 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2013/ia/paper-2/1a/solution

Codex  0 Created 2026-09-24 Updated 2026-10-07

(r - 2) (r + 1)

A e^{2 t} + B e^{- t}

. Each exponential forcing resonates with a homogeneous mode; the method of undetermined coefficients therefore uses

t e^{2 t}

and

t e^{- t}

. Substitution gives the particular solution

t e^{2 t} - t e^{- t} - 3 t

. Thus

y = (A + t) e^{2 t} + (B - t) e^{- t} - 3 t .

A + B = 0

and

2 A - B - 3 = 0

. Hence

y (t) = (t + 1) e^{2 t} - (t + 1) e^{- t} - 3 t .

The factors of

t

in the particular solution are essential: an unmodified exponential trial lies in the kernel of the differential operator.

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