 Solution

ID: past-exam-of-the-mathematical-methods-ii-course-of-the-university-of-oxford/2016-resit/1/b/solution

2 y^{'} = 15 x^{2} + (cos y) y^{'}

. Therefore

y^{'} = \frac{15 x ^{2}}{2 - cos y} .

For real

y

, the denominator is at least one, so division introduces no singularity. In fact the function

2 y - sin y

is strictly increasing, consistent with a unique real branch

y (x)

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