Solve for particular solution for $y''+9y=-6sin(3t)$

The general solution will be the sum of the $color{blue}{text{complementary solution}}$ and $color{blue}{text{particular solution}}$.

Complementary solution: Let $$y''+9y=0 quad overbrace{implies}^{y=e^{rt}} quad r^{2}+9=0 implies r=pm 3i $$So, fundamental set of solutions will be $C.F.S={e^{3it},e^{-3it}}$ and therefore the complementary solution for ODE is $$boxed{y_{c}(t)=Ae^{3it}+Be^{-3it} quad overbrace{implies }^{text{apply Euler's indentity}} quad y_{c}(t)=Acos(3t)+Bsin(3t) }$$

Particular solution (undetermined coefficients): The particular solution to ODE is of the form $$y_{p}(t)=t^{s}(acos(3t)+bsin(3t)) quad text{where} quad s=1 quad (text{see the form of ODE}).$$So, we have that $$boxed{y''_{p}+9y_{p}=-6sin(3t) overbrace{implies}^{y_{p}=atcos(3t)+btsin(3t)} a=1 wedge b=0 implies boxed{y_{p}=tcos(3t)}}.$$

General solution: Is the form $$boxed{y=y_{c}+y_{p}}$$Therefore $$boxed{y(t)=Acos(3t)+Bsin(3t)+color{blue}{tcos(3t)}, quad A,B in mathbb{F}}.$$where $mathbb{F}$ is scalar field.