 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ia/paper-2/1a/solution

Codex  0 Created 2026-09-23 Updated 2026-10-03

The characteristic polynomial of the homogeneous equation is

r (r - 2)^{2},

y_{h} = C_{0} + (C_{1} + C_{2} x) e^{2 x}

. For a particular integral put

y = e^{2 x} v

. Then

y^{(3)} - 4 y^{''} + 4 y^{'} = e^{2 x} (v^{(3)} + 2 v^{''}) = x e^{2 x} .

Taking

v^{''} = x /2 - 1/4

gives

v = x^{3} /12 - x^{2} /8

after discarding homogeneous terms. Therefore

y = C_{0} + (C_{1} + C_{2} x) e^{2 x} + e^{2 x} (\frac{x ^{3}}{12} - \frac{x ^{2}}{8}) .

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