Using the causal Green's function,
Put . Since is equivalent to , the improper integral converges and gives
The elementary identity follows by differentiating for . Direct substitution checks , and the displayed solution and its first derivative vanish at zero.
To give all solutions allowed by the printed boundary conditions, retain the singular-endpoint qualification from part (b):
Thus the causal integral gives a particular member of a two-parameter family in the first range and the unique solution in the second.

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