Past exam of the mathematical methods I course of the University of Oxford / 2019 resit / 12 / d / Solution

The forcing $\sec (2x)$ is undefined at $x=\pm \pi /4$, both inside the printed interval. Therefore there is no classical solution on the whole stated interval. Solve separately on the three connected components obtained by removing those two points.

Choose $u_1=\cos (2x)$ and $u_2=\sin (2x)$, whose Wronskian is two. Variation of parameters gives $t_1^{\prime }=-(1/2)\tan (2x)$ and $t_2^{\prime }=1/2$, so a particular solution is $\frac{1}{4}\cos (2x)\log |\cos (2x)|+\frac{x}{2}\sin (2x)$. The full answer on each regular component isThe constants $A,B$ can differ between $(-\pi /2,-\pi /4)$, $(-\pi /4,\pi /4)$ and $(\pi /4,\pi /2)$. Differentiation gives $y^{\prime \prime }+4y=1/\cos (2x)$ wherever the forcing is defined. Although the particular solution has finite one-sided values at the singularities, its derivative has a logarithmic divergence, and the equation itself is undefined there. It cannot be promoted to one classical solution across the poles.