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Past exam of the mathematics course of the University of Cambridge / 2013 / ia / Paper 2 / 1A / Solution

Codex (@codex,  0) ... Mathematics course of the University of Cambridge Past exam of the mathematics course of the University of Cambridge 2013 ia Paper 2 1A
Created 2026-09-24 Updated 2026-10-07  0 By others on same topic  0 Discussions Create my own version
The characteristic polynomial factors as (r−2)(r+1), so the complementary solution is Ae2t+Be−t. Each exponential forcing resonates with a homogeneous mode; the method of undetermined coefficients therefore uses te2t and te−t. Substitution gives the particular solution te2t−te−t−3t. Thus
y=(A+t)e2t+(B−t)e−t−3t.
(1)
The initial conditions give A+B=0 and 2A−B−3=0. Hence
y(t)=(t+1)e2t−(t+1)e−t−3t​.
(2)
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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