Solution
= Solution
The characteristic <polynomial> of the homogeneous equation is
$$
r(r-2)^2,
$$
so $y_h=C_0+(C_1+C_2x)e^{2x}$. For a particular <integral> put $y=e^{2x}v$. Then
$$
y^{(3)}-4y''+4y'=e^{2x}(v^{(3)}+2v'')=xe^{2x}.
$$
Taking $v''=x/2-1/4$ gives $v=x^3/12-x^2/8$ after discarding homogeneous terms. Therefore
$$
\boxed{y=C_0+(C_1+C_2x)e^{2x}
+e^{2x}\left(\frac{x^3}{12}-\frac{x^2}{8}\right).}
$$