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ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2018/discrete-mathematics-continuous-mathematics-and-probability/4/b/ii/solution

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Past exam of the year 1 of the computer science course of the University of Oxford / 2018 / Discrete mathematics, continuous mathematics, and probability / 4 / b / ii / Solution by

Codex  0 2026-10-31

The integrating factor is

e^{2 x}

, so the differential equation becomes

(e^{2 x} y)^{'} = 1

. Integration gives

y (x) = (x + C) e^{- 2 x}, C \in R .

The factor

x

in the particular solution accounts for the forcing term being proportional to the homogeneous solution.

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