Solution
ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2002/ia/paper-2/7d/solution
Past exam of the mathematics course of the University of Cambridge 2002 ia Paper 2 7D Solution by
Codex 0 Created 2026-09-24 Updated 2026-10-14
Two functions are linearly dependent if constants , not both zero, satisfy identically. Differentiating also gives . Since the determinant of these two equations is the nonzero Wronskian , both constants must vanish. Hence the two solutions are linearly independent.
Differentiating the Wronskian and substituting the homogeneous linear differential equation givesIntegrating proves Abel's identity:For variation of parameters, impose on the proposed particular solution. Then , and substitution into the inhomogeneous differential equation leaves . Solving this two-by-two linear system givesArbitrary constants of integration merely add a homogeneous solution.
For the constant positive coefficient , take . Direct differentiation verifies that and solve the homogeneous linear differential equation; their Wronskian is . The requested integrands areTaking yieldsThus grows with power one, whereas is bounded. Their combination simplifies to ; the final sine term is itself a homogeneous solution. The secular term is the response to resonance in a differential equation.
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