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?
 A: 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
$$\mathbf{A} = \mathbf{U}^{-1}\mathbf{V}$$
