= Solution
The second equation gives $y=t+x'-x$. Differentiating it gives $y'=1+x''-x'$. Substitution into the first equation cancels the $x'$ terms and the constant forcing, leaving $x''+x=0$. Hence
$$
\boxed{x=A\cos t+B\sin t,\qquad
y=t-A(\sin t+\cos t)+B(\cos t-\sin t).}
$$
This has two independent constants, as expected for the first-order <linear system of differential equations>. At zero, the conditions give $A=0$ and $B=1$. Thus the required particular solution is
$$
\boxed{x=\sin t,\qquad y=t+\cos t-\sin t.}
$$
Checking directly, $x'+y'+x=\cos t+(1-\sin t-\cos t)+\sin t=1$, while $-x'+x+y=t$. This also verifies the signs introduced during elimination.
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