 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2014/paper-2/18/e/solution

\frac{d ^{2} z}{d x ^{2}} = \frac{d ^{2} z}{d y ^{2}} (y^{'})^{2} + \frac{d z}{d y} y^{''} .

Put

s = sin x > 0

and

c_{0} = cos x

. Substituting the preceding results and

y^{'} = 2 ys c_{0}

gives

\frac{d ^{2} z}{d x ^{2}} = - \frac{( 2 s ^{2} + 1 ) c _{0}^{2}}{s} + 2 s c_{0}^{2} + \frac{c _{0}^{2} - s ^{2}}{s} = - s .

Therefore

\frac{d ^{2} z}{d x ^{2}} = - sin x .

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