= Solution
The <characteristic polynomial> of the homogeneous <linear ordinary differential equation> is
$$
r^2-2r-3=(r-3)(r+1),
$$
so its <complementary solution> is $Ae^{3x}+Be^{-x}$. Because the forcing contains the resonant factor $e^{-x}$, set $y=e^{-x}v$. Substitution gives
$$
y''-2y'-3y=e^{-x}(v''-4v')=-16xe^{-x}.
$$
Writing $w=v'$, a polynomial particular solution of $w'-4w=-16x$ is $w=4x+1$, and hence $v=2x^2+x$. Thus
$$
y=Ae^{3x}+(B+2x^2+x)e^{-x}.
$$
Boundedness as $x\to\infty$ forces $A=0$, while the <initial condition> $y(0)=1$ gives $B=1$. Therefore
$$
\boxed{y(x)=(2x^2+x+1)e^{-x}}.
$$
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