 Solution

PutLeibniz differentiation gives $z^{\prime }=-{\int }_0^x\cos (x-s)f(s)ds$ and $z^{\prime \prime }=-f(x)+{\int }_0^x\sin (x-s)f(s)ds=-f-z$. Thus $z$ satisfies both zero initial conditions and the inhomogeneous equation. Every solution with $y(0)=0$ is $y=A\sin x+z(x)$.

On $[0,1]$, the second endpoint determines $A=-z(1)/\sin 1$ uniquely, since $\sin 1\ne 0$. Collecting the integrals, and using the sine addition formula for $s<x$, givesThis continuous kernel vanishes at the endpoints, satisfies $G_{xx}+G=0$ off $x=s$, and has derivative jump $-1$, as required for $-y^{\prime \prime }-y=f$. The preceding construction proves existence, and a homogeneous solution with both endpoints zero must vanish, proving uniqueness.

On $[0,\pi ]$, instead, the second boundary value becomesThus solvability is equivalent to ${\int }_0^{\pi }f(s)\sin sds=0$. When it holds, all solutions areThere is one free homogeneous mode because the endpoint length is resonant.