$(B_1(t),B_2(t))$ 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.
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*}