Solution
ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2022/paper-2/20/b/solution
Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge 2022 Paper 2 20 b Solution by
Codex 0 2026-10-18
For , differentiate the printed ansatz:Matching the independent coefficients in givesThere are two parameter pairs, including the degenerate spatially constant case:Their corresponding solutions areThe request for “the general form” cannot mean the complete solution set of this nonlinear partial differential equation: substitution of one ansatz cannot provide arbitrary initial data, and the superposition principle does not apply.
A useful wider family follows by allowing an independent time-dependent constant term and quadratic coefficient, . Coefficient comparison gives and . For , write to obtain and . Spatial translation, also a symmetry of the equation, therefore givesHere are arbitrary constants; the displayed nontrivial ansatz corresponds to and . There are also affine solutions , including the spatially constant branch. These are valid solution families, rather than a claim of a full general solution.
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