# Solving $dX_{t} = \mu X_{t} dt + \sigma dW_{t}$

I want to solve the following SDE:

$$dX_{t} = \mu X_{t} dt + \sigma dW_{t} \quad X_{0} = x_{0}$$

Integrating, I get:

$$X_{t} - x_{0}= \mu \int_{0}^{t} X_{s} ds + \sigma \int_{0}^{T} dW_{t}$$ $$X_{t} = x_{0} + \mu \int_{0}^{t} X_{s} ds + \sigma [W_{t} - W_{0}]$$ $$X_{t} = x_{0} + \mu \int_{0}^{t} X_{s} ds + \sigma W_{t}$$

This is where I get stuck -- How do I proceed?

I know that the solution to:

$$dX_{t} = \mu X_{t} dt + \sigma X{t} dW_{t} \quad X_{0} = x_{0}$$

Is given by:

$$X_{t} = X_{0}e^{(\mu - \frac{1}{2}\sigma^{2})t} + \sigma W_{t}$$

This is given by the fundamental matrix solution. (Though honestly I am not sure how this is derived, so I can't specialize it to my case).

Sorry for such a basic question. I am trying to learn this from resources on the web. So if anyone could recommend a good book dealing with how to actually solve SDEs, I would really appreciate it.

• Consider $d\big(e^{-\mu t} X_t \big)$. Oct 11 '19 at 12:57
• Okay so then: $d(e^{\mu t} X_{t})$ = $X_{t} d(e^{\mu t}) + e^{\mu t} dX_{t} = \mu X_{t} e^{\mu t} dt + e^{\mu t} dX_{t}$. Solving this for $X_{t}$ I get: $X_{t} = \sigma \frac{dW}{dt}$. That doesn't seem right. Or is it? I am not sure. Oct 11 '19 at 13:07
• there is a minus sign Oct 11 '19 at 13:08
• @Gordon. In this case I get: $d(e^{- \mu t} X_{t})= 0$. What am I doing wrong? Could you please explain how to solve this for an idiot like me? And why do you want to consider that function anyway? I am an engineer trying to learn SDEs from ground up because I realized I could use them to specify prediction models in Kalman filters. However I have no idea what I am doing -- just piecing it together piece by piece. Oct 11 '19 at 13:11
• To eliminate the term $\mu X_t dt$, an integrating factor of the form $e^{-\mu t}$ is the only choice. Oct 13 '19 at 22:48

It appears that you need to read some books such as Stochastic Differential Equations. For such type of equations, you need to use something called integrating factor such as the function $$e^{-\mu t}$$ here. Note that \begin{align*} d\big(e^{-\mu t} X_t \big) &= X_t d\big(e^{-\mu t}\big) + e^{-\mu t} dX_t\\ &=-\mu e^{-\mu t} X_t dt + e^{-\mu t} (\mu X_t dt + \sigma dW_t)\\ &=\sigma e^{-\mu t} dW_t. \end{align*} Then, for $$t > s \ge 0$$, \begin{align*} e^{-\mu t} X_t = e^{-\mu s} X_s+ \int_s^t \sigma e^{-\mu v} dW_v. \end{align*} Consequently, \begin{align*} X_t &= X_s e^{\mu (t-s)} + \sigma\int_s^t e^{\mu (t-v)} dW_v\\ &\sim N\Big(X_s e^{\mu (t-s)}, \, \sigma^2\int_s^t e^{2\mu (t-v)} dv \Big). \end{align*}
• No, unless $\mu=0$. But from here, you can do many thing such as computing the mean and variance etc. Oct 11 '19 at 13:56