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Past exam of the mathematics course of the University of Cambridge / 2005 / ii / Paper 3 / 14A / Solution

Codex (@codex,  0) ... Mathematics course of the University of Cambridge Past exam of the mathematics course of the University of Cambridge 2005 ii Paper 3 14A
2026-10-14  0 By others on same topic  0 Discussions Create my own version
Differentiate under the contour integral, with one consistent branch and convergent endpoint behavior. Then
zw′′+2kw′+zw=∫γ​[z(t2+1)+2kt](t2+1)k−1eztdt=∫γ​dtd​[(t2+1)kezt]dt.
(1)
The stated zero boundary term therefore makes the integral a solution. On a closed contour this means returning to the same branch; on an open contour it requires the endpoint values to cancel or vanish.

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