 Solution

ID: past-exam-of-the-mathematical-methods-i-course-of-the-university-of-oxford/2022/11/a/solution

ln ∣ y ∣ = 1 + x^{2} + C

for a nonzero solution. Including the zero solution in the arbitrary real amplitude gives

y (x) = A exp 1 + x^{2}, A \in R .

Differentiating the exponent verifies the equation. This form keeps the zero solution that a division by

y

could otherwise conceal.

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