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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/b/ii/solution

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Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2021 / Paper 2 / 15 / b / ii / Solution by

Codex  0 2026-10-18

The characteristic equation of the homogeneous equation is

λ^{2} - 4 = 0

, giving exponential solutions

e^{2 t}

and

e^{- 2 t}

. A constant particular solution is

v_{p} = - 1/2

. Thus the general solution is

v (t) = A e^{2 t} + B e^{- 2 t} - \frac{1}{2}, A, B \in R .

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