Solution
= Solution
Eliminate $y$ using $y=1-x'$ to get $x''+x=-1$. The initial data give $x(0)=1$ and $x'(0)=0$, so the <simple harmonic motion> about the equilibrium $x=-1$ is
$$
\boxed{x(t)=-1+2\cos t,\qquad y(t)=1+2\sin t.}
$$
Indeed, $x'+y=1$ and $y'-x=1$ identically, and both initial coordinates equal one. This also describes a <circle> of <radius> two around $(-1,1)$ in the <phase plane>.