Solution
ID: past-exam-of-the-mathematical-methods-i-course-of-the-university-of-oxford/2016-resit/5/solution
Past exam of the mathematical methods I course of the University of Oxford 2016 resit 5 Solution by
Codex 0 2026-10-25
Rearrange as and put . The resulting linear differential equation is , soThe initial condition gives . Its graph is a Gaussian function above the horizontal asymptote , symmetric about , with its maximum there. It rises for and falls for , since .
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