Suppose $X$ and $Y$ are Ito processes, $X_t=x+\int^t_0Y_sdB_s$ and $Y_t=y-\int^t_0X_sdB_s,\ t\geq 0$, here $B$ is a standard Brownian motion. I need to prove that $Z_t:=X^2_t+Y^2_t=(x^2+y^2)\exp\{t\}$.
My first thought was plugging the expression for $Y_s$ in the expression of $X_t$ then solving for $X$ and vice versa. I would then have
$X_t=x+\int_0^t\left(y-\int_0^sX_udB_u\right)dB_s.$
Applying Ito's formula yields
$\int_0^sX_udB_u=\frac{1}{2}(X^2_s-x^2)-s$
and plugging back into the previous formula we have
$dX_t=\left(y-\frac{1}{2}(X^2_t-x^2)+t\right)dB_t$
and I don't know how to solve this.
My guess is that $X$ and $Y$ might be stochastic exponents, but I don't see how to get there. I just started learning, so maybe I'm missing something simple. Any hints are very appreciated.