\begin{align*}
y^"+y &= 3sin(2t)+tcos(st)
\end{align*}
First consider the homogeneous differential equation for finding complimentary solution
\begin{align*}y^"+y &= 0
\end{align*}
{Assume} y=e$^{rx}$ \text {be the solution of the equation,then}
\begin{align*}r^2+1&=0\\
r^2&=-1\\
r &=\pm i\end{align*}
Therefore, the roots are r $=\pm i$\\
Therefore, the complementary solution of the given differential equation is
\begin{align*}
y_c(t)&=c_{1}cos(t)+c_{2}sin(t)\end{align*}
the particular solution for the given differential equation is of the following form:
y$_{p}$ = (At+B)cos(2t)+(Ct+D)sin(2t)\\
\text {Differentiate} y$_{p}$ \text {with respect to t as follows:}\\
\begin{align*}
y^{'}_{p} &= -2(At+B)sin(2t)+Acos(2t)+2(Ct+D)cos(2t)+Csin(2t)\\
y^{'}_{p}&=(-2At-2B+C)sin(2t)+(2Ct+2D+A)cos(2t)\\
y^{"}_{p}&=(-4At-4B+4C)cos(2t)-(4Ct+4D+4A)sin(2t)\\
\end{align*}
Substitue y$_{p}$ and y$^{"}_{p}$ in equation\\
\begin{align*}y^{"}+y^{'} &= 3sin2t + t cos2t\\
-2(At+B)sin(2t)+Acos(2t)+2(Ct+D)cos(2t)+Csin(2t)+-2(At+B)sin(2t)+Acos(2t)+2(Ct+D)cos(2t)+Csin(2t) = 3sinwt + t cos2t\\
-Atcos(2t)+(-3B+4C)cos(2t)-3Ctsin(2t)+(-3D-4A)sin(2t) &= 3sin2t+tcos2t\\\end{align*}
Compare the coefficients of tcos(2t) on both sides\\
\begin{align*}-3A&=1\\
A=-\frac{1}{3}\\\end{align*}
Compare the coefficients of sin(2t) on both sides\\
\begin{align*}-3D-4A&=3\\
-3D-4(-\frac{1}{3})&=3\\
D&=-\frac{5}{9}\end{align*}
Compare the coefficients of tsin(2t) on both sides\\
\begin{align*}-3C&= 0\\
C&=0
\end{align*}
