Past exam of the algebra, analysis and differential equations II course of the University of Oxford / 2012 / 8 / b / Solution

For this forcing the integral in part (a) simplifies toConsequently the unique boundary value problem solution isIt vanishes at zero and at $\pi /2$. To check the differential equation directly, set $y=e^{-2x}w$. Then $Ly=e^{-2x}(w^{\prime \prime }+w)$. For $w=1-\sin x-\cos x$, one has $w^{\prime \prime }+w=1$, as required.