Find the general solution of the given differential equation.
$y''+4y=3csc(2t), o<t<\frac{\pi}{2}$
For homogeneous equation: y''+4y=0
$$r^2+4=0$$
$$r=\pm2i$$
$$y_c(t)=c_1cos2t+c_2sin2t$$
For non-homogeneous equation:y''+4y=3csc(2t)
$$p(t)=0,q(t)=4,g(t)=3csc(2t)$$
$$W=2cos^2t+2sin^2t=2$$
$$u_1=-\int\frac{y_2(t)g(t)}{W}=-\int\frac{sin2t*3csc2t}{2}dt=-\frac{3}{2}t$$
$$u_2=\int\frac{y_1(t)g(t)}{W}=\int\frac{cos2t*3csc2t}{2}dt=\frac{3}{4}ln|2t|$$
$$y_p(t)=u_1y_1(t)+u_2y_2(t)
=cos2t*(-\frac{3}{2}t)+sin2t*(\frac{3}{4}ln|2t|)$$
$$\therefore y=c_1cos2t+c_2sin2t+cos2t*(-\frac{3}{2}t)+sin2t*(\frac{3}{4}ln|2t|)$$
