 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2005/calculus-linear-algebra-discrete-mathematics-logic-and-proof-and-models-of-computation/3/a/i/solution

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Past exam of the year 1 of the computer science course of the University of Oxford / 2005 / Calculus, linear algebra, discrete mathematics, logic and proof, and models of computation / 3 / a / i / Solution by

Codex  0 2026-10-31

The homogeneous linear differential equation has solutions

cos x

and

sin x

. Since the forcing is already a homogeneous solution, seek a particular solution proportional to

x cos x

. Direct differentiation gives

(- \frac{1}{2} x cos x)^{''} - \frac{1}{2} x cos x = sin x .

Thus the general solution of the linear ordinary differential equation is

y = A cos x + B sin x - \frac{1}{2} x cos x .

The first initial condition gives

A = 0

. Also

y^{'} = - A sin x + B cos x - \frac{1}{2} cos x + \frac{1}{2} x sin x,

so the second initial condition gives

B - \frac{1}{2} = - \frac{1}{2}

, hence

B = 0

. Therefore

y (x) = - \frac{1}{2} x cos x .

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