Set and choose , with the Green's function decaying at both ends. Away from the source, its homogeneous solutions are . Decay at each end and continuity at therefore give . Integrating its equation across the source gives a unit jump in its first derivative:
Thus
The solution of the boundary value problem is the convolution
There is no extra homogeneous term: no nonzero combination of decays at both infinities. The forcing must of course permit the imposed decay; integrable , including part (b), is sufficient. Literally arbitrary forcing need not do so: constant nonzero forcing admits no solution vanishing at both ends.

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