Hello Alvin,
We are given to solve:
$$y''+y=\csc(x)$$
I will use variation of parameters because I don't see how the annihilate the right side of the equation.
We see that a fundamental solution set for the corresponding homogeneous equation are:
$$y_1(x)=\cos(x)$$
$$y_2(x)=\sin(x)$$
And so we take as our particular solution:
$$y_p(x)=v_1(x)\cos(x)+v_2(x)\sin(x)$$
Next, we want to solve the system:
$$\cos(x)v_1'+\sin(x)v_2'=0$$
$$-\sin(x)v_1'+\cos(x)v_2'=\csc(x)$$
From the first equation, we find:
$$v_1'=-\tan(x)v_2'$$
And so substituting into the second equation, we obtain:
$$\frac{\sin^2(x)}{\cos(x)}v_2'+\cos(x)v_2'=\csc(x)$$
Multiply through by $\cos(x)$:
$$\sin^2(x)v_2'+\cos^2(x)v_2'=\cot(x)$$
Apply a Pythagorean identity on the left:
$$v_2'=\cot(x)$$
Hence, this implies:
$$v_1'=-1$$
Integrating, we obtain:
$$v_1=-x$$
$$v_2=\ln|\sin(x)|$$
And so out particular solution is:
$$y_p(x)=-x\cos(x)+\sin(x)\ln|\sin(x)|$$
And then by superposition, we obtain the general solution:
$$y(x)=y_h(x)+y_p(x)=A\cos(x)+B\sin(x)-x\cos(x)+\sin(x)\ln|\sin(x)|$$
