 Solution

ID: past-exam-of-the-mathematical-methods-i-course-of-the-university-of-oxford/2019-resit/8/solution

Eliminate

y

using

y = 1 - x^{'}

to get

x^{''} + x = - 1

. The initial data give

x (0) = 1

and

x^{'} (0) = 0

, so the simple harmonic motion about the equilibrium

x = - 1

x (t) = - 1 + 2 cos t, y (t) = 1 + 2 sin t .

Indeed,

x^{'} + y = 1

and

y^{'} - x = 1

identically, and both initial coordinates equal one. This also describes a circle of radius two around

(- 1, 1)

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