Can I please get some feedback on my work for the following problem? Is there a more simple approach to arrive to the solutions for part a? Thank you for your time and consideration!
Let $X$ be a standard Brownian motion, let $Y_1$ and $Y_2$ satisfy the following system of SDEs: $$dY_1(t) = \alpha Y_1(t) dt - Y_2(t)dX(t), \hspace{10 pt} Y_1(0) = y_{10}$$ $$dY_2(t) = \alpha Y_2(t) dt + Y_1(t)dX(t), \hspace{10 pt} Y_2(0) = y_{20},$$ where $y_{10}. y_{20}$ are given constants.
(a) Show the sum of the squares process $S(t) = Y_1^2(t) + Y_2^2 (t)$ is deterministic. For what $\alpha$ does $S$ become a constant?
$\textit{Solution.}$ Suppose $Y_1 = f(t,X_t)$ and $Y_2 = g(t,X_t)$ for some smooth $f,g.$ By Itô's lemma: $$Y_1 = Y_1(0) + \int_0^t [f_1 (s,X_s) + f_{22}(s, X_s)]ds + \int_0^t f_2(s,X_s)dX_s$$ $$Y_2 = Y_2(0) + \int_0^t [g_1 (s,X_s) + g_{22}(s, X_s)]ds + \int_0^t g_2(s,X_s)dX_s.$$ By matching integrands, we get the following SDEs $$\alpha f(t,x) = f_1(t,x) + \frac{1}{2} f_{22} (t,x)$$ $$-g(t,x) = f_2(t,x)$$ $$\alpha g(t,x) = g_1(t,x) + \frac{1}{2} g_{22}(t,x)$$ $$f(t,x) = g_2(t,x)$$ however, $g=f_2 \implies -g_2 = f_{22} \implies f = -f_{22}$ and $f= g_2 \implies f_2 = g_{22}\implies g = -g_{22}.$ Thus, we can rewrite our system of differential equations as $$(\alpha + \frac{1}{2} f(t,x) = f_1(t,x)$$ $$f(t,x) = -f_{22}(t,x)$$ $$(\alpha + \frac{1}{2}) g(t,x) = g_1(t,x)$$ $$g(t,x) = -g_{22}(t,x)$$ and assume $f,g$ are separable: $$f(t,x) = m(t)n(x)$$ $$g(t,x) = \tilde m(t) \tilde n(x)$$ where $n(0) = \tilde n(0) = 1.$ Thus, $$(\alpha + \frac{1}{2}) m(t) = m'(t)$$ $$n(x) = -n''(x)$$ $$(\alpha + \frac{1}{2} \tilde m(t) = \tilde m'(t)$$ $$\tilde n (x) = -\tilde n ''(x)$$ and solving these ODEs gives us $$m(t) = m(0)e^{(\alpha + 1/2)t}$$ $$n(t) = n(0) \cos(x) + c\sin(x) = \cos(x) + c\sin(x)$$ $$\tilde m (t) = \tilde m (0) e^{(\alpha + 1/2)t}$$ $$\tilde n(t) = \tilde n(0) \cos(x) + \tilde c \sin(x) = \cos(x) + \tilde c \sin(x),$$ with $c$ and $\tilde c$ as arbitrary constants. However, $f=g_2$ so $$mn = \tilde m \tilde n' = \frac{\tilde m (0)}{m(0)} m\tilde n'$$ so $n=\frac{\tilde m(0)}{m(0)} \tilde n'.$ Thus, $\cos(x) + c\sin(x) = -\frac{\tilde m(0)}{m(0)} \sin(x) + \frac{\tilde m(0)}{m(0)} \tilde \cos(x)$ where $c = -\frac{\tilde m(0)}{m(0)}$ and $\tilde c = \frac{m(0)}{\tilde m(0)}$ and $$n(x) = \cos(x) - \frac{\tilde m(0)}{m(0)} \sin(x), \text{ and } \tilde n(x) = \cos(x) - \frac{m(0)}{\tilde m(0)} \sin(x).$$ Thus, $$f(t,x) = m(0)e^{(\alpha + 1/2)t}(\cos(x) - \frac{\tilde m(0)}{m(0)}\sin (x))$$ and $$g(t,x) = \tilde m(0)e^{(\alpha + 1/2)t}(\cos(x) + \frac{m(0)}{\tilde m(0)}\sin (x)).$$ Applying initial conditions gives us $$Y_1(t) = y_{10} e^{(\alpha + 1/2)t}(\cos(x) - \frac{y_{20}}{y_{10}}\sin (x))$$ $$Y_2(t) = y_{20} e^{(\alpha + 1/2)t}(\cos(x) + \frac{y_{10}}{y_{20}}\sin (x)).$$ The sum of squares process $S(t)$ is denoted as $$S(t) = Y_1^2(t) + Y_2^2 (t)$$ $$= e^{(\alpha + 1/2)t}\left[ y_{10}^2 \left(\cos(x) - \frac{y_{20}}{y_{10}}\sin (x)\right)^2 + y_{20}^2 \left(\cos (x) + \frac{y_{10}}{y_{20}} \sin(x)\right)^2\right]$$ $$=e^{(\alpha + 1/2)t}(y_{10}^2 + y_{20}^2) [\cos^2(x) + \sin^2(x)]$$ $$=e^{(\alpha + 1/2)t}(y_{10}^2 + y_{20}^2).$$ Thus, $S(t)$ is deterministic and $S(t)$ is a constant for $\alpha = -\frac{1}{2}.$
