A particular solution to $\frac{d^2y}{dx^2}+y =\csc x$ 
The solution to the BVP $\frac{d^2y}{dx^2}+y =\csc x$, $0 < x < \frac{\pi}{2}$ $y(0)=0$, $y(\pi/2)=0$ is
$(A)$ Concave
$(B)$ Convex
$(C)$ Negative
$(D)$ Positive

For the homogeneous problem of course $\sin x$ and $\cos x $ are linearly independent solutions. I have trouble finding a particular solution to the non-homogeneous problem. Any help would be much appreciated.
PS: There could be multiple correct options.
 A: You have: 
$$y''(x)+y(x)=\csc(x)$$
Solving an ODE like this can be solved by looking for the complementary and particular solution. For further information look at https://en.wikipedia.org/wiki/Variation_of_parameters in the paragraph "General second order equation". It is a good explanation how to approach this.
First solve the homogeneous part $$y_c''(x)+y_c(x)=0.$$ This can by substituting $y_c(x)=e^{\lambda x}$ into the equation and solving for $\lambda$. You get (do this) $y_c(x)=A \cos(x)+B \sin(x)$ as the complementary solution.
Now the particular solution by variation of parameters.  The basic solution of $y_c$ were $\cos(x)$ and $\sin(x)$ and the particular solution will be something like 
$$y_p(x)=v_1(x) \cos(x) + v_2(x) \sin(x).$$
You get the coefficients via $v_1(x)=-\int \frac{\csc(x) \sin(x)}{W(x)} \text{d}x$ and $v_2(x)=\int \frac{\csc(x) \cos(x)}{W(x)} \text{d}x$
where $W$ indicates the Wronskian
$$W(x)=\begin{vmatrix} \cos(x) & \sin(x) \\ -\sin(x) & \cos(x)\end{vmatrix}=1.$$
You get (do this) $v_1(x)=-x$ and $v_2(x)=\log(\sin(x))$ and therefore we got the solution

$$y(x)=y_c(x)+y_p(x)=A \cos(x) + B \sin(x) - x\cos(x)+\log(\sin(x))\sin(x)$$

Plug in your initial conditions to get $A$ and $B$. Now you can check for convexity and the sign.
A: Here is an answer without explicitly solving the equation. It's a multiple choice question, so I will make an assumption that there is exactly one correct answer. [Which turns out to be a false assumption, making this a poorly written question. See below at the end.]
The solution is not (A) concave. For a twice differentiable function, that would mean $\frac{d^2y}{dx^2}<0$. But near $x=\pi/2$ you have $\frac{d^2y}{dx^2}\approx1>0$.
Suppose the solution were (B) convex. For a twice differentiable function, that would mean $\frac{d^2y}{dx^2}>0$. Suppose further that $y(x_1)>0$ for some $x_1\in(0,\pi/2)$. Then by the MVT, there is a positive slope achieved at $c_1$ in $(0,x_1)$ and a negative slope achieved at $c_2$ in $(x_1,\pi/2)$. Again by the MVT this would mean that the second derivative is negative somewhere in $(c_1,c_2)$, a contradiction. So if the curve is convex, it would also have to be (C) negative. And then two answers are correct. With the assumption that only one answer is correct, the curve cannot be (B) convex.
Suppose the curve were (C) negative on $(0,\pi/2)$. Then since $\csc(x)$ is positive on $(0,\pi/2)$, the differential equation tells is that $\frac{d^2y}{dx^2}$ must be positive on $(0,\pi/2)$. So both (C) negative and (B) convex would be correct answers. Again with the assumption that there is only one correct answer, this implies the answer cannot be (C) negative.
The only answer available is (D) positive.

It can be established (see Marvin's answer) that the actual solution curve is: $$y=\ln(\sin(x))\sin(x)-x\cos(x)$$ This satisfies both endpoint conditions and the differential equation. This curve is both (C) positive and (B) convex on $(0,\pi/2)$. So two of the four choices are in fact correct. Perhaps the question was intended to be four separate Yes/No questions, but the instructions could make that more clear.
