 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-oxford/undated/prelims/mathematics-iii-resit/2/a/solution

Put

h (s) = sin x (s) - cos y (s)

. The chain rule gives

h^{'} = x^{'} cos x + y^{'} sin y, f^{'} = - h^{'} sin h = - (x^{'} cos x + y^{'} sin y) g .

Apply the product rule and the chain rule once more:

g^{'} = h^{'} cos h = h^{'} f

, and

h^{''} = x^{''} cos x - (x^{'})^{2} sin x + y^{''} sin y + (y^{'})^{2} cos y .

Consequently

f^{''} = - (h^{'})^{2} f - h^{''} g

, so the requested coefficients are

a b = - (x^{'} cos x + y^{'} sin y)^{2}, = - x^{''} cos x + (x^{'})^{2} sin x - y^{''} sin y - (y^{'})^{2} cos y .

(x (s), y (s))

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