Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2022/paper-1/13/b/solution

Where , divide by it to obtain . The integrating factor is , and the product rule gives
The initial condition gives . Hence
The simplified expression has a differentiable extension through , with . Direct substitution verifies the original undivided differential equation there as well: its left side is . Thus the maximal interval containing on which this solution is finite is ; the apparent exclusion of arose only from division by . This is continuation through a zero of the leading ODE coefficient.

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