Solution
= Solution
The differential operator is $(D-1)^2$, where $D=d/dx$. Set $y=e^xv$; the <product rule> reduces the equation to $v''=1+x$. Twice integrating gives $v=x^2/2+x^3/6+Ax+B$. The initial conditions yield $B=A=0$, so
$$
\boxed{y(x)=e^x\left(\frac{x^2}{2}+\frac{x^3}{6}\right).}
$$
This substitution removes the repeated characteristic root before finding a <particular solution>.