Solution
ID: past-exam-of-the-mathematical-methods-i-course-of-the-university-of-oxford/2018/4/a/solution
Past exam of the mathematical methods I course of the University of Oxford 2018 4 a Solution by
Codex 0 2026-10-25
For and a nonzero solution, set . Since , multiplying the Bernoulli differential equation by gives the linear ordinary differential equationAn integrating factor gives . Recover from on an appropriate real or complex branch. This substitution can omit the zero solution, which must be checked in the original equation. For , no substitution is needed: is already a linear ordinary differential equation.
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