solving second order non homogeneous differential equation I am given a non homogeneous differential equation
$$
y''+4y = 3 \csc 2x.
$$ 
When I try to find value of the Wronskian $W(y_{1},y_{2}),$ the result is zero.  I  can't solve it the differential equation because of division by zero.

Start by writing down the charactersitic equation of the associated homogeneous equation and solving for the roots.
$$m^2 + 4 = 0$$
$$m^2 = -4 = 4i^2$$
$m_1= 2i$ and $m_2=-2i$
Thus $y_h=c_1\cos2x +c_2\sin2x$
Now calculate the Wronskian and determine the particular solution.
$w= \left[
  \begin{array}{ c c }
    \cos(2x) & \sin(2x) \\
     -2\sin(2x) & 2\cos(2x)
  \end{array} \right] = 2\cos^2(2x)+2\sin^2(2x) = 2(\cos^2(2x)+ \sin^2(2x)) = 2 \times 1 = 2
$
$w_1= \left[
  \begin{array}{ c c }
     0 & \sin(2x) \\
     3\csc(2x) & 2\cos(2x)
  \end{array} \right] = -3\csc(2x)\sin(2x) = -3\times\frac{\sin(2x)}{\sin(2x)} = -3
$
$w_2= \left[
  \begin{array}{ c c }
     \cos(2x) & 0 \\
     -2\sin(2x) & 3\csc(2x)
  \end{array} \right] = 3\cos(2x)\csc(2x) = 3\times\frac{\cos(2x)}{\sin(2x)} = 3\times{\cot(2x)}
$
Thus 
$u_1'=\frac{w_1}{w} = \frac{-3}{2} \implies u_1= \frac{-3}{2}x  $
$u_2'= \frac{w_2}{w} = \frac{3}{2}\cot(2x) \implies u_2= \frac{3}{4}\ln|\sin(2x)|$
Then $$y = y_h + y_p = y_h + uy_1 + uy_2$$
$$y= c_1\cos(2x) +c_2sin(2x) - \frac{3}{2}x\cos(2x) +\frac{3}{4}\sin(2x)\ln|\sin(2x)|$$
