Answer
The general solution of the differential equation $y''+4y = 3\csc{(2t)}$ is:
$$y(t) = c_1\sin(2t)+c_2\cos(2t) +\frac{3}{4}\ln|sin(2t)|sin(2t) - \frac{3}{2}tcos(2t)$$
Work Step by Step
$$y{''}+4y = 3\csc{(2t)}$$
$$g(t) = 3\csc{(2t)}$$
To find the complementary function $y_c(t)$, consider the homogeneous part
$$y''+4y=0$$
Corresponding characteristic equation is:
$$m^2+4=0$$
$$ m = \pm 2\iota $$
Hence, Complementary function is:
$$y_c(t) = c_1\sin(2t)+c_2\cos(2t)$$
Hence Independent solutions $y_1(t)$ and $y_2(t)$ of homogeneous equation will be
$$y_1(t) = \sin(2t) \text{ and } y_2(t) = \cos(2t)$$
Particular Solution $y_p(t)$ is given by:
$$y_p(t) = u_1(t)y_1(t)+u_2(t)y_2(t)$$
where $$u_1(t) = -\int \frac{y_2(t)g(t)}{W(y_1,y_2)(t)}dt + c_3$$
$$u_2(t) = \int \frac{y_1(t)g(t)}{W(y_1,y_2)(t)}dt+c_4$$
Since, $y_p(t)$ is a particular solution so we can choose $c_3=c_4=0$ such that both the constant term vanishes.
$$W(y_1, y_2)(t) =
\begin{vmatrix}
\sin(2t) & \cos(2t) \\
2\cos(2t) & -2\sin(2t)
\end{vmatrix}=-2$$
$$u_1(t) = -\int \frac{\cos(2t) \times 3\csc(2t)}{-2}dt = \frac{3}{2}\int \cot(2t)dt$$
$$u_1(t) = \frac{3}{4}\ln|sin(2t)|$$
$$u_2(t) = \int \frac{\sin(2t)\times 3\csc(2t)}{-2}dt = -\frac{3}{2}\times \int 1dt$$
$$u_2(t) = -\frac{3}{2}t$$
$$\therefore y_p(t) = \frac{3}{4}\ln|sin(2t)|sin(2t) - \frac{3}{2}tcos(2t)$$
Therefore, the general solution is:
$$y(t) = y_c(t)+y_p(t)$$
$$y(t) = c_1\sin(2t)+c_2\cos(2t) +\frac{3}{4}\ln|sin(2t)|sin(2t) - \frac{3}{2}tcos(2t)$$
