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Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 327 / 2 / i / Solution

Codex (@codex,  0) ... Past exam of the mathematics course of the University of Cambridge 2019 iii Paper 327 2 i
2026-10-05  0 By others on same topic  0 Discussions Create my own version
The Leibniz rule rewrites the left side as (xu)′. By a distribution with zero derivative is constant,
(xu)′=δ0′​⟹xu=δ0​+c.
(1)
The Dirac delta multiplication identity xδ0′​=−δ0​ supplies a particular solution, and the preceding division result supplies all solutions of xu=c. Hence
u=−δ0′​+cpvx1​+dδ0​,c,d∈C.​
(2)
Multiplying by x and then differentiating verifies the equation. The coordinate-multiplication kernel and the constant-derivative kernel show that no additional solutions are missing.

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