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

The same variation of parameters construction works on $[0,\pi ]$. All solutions satisfying the boundary condition at zero have the formAt $x=\pi$, the homogeneous term vanishes for every $C$; it can no longer be chosen to correct the final boundary value. Since $\sin (\pi -t)=\sin t$, that boundary value is zero exactly whenThis condition is necessary by the displayed general solution, and sufficient by the same formula. When it holds, the full answer isFor complex-valued forcing, allow $C\in \mathbb{C}$ instead. The freedom is addition of the nonzero homogeneous Dirichlet solution $e^{-2x}\sin x$. At this endpoint separation, the homogeneous boundary problem has a nonzero solution. The resulting Fredholm solvability condition restricts the forcing to the compatible functions described above. In the language of the Fredholm alternative, the condition is orthogonality to the adjoint null solution $e^{2x}\sin x$ of $D^2-4D+5$ with zero endpoint values.