3# zhangpeng95
Thanks for posting the paper.
The trick is to find the integrating factor. But first you need to change the variable from t to \tau.
the SDE can be rewritten as:
dX(\tau) - \frac{\mu X}{ \tau} d \tau = - \gamma dz(\tau)
Multiply both sides by the integrating factor:
\tau ^{ - \mu }
The LHS now becomes exact differential form and you are done.
Good luck with the paper.