Solving a simultaneous (closed) stochastic differential equation I have the two following stochastic differential equations in differential/Ito form which closes onto each other such that
$$
dA_{s}(t)=\alpha A_{s}(t)dt +\beta A_{i}(t)dt+dW_{s}(t) \\
dA_{i}(t)=\alpha^{*} A_{i}(t)dt +\gamma A_{s}(t)dt+dW_{i}(t) 
$$
where $\alpha,\beta,\gamma$ are constants and $dW_{s,i}$ denote Wiener increments for $A_{s}$ and $A_{i}$ respectively. How does one go about solving this? Naively, I would divide both sides of the two equations with $dt$ and form a system of non-homogenous differential equations, and proceed to solve them conventionally. But the derivative of the Wiener process is not defined. How should one approach this?

Let's write the system in matrix form
$$d\mathbf{A}=\mathbf{M}\mathbf{A}dt+d\mathbf{W} \tag{1}$$
with
$$\mathbf{A}=\begin{pmatrix} A_s(t)\\ A_i(t)  \end{pmatrix} $$
$$\mathbf{M}=\begin{pmatrix} \alpha & \beta\\ \alpha^* &\gamma   \end{pmatrix} $$
$$\mathbf{W}=\begin{pmatrix} W_s(t)\\ W_i(t)  \end{pmatrix} $$
Diagonalize $\mathbf{M}$: $\mathbf{M}=\mathbf{U}^{-1}\mathbf{D}\mathbf{U}$ with $\mathbf{D}$ -diagonal matrix, then
$$\begin{align}
(1)&\Longleftrightarrow d(\mathbf{A}) = \mathbf{U}^{-1}\mathbf{D}\mathbf{U}\mathbf{A}dt+d\mathbf{W}\\
&\Longleftrightarrow d(\mathbf{U}\mathbf{A}) = \mathbf{D}\mathbf{U}\mathbf{A}dt+\mathbf{U}d\mathbf{W}\\
 &\Longleftrightarrow  d(\mathbf{V}) = \mathbf{D}\mathbf{V}dt+\mathbf{U}d\mathbf{W}\tag{2}
\end{align}$$
with $\mathbf{V} :=\mathbf{U}\mathbf{A}$
As $\mathbf{D}$  is a diagonal matrix, $(2)$ can be solved easily for $\mathbf{V}$.
(For example, the first equation of $(2)$ will be
$$dV_1(t) = d_1 V_1(t)dt + \eta_{11} dW_s(t) +\eta_{12} dW_i(t) \tag{3}$$
and the SDE $(3)$ can be solved easily. Same for the second equation of $(2)$
)
and so
$$\mathbf{A} = \mathbf{U}^{-1}\mathbf{V}$$
