Solution
= Solution
The <characteristic equation> has roots zero and three, giving $y_h=C_1+C_2e^{3x}$. Since the forcing is quadratic and the <homogeneous solution> contains a constant, use the <polynomial> trial $y_p=Ax^3+Bx^2+Cx$. Substitution gives
$$
y_p''-3y_p'=-9Ax^2+(6A-6B)x+2B-3C.
$$
Matching coefficients gives $A=-1/3$, $B=-1/3$, and $C=-8/9$. Therefore
$$
\boxed{y=C_1+C_2e^{3x}-\frac{x^3}{3}-\frac{x^2}{3}-\frac{8x}{9}}.
$$