Solving a 2-D SDE Need to solve the system of SDE
\begin{align*}
dX^1&=-X^2dt+dW^1\\
dX^2&=X^1dt+dW^2
\end{align*}
I wrote the first eq. in integral form as $$X_t^1=X_0^1-\int_0^tX^2(s)ds+W_t^1$$
Then I plugged it into the second eq. and took the integral form of that which turned out to be
$$X_t^2=X_0^2+X_0^1t-\int_0^t\left(\int_0^tX^2(s)ds\right)ds+\int_0^tW^1(s)ds+W_t^2$$
Is this going in the right direction? If so, how much more can it be simplified?
 A: Good question!
It may first be helpful to rewrite as
\begin{align*}
\begin{pmatrix} dX_{t}^{(1)} \\ dX_{t}^{(2)} \end{pmatrix}
= 
\begin{pmatrix} 0 & - 1 \\ 1 & 0 \end{pmatrix}
\begin{pmatrix} X_{t}^{(1)} \\ X_{t}^{(2)} \end{pmatrix}\, dt 
+ \begin{pmatrix} dW_{t}^{(1)} \\ dW_{t}^{(2)} \end{pmatrix}.
\end{align*}
This is similar to vector ODE: $\vec{X}'(t) = A\vec{X}(t) + \vec{b}(t)$.

*

*First diagonalize

\begin{align*}
A = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} :&= PDP^{-1} = \begin{pmatrix} 1 & 1 \\ i & -i \end{pmatrix} \begin{pmatrix} i & 0 \\ 0 & -i \end{pmatrix} \begin{pmatrix} \frac{1}{2} & -\frac{i}{2} \\ \frac{1}{2} & \frac{i}{2} \end{pmatrix}.
\end{align*}


*Next compute

\begin{align*}
e^{At} &= Pe^{Dt}P^{-1} \\
&= \begin{pmatrix} 1 & 1 \\ i & -i \end{pmatrix} \begin{pmatrix} e^{it} & 0 \\ 0 & e^{-it} \end{pmatrix} \begin{pmatrix} \frac{1}{2} & -\frac{i}{2} \\ \frac{1}{2} & \frac{i}{2} \end{pmatrix} \\
&= \begin{pmatrix} e^{it} & e^{-it} \\ ie^{it} & -ie^{-it} \end{pmatrix} \begin{pmatrix} \frac{1}{2} & -\frac{i}{2} \\ \frac{1}{2} & \frac{i}{2} \end{pmatrix} \\
&= \begin{pmatrix} \cos(t) & \sin(t) \\ -\sin(t) & \cos(t) \end{pmatrix}
\end{align*}


*Solve using d'Alembert's principle.

\begin{align*}
X_{t} &= e^{-At}\vec{X}_{0} + \int_{0}^{t} e^{-A(t-s)} \begin{pmatrix} dW_{s}^{(1)} \\ dW_{s}^{(2)} \end{pmatrix}
\end{align*}
This yields
\begin{align*}
&X_{t}^{(1)} = \cos(t)X_{0}^{(1)} - \sin(t)X_{0}^{(2)} + \int_{0}^{t} \cos(t-s)\, dW_{t}^{(1)} - \int_{0}^{t} \sin(t-s)\, dW_{t}^{(2)} \\
&X_{t}^{(2)} = \sin(t)X_{0}^{(1)} + \cos(t)X_{0}^{(2)} + \int_{0}^{t} \sin(t-s)\, dW_{t}^{(1)} + \int_{0}^{t} \cos(t-s)\, dW_{t}^{(2)}.
\end{align*}
You may wish to check, for example, that
\begin{align*}
dX_{t}^{(1)} &= -\sin(t)X_{0}^{(1)}\, dt - \cos(t)X_{0}^{(2)}\, dt + \left (\int_{0}^{t} \partial_{t}(\cos(t-s)) \, dW_{s}^{(1)}\right )\, dt \\ 
&\quad+ \cos(t-t)\, dW_{t}^{(1)} - \left (\int_{0}^{t} \partial_{t}(\sin(t-s)) \, dW_{s}^{(2)} \right )\, dt - \sin(t-t)\, dW_{t}^{(2)} \\
&= -X_{t}^{(2)}\, dt + dW_{t}^{(1)}
\end{align*}
as desired.
