Solution
= Solution
The repeated factor in the <differential operator> suggests $y=e^{-x}u(x)$. Direct differentiation gives $y''+2y'+y=e^{-x}u''$, so $u''=4$ and
$$
u=2x^2+Bx+C.
$$
The first value gives $C=1$. The second gives $u(1)=ey(1)=1$, hence $2+B+1=1$ and $B=-2$. Therefore
$$
y(x)=e^{-x}(2x^2-2x+1),\qquad \boxed{y(2)=5e^{-2}}.
$$