 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2022/paper-2/15/a/ii/solution

Again

C = 0

. Since the initial sine is nonzero, the local branch must satisfy

sin y = x

. Selecting the inverse-sine branch through

y = π /6

gives

y (x) = arcsin x, - 1 < x < 1 .

This passes through the specified point and has

y^{'} = 1/ 1 - x^{2}

, which verifies the original equation on this interval.

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