Solving a stochastic differential equation with trigonometric functions Consider the stochastic differential equations
$$dX_t = -\frac{1}{2} X_t dt - Y_t d W_t,$$
$$dY_t = -\frac{1}{2} Y_t dt + X_t d W_t,$$
with $X_0 = 1$ and $Y_0 = 0.$ My goal is to transform this into polar coordinates and solve using $R_t$ and $\Theta_t.$ After about 12 pages of work (front and back), I got this:
$$d R_t = 0,$$
$$d \Theta_t = 2R_t \tan \Theta _t dt + 2R_t \sec^2 \Theta _t d W_t,$$
with $R_0 = 1$ and $\Theta _0 = 0.$ Now, I know the $dR_t = 0$ part is correct, but like I said, I had about 12 pages of work both front and back to get all of this, and it was very messy. Luckily, it simplified surprisingly nicely, so I am hoping this is correct. But now, I have no idea how to solve this. Can anyone help?
Thanks.
 A: From the two initial SDE, by multiply the first one with $X_t$ and the second one with $Y_t$, we have
$$X_tdX_t = -\frac{1}{2}X_t^2dt-X_tY_tdW_t\tag{1}$$
$$Y_tdY_t = -\frac{1}{2}Y_t^2dt+Y_tX_tdW_t\tag{2}$$
Take $(1)+(2)$, we obtain
$$X_tdX_t+Y_tdY_t=-\frac{1}{2}\left( X_t^2+Y_t^2\right)dt \tag{3}$$
Let us denote $$X_t = R_t\sin(\Theta_t)$$
$$Y_t = R_t\cos(\Theta_t)$$
With Ito lemma and $(3)$, we have
$$\begin{align}
d(R_t^2) &= d(X_t^2+Y_t^2) \\
&=2X_tdX_t+\frac{1}{2}\cdot2 \langle dX_t,dX_t\rangle +2Y_tdY_t+\frac{1}{2}\cdot \langle 
 dY_t,dY_t\rangle\\
&=2(X_tdX_t+Y_tdY_t)+(\langle dX_t,dX_t\rangle+\langle dY_t,dY_t\rangle)\\
&=-( X_t^2+Y_t^2)dt+(Y_t^2+X_t^2)dt\\
&=0
\end{align}
$$
Then, $$R_t = R_0 = 1 \tag{4}$$
Besides, from the two intial SDE, by multiply the first one with $Y_t$ and the second one with $X_t$, we have also
$$Y_tdX_t = -\frac{1}{2}X_tY_tdt-Y_t^2dW_t\tag{5}$$
$$X_tdY_t = -\frac{1}{2}X_tY_tdt+X_t^2dW_t\tag{6}$$
Take $(6)-(5)$ and by knowing $(4)$ we obtain
$$X_tdY_t-Y_tdX_t = dW_t $$
$$
\iff \sin(\Theta_t) d\cos(\Theta_t) - \cos(\Theta_t) d\sin(\Theta_t) = dW_t \tag{7}\\
$$
By Ito lemma, we know that
$$d\cos(\Theta_t) = -\sin(\Theta_t) d\Theta_t+\frac{1}{2}(-\cos(\Theta_t))\langle d\Theta_t,d\Theta_t \rangle$$
$$d\sin(\Theta_t) = \cos(\Theta_t) d\Theta_t+\frac{1}{2}(-\sin(\Theta_t))\langle d\Theta_t,d\Theta_t\rangle$$
Then, from $(7)$, we have
$$
\begin{align}
dW_t &= \sin(\Theta_t) d\cos(\Theta_t) - \cos(\Theta_t) d\sin(\Theta_t) \\
&= -\sin^2(\Theta_t)d\Theta_t - \frac{1}{2}\sin(\Theta_t)\cos(\Theta_t)\langle d\Theta_t,d\Theta_t\rangle -\cos^2(\Theta_t)d\Theta_t+\frac{1}{2}\sin(\Theta_t)\cos(\Theta_t)\langle d\Theta_t,d\Theta_t\rangle \\
&=d\Theta_t \tag{8}
\end{align}
$$
From $(8)$, we deduce easily that, with $c\in \Bbb R$ $$\Theta_t = W_t +c$$
Then
$$X_t = \sin(W_t+c)$$
$$Y_t = \cos(W_t+c)$$
Q.E.D
A: At least part of what you have is correct: $dR_t = 0$.  Unfortunately, I suspect that your equation for $d\Theta_t$ is wrong, but without seeing the work it would be hard to say why.
The key here is to note that since $dR_t = 0$, we know $R_t = R_0 = X_0^2 + Y_0^2 = 1$.  The fact that $X_t^2 + Y_t^2 = 1$ suggests we should try guessing that the solution is $X_t = \cos(W_t)$, $Y_t = \sin(W_t)$.  Then Ito's formula gives \begin{align*}
dX_t &= -\sin(W_t) dW_t + \frac 12 \cos(W_t) dt = -Y_t dW_t + \frac 12 X_t dt \\
dY_t &= \cos(W_t) dW_t - \frac 12 \sin(W_t) dt = X_t dW_t - \frac 12 Y_t dt
\end{align*}
so this is indeed the solution.
