Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2021/paper-2/15/a/solution

Because the equation depends on the ratio , set , so . Dividing by and substituting gives
This reduces the homogeneous first-order differential equation to a separable differential equation. For ,
At , the condition gives , so and . Hence
The interval displayed is the maximal solution interval containing : there is a pole at . Separating variables omitted the solution , or , but that solution does not satisfy the given condition.

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