I'm assuming this is question 10 from 3.5, in which case you've copied the question wrong. The given DE should read:
$$y''+y=3\sin{2t}+t\cos{2t}\tag{1}.$$
The corresponding homogeneous equation is
$$y''+y=0$$
with characteristic equation
$$r^{2}+1=0$$
and roots
$$r_{1,2}=\pm i$$.
The solution to the corresponding homogenous equation is
$$y_h=c_1 \cos{t}+c_2 \sin{t}.$$
Particular Solution (using method of undetermined coefficients):
Splitting up the right side of $(1)$, we get two equations
$$y''+y=3\sin{2t},$$
$$y''+y=t\cos{2t}.$$
Since $\sin{t}$ and $\cos{t}$ are not solutions to the corresponding homogenous equation, we set our particular solution to be of the form
$$Y(t)=(A_0 t+A_1)\sin{2t}+(B_0 t+B_1)\cos{2t}.$$
$$Y'(t)=(A_0-B_1)\sin{2t}+(2A_1+B_0)\cos{2t}-2B_0 t\sin{2t}+2A_1 t\cos{2t}$$
$$Y''(t)=2(A_0 -2 B_1)\cos{2t}+2A_0\cos{2t}-4A_0 t\sin{2t}-2(2A_1+B_0)\sin{2t}-2B_0\sin{2t}-4B_0t\cos{2t}$$
Plugging $Y$, and $Y''$ into $(1)$, we have
$$[(A_0 t+A_1)\sin{2t}+(B_0 t+B_1)\cos{2t}]''+(A_0 t+A_1)\sin{2t}+(B_0 t+B_1)\cos{2t}=3\sin{2t}+t\cos{2t}$$
$$2(A_0 -2 B_1)\cos{2t}+2A_0\cos{2t}-4A_0 t\sin{2t}-2(2A_1+B_0)\sin{2t}-2B_0\sin{2t}-4B_0t\cos{2t}+(A_0 t+A_1)\sin{2t}+(B_0 t+B_1)\cos{2t}=3\sin{2t}+t\cos{2t}$$
$$(4A_0-3B_1)\cos{2t}+(-3A_0)t\sin{2t}+(-3A_1-4B_0)\sin{2t}+(-3B_0)t\cos{2t}=3\sin{2t}+t\cos{2t}$$
$$\left\{\begin{aligned} &4A_0-3B_1=0\\ &-3A_0=0\\& -3A_1-4B_0=3\\&-3B_0=1\end{aligned}\right. \label{eq4} $$
$$\Rightarrow A_0=0, A_1=-\frac{5}{9}, B_0=-\frac{1}{3}, B_1=0$$
Thus, the particular solution of the nonhomogenous equation is
$$Y(t)=-\frac{5}{9} \sin{2t}-\frac{1}{3} t\cos{2t}.$$
The solution to $(1)$ is
$$y(t)=y_h(t)+Y(t)=c_1\cos{t}+c_2\sin{t}-\frac{5}{9} \sin{2t}-\frac{1}{3} t\cos{2t}.$$
typing that out was brutal, so I hope this helps lol