(b) Calculate $\mathbb{E}[Y_1(t)]$ and $\text{cov}(Y_1(t),Y_2(t)).$
$\textit{Solution.}$ Let $X \sim \mathcal{N}(\mu, \sigma^2).$ Then we have $$\mathbb{E}(e^{iX}) = \mathbb{E}[\cos(X)] + i\mathbb{E}[\sin(X)]$$ $$=e^{i\mu} e^{-\sigma^2/2}$$ $$=[\cos(\mu) + i\sin(\mu)]e^{-\sigma^2/2}.$$ Matching real and imaginary component gives $$\mathbb{E}[\cos(X)] = \cos(\mu)e^{-\sigma^2/2}$$ $$\mathbb{E}[\sin(X)] = \sin(\mu)e^{-\sigma^2/2}.$$ Thus, we have $$\mathbb{E}[Y_1(t)] = \mathbb{E}\left[y_{10} e^{(\alpha + 1/2)t} \left(\cos(x) - \frac{y_{20}}{y_{10}} \sin(x) \right)\right]$$ $$=y_{10} e^{(\alpha + 1/2)t} \mathbb{E} \left[\cos(x) - \frac{y_{20}}{y_{10}} \sin(x) \right]$$ $$= y_{10} e^{(\alpha + 1/2)t} \left( \cos(0) e^{-t/2} - \frac{y_{20}}{y_{10}}\sin(0)e^{-t/2}\right)$$ $$= y_{10}e^{\alpha t}.$$
Now, $\text{cov}(Y_1(t),Y_2(t)) = \mathbb{E}[(Y_1 - \mathbb{E}(Y_1))(Y_2-\mathbb{E}(Y_2))]$ $$=\mathbb{E}[Y_1Y_2 + Y_1y_{20}e^{\alpha t} - y_{10}e^{\alpha t} Y_2 + y_{10}y_{20}e^{2\alpha t}]$$ $$=y_{10}y_{20}e^{(2\alpha + 1)t}\mathbb{E} \left[ \left( \cos(x) - \frac{y_{20}}{y_{10}}\sin(x)\right) \left( \cos(x) + \frac{y_{10}}{y_{20}}\sin(x)\right) \right] + y_{10}y_{20}e^{2\alpha t}$$ $$=y_{10}y_{20}e^{(2\alpha + 1)t}\mathbb{E} \left[ \cos^2(x) + \frac{y_{10}}{y_{20}} \cos(x)\sin(x) - \frac{y_{20}}{y_{10}}\cos(x) \sin(x) -\sin^2(x) \right] + y_{10}y_{20}e^{2\alpha t}$$ $$=y_{10}y_{20}e^{(2\alpha + 1)t}\mathbb{E} \left[ \cos^2(x) -\sin^2(x) + \frac{y_{10}^2-y_{20}^2}{y_{10}y_{20}} \cos(x)\sin(x)\right] + y_{10}y_{20}e^{2\alpha t}$$ $$=y_{10}y_{20}e^{(2\alpha + 1)t}\mathbb{E} \left[ \cos(2x) + \frac{y_{10}^2-y_{20}^2}{2y_{10}y_{20}} \sin(2x)\right] + y_{10}y_{20}e^{2\alpha t}$$ $$=y_{10}y_{20}e^{(2\alpha + 1)t} \left(\cos(0)e^{-2t} + \frac{y_{10}^2 - y_{20}^2}{2y_{10}y_{20}} \sin(0)e^{-2t} \right) + y_{10}y_{20}e^{2\alpha t}$$ $$ = y_{10}y_{20}(e^{(2\alpha - 1)t} + e^{2\alpha t}).$$
There is a much simpler solution for part (a), but this method doesn't solve for $Y_1$ and $Y_2$ as explicit functions of $t$ and $X_t$ so it doesn't help as much for part (b):
By Ito's formula, \begin{align*}d(Y_1(t)^2) &= 2 Y_1(t)dY_1(t) + dY_1(t)dY_1(t) \\ &= 2 Y_1(t)(\alpha Y_1(t)dt - Y_2(t)dX(t)) + Y_2(t)^2 dt \\ &= (2 \alpha Y_1(t)^2 + Y_2(t)^2) dt - 2 Y_1(t)Y_2(t)dX(t) \end{align*} and similarly \begin{align*} d(Y_2(t)^2) &= (Y_1(t)^2 + 2 \alpha Y_2(t)^2)dt + 2Y_1(t)Y_2(t)dX(t). \end{align*} Hence \begin{align*} dS(t) &= d(Y_1(t)^2) + d(Y_2(t)^2) \\ &= (2\alpha (Y_1(t)^2 + Y_2(t)^2) + Y_1(t)^2 + Y_2(t)^2)dt \\ &= (2\alpha+1)S(t)dt. \end{align*} This shows $S$ is differentiable with $\frac{dS(t)}{dt} = (2\alpha+1) S(t)$, so $S(t) = S(0)e^{(2\alpha+1)t}$, which is constant when $\alpha = -\frac 12$.
I noticed that what I found for the $S(t)$ process is slightly different from what you found. I think that's because you forgot a factor of $\frac 12$ on the second derivative terms when applying Ito's formula at the start.