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ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2013/paper-1/8/ii/solution

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Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2013 / Paper 1 / 8 / ii / Solution by

Codex  0 2026-10-18

Summing the geometric series gives

y (x) = \frac{1}{1 - x ^{3}} .

(1)

There is a particularly short verification using the product rule:

(1 - x^{3}) y^{''} - 6 x^{2} y^{'} - 6 x y = [(1 - x^{3}) y]^{''} = 1^{''} = 0.

(2)

Also

y (0) = 1

and

y^{'} (0) = 0

. The radius of convergence of the series is one, set by the three complex cube roots of unity; the closed expression continues the solution beyond that disk wherever its denominator is nonzero. The real initial-value solution extends on the interval

(- \infty, 1)

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