$Question: y^{\prime \prime}+2 y^{\prime}=3+4 \sin (2 t)$
$$
\begin{array}{l}{r^{2}+2 r=0} \\ {r(r+2)=0} \\ {r_{1}=0, r_{2}=-2}\end{array}
$$
$$
y_{c}(t)=c_{1}+c_{2} e^{-2 t}
$$
$$
y^{\prime \prime}+2 y^{\prime}=3 \quad y_p(t)=3 a+b \quad y^{\prime} _p(t)=a \quad y^{\prime \prime}_p(t)=0
$$
$$
\begin{array}{r}{2 a=3} \\ {a=\frac{3}{2}}\end{array}
$$
$$
\therefore y_{p(t)}=\frac{3}{2} t
$$
$$
\begin{array}{l}{y^{\prime \prime}+2 y^{\prime}=4 \sin (2 t)} \\ {y_{P}(t)=A \cos (2 t)+B \sin (2 t)} \\ {y^{\prime} p(t)=-2 A \sin (2 t)+2 B \cos (2 t)} \\ {y^{\prime \prime} p(t)=-4 A \cos (2 t)-4 B \sin (x t)}\end{array}
$$
$$
-4 A \cos (2 t)-4 B \sin (2 t)-4 A \sin (2 t)+4 B \cos (2 t)=4 \sin 2 t
$$
$$
\left\{\begin{array}{ll}{-4 A+4 B=0}\\{-4 B-4 A=4}\end{array}\right.\qquad{\left\{\begin{array}{l}{A=-\frac{1}{2}} \\ {B=-\frac{1}{2}}\end{array}\right.}
$$
$$
y_{p}(t)=\frac{1}{2} \cos (2 t)-\frac{1}{2} \sin (2 t)
$$
$$
\therefore y(t)=y_{c}(t)+y_{p}(t)=c_{1}+c_{2} e^{-2 t}+\frac{3}{2} t-\frac{1}{2} \cos (2 t)-\frac{1}{2} \sin (2 t)
$$
