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/a/solution

Because the original differential equation divides by , its domain requires and . Set , so . The equation becomes the first-order linear differential equation
An integrating factor is , since . Thus
Returning to gives the explicit general real solutions
Each sign is fixed on a connected interval where and the radicand is strictly positive. The strict condition excludes , where the original differential equation is undefined. There is no additional zero solution.

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