$dX_t=-\mu X_tdt + \sigma dW_t$. Prove that $X_t = e^{-\mu t}X_0 + \sigma \int_0^t e^{-\mu(t-u)}dW_u $

Question

So the solution says use Ito-s formula, taking $Y_t:= e^{\mu t}X_t$ to obtain $dY_t = [\mu e^{\mu t}X_t - e^{\mu t}\mu X_t + e^\mu t \sigma dW_t] $.

As far as I can see though, Ito's formula says that $dY_t = [ - e^{\mu t}\mu X_t + e^\mu t \sigma dW_t] $.

Where is the extra $e^{\mu t}\mu X_t$ coming from?

$\{W_t\}_{t\geq0}$ is a standard brownian motion. $\mu, \sigma, X_0$ are all constants

Answer 1

Suppose that

$$dX_t = - \mu X_t \, dt + \sigma \, dW_t \tag{1}$$

and set $Y_t := e^{\mu t} X_t$. Itô's formula states that

$$df(t,X_t)= \frac{\partial}{\partial x} f(t,X_t) \, dX_t + \frac{\partial}{\partial t} f(t,X_t) \, dt + \frac{1}{2} \frac{\partial^2}{\partial x^2} f(t,X_t) \, d\langle X \rangle_t. \tag{2}$$

Here, we have $f(t,x) = e^{\mu t} \cdot x$, i.e.

$$\begin{align*} \frac{\partial}{\partial x} f(t,x) = e^{\mu t} \qquad \quad \frac{\partial^2}{\partial x^2} f(t,x) = 0 \qquad \quad \frac{\partial}{\partial t} f(t,x) = \mu e^{\mu t} x. \end{align*}$$

Plugging this into $(2)$ yields

$$\begin{align*} dY_t &= e^{\mu t} dX_t + \mu e^{\mu t} X_t \, dt \stackrel{(1)}{=} - \mu e^{\mu t} X_t \, dt + \sigma e^{\mu t} dW_t + \mu e^{\mu t} X_t\end{align*}$$

... and that's exactly the claim.