How can I solve this stochastic system of equation? $(B_1(t),B_2(t))$ is a 2-dimensional standard Brownian motion.
$\alpha , \beta$ are constant.
The system of equations is:
$$dX_1(t)=X_2(t)dt+\alpha dB_1(t)\\dX_2(t)=-X_1(t)dt+\alpha dB_2(t)$$
I tried many 1-dimensional Brownian SDEs and solved them, but I am stuck on this problem.
 A: Let $A = \left(\begin{array}{rr}
0 & 1 \\
-1 & 0
\end{array}\right)$, $X_t = \left(\begin{array}{r}
X_1(t)\\
X_2(t)
\end{array}\right)$, and $B_t = \left(\begin{array}{r}
B_1(t)\\
B_2(t)
\end{array}\right)$. Then
\begin{align*}
dX_t = AX_t\,dt + \alpha\, dB_t.
\end{align*}
Note that 
\begin{align*}
d\left(e^{-At} X_t \right) &= -Ae^{-At} X_t\, dt + e^{-At} dX_t\\
&=\alpha e^{-At} dB_t.
\end{align*}
This implies that
\begin{align*}
X_t = e^{At} X_0  + \alpha \int_0^t e^{-A (s-t)} dB_s.
\end{align*}
Note that 
\begin{align*}
e^{At} &= \sum_{n=0}^{\infty} \frac{A^n t^n}{n!}\\
&=\sum_{n=0}^{\infty} \left(\frac{A^{2n} t^{2n}}{(2n)!} + \frac{A^{2n+1} t^{2n+1}}{(2n+1)!}\right)\\
&=\sum_{n=0}^{\infty} \left(\frac{(-1)^n t^{2n}}{(2n)!}I + \frac{(-1)^n t^{2n+1}}{(2n+1)!} A\right)\\
&= \cos t\, I + \sin t\, A\\
&= \left(\begin{array}{rr}
\cos t & \sin t \\
-\sin t & \cos t
\end{array}\right).
\end{align*}
That is,
\begin{align*}
X_t =  \left(\begin{array}{rr}
\cos t & \sin t \\
-\sin t & \cos t
\end{array}\right)X_0 + \alpha \int_0^t \left(\begin{array}{rr}
\cos (s-t) & -\sin (s-t) \\
\sin (s-t) & \cos (s-t)
\end{array}\right) dB_s,
\end{align*}
or, more explicitly,
\begin{align*}
X_1(t) &= X_1(0) \cos t + X_2(0) \sin t + \alpha \int_0^t \Big[\cos (s-t) \, dB_1(t) - \sin (s-t) \, dB_2(t) \Big],\\
X_2(t) &= -X_1(0) \sin t + X_2(0) \cos t + \alpha \int_0^t \Big[\sin (s-t) \, dB_1(t) + \cos (s-t) \, dB_2(t) \Big].
\end{align*}
