 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ia/paper-2/8a/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

Termwise differentiation of the matrix exponential gives

\frac{d}{d t} e^{t A} = m = 1 \sum \infty \frac{m t ^{m - 1} A ^{m}}{m !} = A j = 0 \sum \infty \frac{t ^{j} A ^{j}}{j !} = A e^{t A} .

Thus

y (t) = e^{t A} y_{0}

satisfies

y^{'} = A y

and

y (0) = y_{0}

. If two solutions existed, multiplying their difference by

e^{- t A}

would give a vector with zero derivative and zero initial value, proving uniqueness.

For the inhomogeneous equation, multiply by the integrating factor

e^{- t A}

\frac{d}{d t} (e^{- t A} x) = e^{- t A} f (t) .

Integration and multiplication by

e^{t A}

x (t) = e^{t A} x_{0} + \int_{0}^{t} e^{(t - s) A} f (s) d s .

For the stated matrix,

A^{2} = 121224 - 4 - 4 - 8 - 4 - 4 - 8, A^{3} = 0,

e^{t A} = I + t A + \frac{t ^{2}}{2} A^{2} .

f (t) = v sin t

with

v = (1, 3, 0)^{T}

, one finds

A x_{0} = 0, A v = (8, 8, 16)^{T}, A^{2} v = 0.

Therefore

x (t) = x_{0} + v (1 - cos t) + A v (t - sin t),

or explicitly

x (t) = 8 t - 8 sin t - cos t + 2 8 t - 8 sin t - 3 cos t + 4 16 t - 16 sin t + 2 .

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