Solution to the linear SDE $dX_t = \alpha X_t \, dt + \sqrt{2} dB_t$ using Itô calculus

Question

So if I have the following generator and an initial condition:

$$A(f)(x) = \alpha x f'(x) + f''(x) \\ X_0 = x \in \mathbb{R}^+$$

I've been asked to find $X_t$ and assume that $\alpha$ is a constant. From the definition of the infinitesimal generator, I get that:

$$dX_t = \alpha X_t dt + \sqrt{2} dB_t$$

How do I go about setting up an application of Ito's Lemma to this diffusion in order to solve for $X_t$? Separation of variables doesn't seem to apply in this situation.

Answer 1

In order to solve the SDE

$$dX_t = \alpha X_t \, dt + \sqrt{2} \, dB_t \qquad X_0 = x \tag{1}$$

we consider the corresponding ordinary differential equation

$$dx(t) = \alpha x(t) \, dt, \qquad x(0)=c.$$

It is not difficult to see that its unique solution equals

$$x(t) = c \, e^{\alpha t}.$$

Now the idea is to use an analogue of the variation of constants-approach: We let the constant $c$ depend on the time $t$ and on $\omega$, i.e. we set

$$C_t(\omega) := X_t(\omega) \cdot e^{-\alpha t}. \tag{2}$$

Applying Itô's formula (to $f(t,x) := x e^{-\alpha t}$), we get

$$C_t - C_0 = \int_0^t e^{-\alpha s} \, dX_s - \alpha \int_0^t X_s e^{-\alpha s} \, ds \stackrel{(1)}{=} \sqrt{2} \int_0^t e^{-\alpha s} \, dB_s.$$

Hence, by $(2)$,

$$X_t = C_t e^{\alpha t} = e^{\alpha t} C_0 + \sqrt{2} \int_0^t e^{\alpha (t-s)} \, dB_s.$$

Finally, note that

$$C_0 = X_0 \cdot 1 = x.$$

Remarks

1. The process $(X_t)_{t \geq 0}$ is an Ornstein-Uhlenbeck process.
2. Another possibility to solve $(1)$ is the following: First solve the homogeneous SDE $$dX_t = \sqrt{2} \, dB_t$$ and then, using again an approach similar to variation of constants, find a solution to the non-homogeneous SDE $(1)$, see e.g. René L. Schilling/Lothar Partzsch: Brownian Motion - An Introduction to Stochastic Processes.