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For the radial Helmholtz Dirichlet Green function, set $H(x,\xi )=xG(x,\xi )$. Multiplication of the radial equation by $x$ givesThe first endpoint follows from the required boundedness of $G$ at zero. For $k\ne 0$ and $\sin k\ne 0$, use the left solution $\sin kx$ and right solution $\sin k(1-x)$. Their Wronskian is $-k\sin k$. Matching continuity and the derivative jump therefore gives, with $x_{<}=\min (x,\xi )$ and $x_{>}=\max (x,\xi )$,This is finite at zero, vanishes at one, is continuous at $x=\xi$, and has $G_x(\xi +,\xi )-G_x(\xi -,\xi )=1$. Thus it has exactly the prescribed delta source, not its negative. Its weighted reciprocity is $x^{2}G(x,\xi )={\xi }^{2}G(\xi ,x)$, as expected for the radial weight.

Integrating the Green function against the constant forcing givesThe integrals are respectively $-x\cos (kx)/k+\sin (kx)/k^2$ and $(1-x\cos (k(1-x)))/k-\sin (k(1-x))/k^2$. After cancellation their weighted sum is $(\sin (kx)-x\sin k)/k$. HenceThe values at zero are interpreted by continuity. At $k=0$ the limit exists: $G(x,\xi )=-\xi x_{<}(1-x_{>})/x$ and $y(x)=(x^2-1)/6$.

The exceptional nonzero values $k\in \pi \mathbb{Z}$ need a genuine qualification. There is then a nonzero bounded homogeneous solution $\sin (kx)/x$ satisfying both endpoint conditions, so an inverse Green function does not exist. Moreover the stated forcing is incompatible, not merely nonunique. If $h=xy$ solved the problem, multiply $h^{\prime \prime }+k^{2}h=x$ by $\sin (kx)$ and integrate by parts using $h(0)=h(1)=0$. The left side is zero, whereasThus there is no solution for nonzero $k\in \pi \mathbb{Z}$. This explains exactly when the generic formula is applicable.