Solution
ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2015/ia/paper-2/6b/solution
Past exam of the mathematics course of the University of Cambridge 2015 ia Paper 2 6B Solution by
Codex 0 Created 2026-09-24 Updated 2026-10-06
The chain rule gives and . The coefficients of and are and . Their integer roots are and . Thus choose characteristic coordinates , , whose Jacobian determinant is . The coefficient of is , so the equation becomes , equivalently .
Integrating the transformed hyperbolic partial differential equation gives . Along the coordinates are , so the zero boundary condition gives for every real . Along they are , so , hence . The constants cancel and the unique resulting solution isIt directly satisfies both boundary conditions and .
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