Solution: We want to solve the differential equation \begin{equation}\label{eq:21June2024-1} \frac{\mathrm{d}^2y}{\mathrm{d}x^2} + y = \cos 2x. \end{equation} The corresponding homogeneous equation is $\frac{\mathrm{d}^2y}{\mathrm{d}x^2} + y =0$ and hence the auxiliary equation will be \[ m^2 + 1 = 0 \implies m = \pm \iota . \] Therefore, the solution corresponding to the homogeneous equation of \eqref{eq:21June2024-1} will be \begin{equation}\label{eq:21June2024-2} y_{\mathsf{cf} } = c_1 \cos x + c_2 \sin x. \end{equation}

Now we can find the particular integral by using the method of variation of parameters. Take \begin{equation}\label{eq:21June2024-3} y_{\mathsf{PI}} = A \cos 2x + B \sin 2x . \end{equation} Substituting \eqref{eq:21June2024-3} in \eqref{eq:21June2024-1}, we obtain \begin{align*} & -4 A \cos 2x - 4B \sin 2x + A \cos 2x + B \sin 2x = \cos 2x \\ \implies & -3A \cos 2x -3B \sin 2x = \cos 2x \\ \implies & -3A = 1 \text{ and } -4B = 0 \\ \implies & A = -\frac{1}{3} \text{ and } B = 0. \end{align*} Thus, \[ y_{\mathsf{PI} } = -\frac{1}{3} \cos 2x . \] Therefore, the solution to the given differential equation will be \[ y(x) = c_1 \cos x + c_2 \sin x - \frac{\cos 2x}{3}. \]