= Solution
A particle <pathline> follows the time-dependent <velocity field>, so its coordinates satisfy $\dot x-x=\sin t$ and $\dot y=-y$. The second equation and the initial data give $y=e^{-t}$. Using the <integrating factor> $e^{-t}$ in the first,
$$
e^{-t}x(t)=x_0+\int_0^t e^{-s}\sin s\,ds
=x_0+\frac12-\frac12e^{-t}(\sin t+\cos t).
$$
Therefore the complete <pathline> is
$$
\boxed{(x(t),y(t))
=\left((x_0+\tfrac12)e^t-\tfrac12(\sin t+\cos t),e^{-t}\right),\quad t\ge0}.
$$
If $x_0+\tfrac12\ne0$, its first coordinate grows without bound in magnitude. If $x_0=-\tfrac12$, the first coordinate remains bounded and periodic while $y\to0$. \b[The unique release coordinate giving a bounded particle path is $x_0=-1/2$.]
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