First look for the fundamental solutions by solving the homogeneous version of the ODE:
[tex]y''+3y'=0[/tex]
The characteristic equation is
[tex]r^2+3r=r(r+3)=0[/tex]
with roots [tex]r=0[/tex] and [tex]r=-3[/tex], giving the two solutions [tex]C_1[/tex] and [tex]C_2e^{-3t}[/tex].
For the non-homogeneous version, you can exploit the superposition principle and consider one term from the right side at a time.
[tex]y''+3y'=2t^4[/tex]
Assume the ansatz solution,
[tex]{y_p}=at^5+bt^4+ct^3+dt^2+et[/tex]
[tex]\implies {y_p}'=5at^4+4bt^3+3ct^2+2dt+e[/tex]
[tex]\implies {y_p}''=20at^3+12bt^2+6ct+2d[/tex]
(You could include a constant term f here, but it would get absorbed by the first solution [tex]C_1[/tex] anyway.)
Substitute these into the ODE:
[tex](20at^3+12bt^2+6ct+2d)+3(5at^4+4bt^3+3ct^2+2dt+e)=2t^4[/tex]
[tex]15at^4+(20a+12b)t^3+(12b+9c)t^2+(6c+6d)t+(2d+e)=2t^4[/tex]
[tex]\implies\begin{cases}15a=2\\20a+12b=0\\12b+9c=0\\6c+6d=0\\2d+e=0\end{cases}\implies a=\dfrac2{15},b=-\dfrac29,c=\dfrac8{27},d=-\dfrac8{27},e=\dfrac{16}{81}[/tex]
[tex]y''+3y'=t^2e^{-3t}[/tex]
[tex]e^{-3t}[/tex] is already accounted for, so assume an ansatz of the form
[tex]y_p=(at^3+bt^2+ct)e^{-3t}[/tex]
[tex]\implies {y_p}'=(-3at^3+(3a-3b)t^2+(2b-3c)t+c)e^{-3t}[/tex]
[tex]\implies {y_p}''=(9at^3+(9b-18a)t^2+(9c-12b+6a)t+2b-6c)e^{-3t}[/tex]
Substitute into the ODE:
[tex](9at^3+(9b-18a)t^2+(9c-12b+6a)t+2b-6c)e^{-3t}+3(-3at^3+(3a-3b)t^2+(2b-3c)t+c)e^{-3t}=t^2e^{-3t}[/tex]
[tex]9at^3+(9b-18a)t^2+(9c-12b+6a)t+2b-6c-9at^3+(9a-9b)t^2+(6b-9c)t+3c=t^2[/tex]
[tex]-9at^2+(6a-6b)t+2b-3c=t^2[/tex]
[tex]\implies\begin{cases}-9a=1\\6a-6b=0\\2b-3c=0\end{cases}\implies a=-\dfrac19,b=-\dfrac19,c=-\dfrac2{27}[/tex]
[tex]y''+3y'=\sin(3t)[/tex]
Assume an ansatz solution
[tex]y_p=a\sin(3t)+b\cos(3t)[/tex]
[tex]\implies {y_p}'=3a\cos(3t)-3b\sin(3t)[/tex]
[tex]\implies {y_p}''=-9a\sin(3t)-9b\cos(3t)[/tex]
Substitute into the ODE:
[tex](-9a\sin(3t)-9b\cos(3t))+3(3a\cos(3t)-3b\sin(3t))=\sin(3t)[/tex]
[tex](-9a-9b)\sin(3t)+(9a-9b)\cos(3t)=\sin(3t)[/tex]
[tex]\implies\begin{cases}-9a-9b=1\\9a-9b=0\end{cases}\implies a=-\dfrac1{18},b=-\dfrac1{18}[/tex]
So, the general solution of the original ODE is
[tex]y(t)=\dfrac{54t^5 - 90t^4 + 120t^3 - 120t^2 + 80t}{405}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-\dfrac{3t^3+3t^2+2t}{27}e^{-3t}-\dfrac{\sin(3t)+\cos(3t)}{18}[/tex]
