $Y_t = \ln(X_t)$ find 
$dY_t = (1-\frac{\sigma^2}{2}-Y_t)dt + \sigma dB_t$ 
then we need to set $Z_t = e^t Y_t$ and solve:
$dZ_t = e^t(1 - \frac{\sigma^2}{2})dt + e^t\sigma dB_t$
Integrate $dZ_t$ to find $Z_t$ and lastly put it back to $X_t$ giving
$X_t = e^{Y_t} = e^{e^{-t}Z_t}$
Question 2: Again how do we know which substitutions to choose?
Question 3: Is there any general rule for how to set the substitutions?
Thanks!
 A: 
$dX_t = (\beta-\alpha X_t) \, dt + \sigma \, dB_t$

Forget a moment about the SDE and consider the associated ordinary differential equation
$$dx_t = (\beta- \alpha x_t) \, dt \tag{1}$$
instead. If I would ask you to solve this ODE, you would (hopefully)


*

*first solve the homogeneous equation $$dx_t = -\alpha x_t \, dt$$ and find that the solution of this equation equals $x_t = c e^{-\alpha t}$,

*use variation of constants to solve the inhomogeneous equation $(1)$, i.e. consider $$x(t) = c(t) e^{-\alpha t}$$ and determine $c(t)$ by plugging $x(t)$ into $(1)$.
Now, since we are interested in the solution to the SDE, we modify the variation of constants approach: we let $c(t)$ depend on $\omega$, i.e. we consider $(c(t))_{t \geq 0}$ as a stochastic process. It remains to determine
$$c(t) = e^{\alpha t} X_t,$$
and to this end, we can use Itô's formula. Note that $c(t) =Y(t)$ (defined in the opening question).

$dX_t = (1-\ln (X_t)) \, X_t \, dt+ \sigma X_t \, dB_t$

Given an SDE of the form
$$dX_t = b(X_t) \, dt + \sigma(X_t) \, dt,$$
there are sufficient and necessary conditions on $b$ and $\sigma$ for the transformation into a linear SDE
$$dY_t = (\alpha + \beta Y_t) \, dt + (\gamma+\delta Y_t) \, dB_t,$$
see this question and the second part of this answer. Applying the criterion gives the transform $Y_t = \ln(X_t)$. The substitution $Z_t := e^t Y_t$ can be motivated using exactly the same reasoning as in the first part of this answer.
