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Choose the left homogeneous solution $v(x)=\sinh x$ and the right one $w(x)=\sinh (1-x)$. Their Wronskian is $v{w}^{\prime }-v^{\prime }w=-\sinh 1$. The Dirichlet modified Helmholtz Green function on an interval, with operator sign $D_x^2-1$, is thereforeIt vanishes at both endpoints, is continuous at $x=\xi$, and its first derivative jumps by $+1$, giving $(D_x^2-1)G=\delta (x-\xi )$. The Green-function representation then givesThe two integrals are $(e^{2x}-1)/4-x/2$ and $e(1-x)/2-(e-e^{2x-1})/4$, respectively. Substituting them and simplifying givesDirect differentiation verifies $u^{\prime \prime }-u=e^x$ and both zero boundary values. The negative kernel also explains why the solution is negative inside the interval despite the positive right-hand side.

For the Green function of $D_x^2-1$, the homogeneous solutions satisfying the left and right endpoint conditions are $u(x)=\sinh x$ and $v(x)=\sinh (1-x)$. Their Wronskian isContinuity of the kernel at $x=\xi$ and integration of $G_{xx}-G=\delta (x-\xi )$ require $[G_x{]}_{\xi -}^{\xi +}=1$. Hence the Dirichlet modified Helmholtz Green function on an interval isHere $x_{<}=\min (x,\xi )$ and $x_{>}=\max (x,\xi )$. The negative sign is essential for this operator convention. Integrating the forcing against the kernel givesThis satisfies both endpoint conditions. It solves the equation classically for continuous $f$, and in the corresponding weak sense for integrable forcing. A homogeneous difference satisfying both endpoint conditions must vanish, so the solution is unique.