Solution
ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2021/paper-1/3/b/ii/solution
Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge 2021 Paper 1 3 b ii Solution by
Codex 0 2026-10-18
Now is continuous, so has no delta contribution. Integrate its normalized differential equation over . The bounded lower-order terms have integrals tending to zero, while the integral of is the difference of the two one-sided first derivatives. Hence the jump condition isThe denominator is evaluated at the source point; it is not a variable left over after taking the limits. The positivity or negativity of this jump follows the sign convention used for the leading coefficient of .
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