I don't understand how to define such integration factor in order to solve SDE, for example, as was shown in Solving $dX_{t} = \mu X_{t} dt + \sigma dW_{t}$ and Solving Stochastic Differential Equation using integrating factor. And in this book Stochastic Differential Equations in some exercises hints recommend to use integrator factor, but how to get it.

Is there some mechanism(algorithm) which allows us to find this integrator factor?

  • $\begingroup$ What you can do is to first find a solution for exponential component, for example, $dX_t=\mu X_t dt$ in your question, and then variate the constant to a stochastic process, and find this stochastic process. This approach is called the method of "variation of constants." $\endgroup$
    – Gordon
    Nov 19, 2021 at 15:14
  • $\begingroup$ This SDE is a special case of this and as such one of the few explicitly solvable SDEs. Unlike for ODEs there isn't a great deal of solution methods for SDEs. Memorizing the cases that have solutions is imho the best approach. $\endgroup$
    – Kurt G.
    Apr 18, 2022 at 8:04
  • $\begingroup$ The bottom of my answer might be useful. $\endgroup$ Dec 14, 2022 at 23:34

2 Answers 2


I believe that Wikipedia gives a fair definition ; see it as a "helper function".

In your example, $e^{-\mu t}$ helps you getting rid of the drift in your SDE, and you end up solving it by mere stochastic integration ; then just "multiply back" by $e^{\mu t}$ and your initial SDE is solved.

  • $\begingroup$ Ok, thanks. Sorry, I wrong asked, I know how it works in ODF, I'm interested in how to understand which function we should choose as integrator factor for SDE, that is, if there is any mechanism that determines it ? For example, for ODE: $$M(x, y)dx + N(x, y)dy = 0$$ and $\frac{dM}{dy} \neq \frac{dN}{dx}$, then $\mu$ is an integration factor if it depends on only from $x$ or olly from $y$ in form of(for case when $\mu$ depends on from x): $$ \mu(x) = e^{ \frac{\frac{\partial M(x, y)}{\partial y} - \frac{\partial N(x, y)}{\partial x}}{M(x, y)} } $$. $\endgroup$
    – Bohdan_
    Nov 18, 2021 at 18:37

Essentially you want to get rid of the drift of the process $\mu X_t d t$ to get it in a solvable form

$$E[d X_t] = E[\mu X_t d t]\\ \implies d E[X_t] = \mu E[X_t] d t\\ \equiv d y(t) = \mu \cdot y(t) d t\\ \implies \frac{d y(t)}{d t} - \mu \cdot y(t)=0$$

Hence solving this ODE via the integrating factor method one has

$$I(t) = e^{\int_0^t -\mu d t} = e^{-\mu t}$$

So getting the SDE

$$d (e^{-\mu t} X_t) = X_t \cdot d(e^{-\mu t}) + e^{- \mu t} \cdot d X_t + (d e^{- \mu t})(d X_t)\\ = X_t \cdot (- \mu e^{- \mu t}dt) + e^{- \mu t}(\mu X_t d t + \sigma d W_t) + 0\\ = - \mu e^{- \mu t}X_t d t + \mu e^{-\mu t}X_t d t + \sigma e^{- \mu t }d W_t\\ = \sigma e^{-\mu t} d W_t$$

Hence we have removed the drift and get $$e^{-\mu t} X_t = X_0 + \sigma \int_0^t e^{-\mu t} d W_t$$

From there it should be obvious. This method should work with any SDE with drift term $\mu(t)X_t d t$


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