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

For , differentiate the printed ansatz:
Matching the independent coefficients in gives
There are two parameter pairs, including the degenerate spatially constant case:
Their corresponding solutions are
The 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 gives
Here 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.

New to topics? Read the docs here!