 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2020/ii/paper-4/31a/iv/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

Multiplying the two Liouville–Green forms by

e^{x /2}

gives

y_{+} (x) \sim x^{2} e^{x}, y_{-} (x) \sim x^{- 2} .

The first expression is in fact exact:

y = x^{2} e^{x}, y^{'} = e^{x} (x^{2} + 2 x), y^{''} = e^{x} (x^{2} + 4 x + 2),

and direct substitution gives

y^{''} - y^{'} - \frac{2 ( x + 1 )}{x ^{2}} y = 0.

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