Solve a system of SDEs

Question

The Question:

If I have the following system of SDEs

$$ dX = (\beta + BX)dt + \sigma dW(t) $$

where $X(t)$ is a $d$-dimensional process, $W(t)$ is a $d$-dimensional Wiener process, $B$ a $d$-by-$d$ matrix and $\beta$ a $d$-dimensional column vector, how do I go about solving it?

My thoughts so far:

Can I break up the SDE into parts like:

$$ \int_0^tdX = \int_0^t\beta ds + \int_0^tBX(s)ds + \int_0^t\sigma dW(s) $$

Now I know that when we have a system of ODEs of the form $dx = Axdt$ for some matrix $A$ we get the solution $x(t) = e^{At}x(0)$, but I can't work out how to use this fact in this case.

Does this imply $X(t) = X(0)(\beta t + e^{Bt} + \sigma W(t))$? Any help would be appreciated.

Answer 1

Let $Y= e^{-Bt}X$. Then

$$ dY = e^{-Bt}(\beta dt+ \sigma dW_t) $$ Integrate

$$Y=Y_0 + \int_0^t\beta e^{-Bs}ds + \int_0^t\sigma e^{-Bs}dW_s $$

Thus, the solution is

$$X = e^{Bt}X_0+ \int_0^t\beta e^{B(t-s)}ds + \int_0^t\sigma e^{B(t-s)} dW_s $$