Solution: The given differential equation is \begin{equation}\label{eq:07Mar2025-1} \frac{\mathrm{d}^2y}{\mathrm{d}x^2} + a^2 y = \csc 3x. \end{equation} The corresponding auxiliary equation will be \[ m^2 + a^2 = 0 \implies m = \pm a \iota. \] Thus, the complementary function will be \[ y_{\mathrm{cf}}(x) = c_1 \cos(ax) + c_2 \sin(ax). \]

Now, we will use the method of variation of parameters to find a particular solution. Let us write $f(x) = \cos ax$ and $g(x) = \sin ax$. The particular integral will be \begin{equation}\label{eq:07Mar2025-2} y_{\mathrm{PI} }(x) = u(x) \cos (ax) + v(x) \sin (ax), \end{equation} where \begin{align*} u(x) = -\int \frac{g(x) \csc x }{W(f,g)}\mathrm{d} x \quad \text{ and } \quad v(x) = \int \frac{f(x) \csc x}{W(f,g)} \mathrm{d} x, \end{align*} where $W(f,g)$ is the Wronskian of the functions $f$ and $g$ which is \[ W(f,g) = \begin{vmatrix} f(x) & g(x) \\ f'(x) & g'(x) \\ \end{vmatrix} = \begin{vmatrix} \cos ax & \sin ax \\ -a\sin ax & a \cos ax \\ \end{vmatrix} = a. \]

Thus, \begin{align*} u(x) & = -\int \frac{\sin ax \csc x}{a}\mathrm{d} x = -\frac{x}{a} \\[1ex] v(x) & = int \frac{\cos ax \csc x}{a}\mathrm{d} x \\[1ex] & = \frac{1}{a} \int \frac{\cos x}{\sin x}\mathrm{d}x \\[1ex] & = \frac{1}{a} \left( \frac{1}{a} \ln \sin ax \right) \\[1ex] & = \frac{1}{a^2} \ln \sin ax. \end{align*} Therefore, \[ y_{\mathrm{PI} }(x) = -\frac{x}{a} \cos ax + \frac{\sin ax}{a^2} \ln \sin ax. \] Hence, the general solution of \eqref{eq:07Mar2025-1} will be \[ \textcolor{blue}{ \boxed{ y(x) = c_1 \cos (ax) + c_2 \sin (ax) -\frac{x}{a} \cos (ax) + \frac{\sin (ax)}{a^2} \ln \sin (ax) } } \]