Prove existence of strong solution for $dX_t = (dX_1(t),dX_2(t))$
Let
$\displaystyle dX_t = (dX_1(t),dX_2(t)) = \begin{cases} dX_1(t) = X_2(t)dt + \alpha dB_1(t) \\ dX_2(t) = -X_1(t)dt + \beta dB_2(t) \end{cases}$
Where $(B_1(t), B_2(t))$ is 2-dimensional Brownian motion and $\alpha, \beta$ are constants.
or in matrix notation
$\displaystyle \bigg [\begin{matrix} dX_1(t) \\ dX_2(t) \end{matrix} \bigg ] = \bigg [ \begin{matrix} 0 && 1 \\ -1 && 0\end{matrix} \bigg ] \bigg[ \begin{matrix} X_1(t) \\ X_2(t)\end{matrix} \bigg]dt + \bigg[ \begin{matrix} \alpha && 0 \\ 0 && \beta \end{matrix} \bigg]\bigg[ \begin{matrix} dB_1(t) \\ dB_2(t) \end{matrix} \bigg]$
According to Oksendal's book, Theorem 5.2.1
Let $T>0$ and $$ \begin{array}{l} b :[0,T]\times\Bbb R^n \to {\mathbb{R}^n};\\ \sigma :[0,T]\times\Bbb R^n\to {\mathbb{R}^{n \times m}}; \end{array} $$ be measurable functions for which there exist constants $C$ and $D$ such that $$ \begin{array}{l} |b(t,x)|+|\sigma (t,x)|\le C(1+|x|);\\ |b(t,x)-b(t,y)|+|\sigma(t,x)-\sigma(t,y)|\le D|x-y|; \end{array} $$ Let $Z$ be a random variable that is independent of the $\sigma$-algebra generated by $B_s$, $s ≥ 0$, and with finite second moment: $$ E[|Z|^2]<\infty $$ Then the stochastic differential equation/initial value problem $$ \begin{array}{l} {\rm{d}}{X_t} =b(t,X_t)\mathrm dt+\sigma(t,X_t)\mathrm dB_t,\quad \text{for } t \in [0,T];\\ X_0 = Z; \end{array} $$ has a Pr-almost surely unique $t$-continuous solution $(t,ω)\mapsto X_t(ω)$ such that $X$ is adapted to the filtration $\mathcal F_t^Z$ generated by $Z$ and $B_s$, $s\leq t$, and $$ E\left[\int_0^T|X_t|^2\,\mathrm dt\right]<\infty. $$
How do I apply this theorem to the above case where there are two differing Brownian motions and matrices involved.
Would $b(t,x) = b(t, (x_1,x_2)) = (x_2, -x_1)$ and $\sigma(t,x) = \sigma(t, (x_1,x_2)) = (\alpha, \beta) $ ?
So I need something like :
$|b(t,x_1, x_2)|+|\sigma (t,x_1,x_2)|\le C(1+|(x_1, x_2)|)$
$\lvert b(t, (x_1,x_2)) - b(t, (y_1,y_2)) \rvert + \lvert \sigma(t, (x_1,x_2)) - \sigma(t, (y_1,y_2)) \rvert \le D \lvert (x_1,x_2) - (y_1,y_2) \rvert $
But the above doesn't quite make sense.
For instance, what is $\lvert (x_1,x_2) \rvert$ supposed to be?
Note you have not prescribed an initial condition for your SDE. Anyway :
Apply the theorem with
$$b(t,x)=b(x)=(x_2,-x_1),~~~~~~\sigma=\begin{pmatrix} \alpha & 0 \\ 0 & \beta \end{pmatrix}.$$
The Brownian motion in the book mentioned is $B_t=(B_t^1,\ldots,B_t^m)$ where $B_i$ are 1 dimensional Brownian motions, i.e $B$ is an $m$ dimensional Brownian motion, in your case $m=2$.
The first condition to be satisfied is a linear growth condition which usually provides existence of a solution, and the second condition is a Lipschitz in space condition usually providing the uniqueness of the solution (see proof in the book you cited).
These two conditions are easy to check, I think your getting confused not seeing that $|(x_1,x_2)|=|x|$...