 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2010/ia/paper-2/7a/c/solution

Codex  0 Created 2026-09-24 Updated 2026-10-14

λ = - 1

the

v_{-}

forcing vanishes and the

v_{+}

- 2

. The initial vector decomposes as

(0, 1, 0)^{T} = \frac{1}{2} v_{-} - \frac{1}{2} v_{+} .

Thus

A = 1/2

B = 0

C = - 1/2

in part (b), and the requested solution is

x y z = \frac{1}{2} e^{- 2 t} v_{-} - (\frac{1}{2} + 2 t) e^{2 t} v_{+} .

Equivalently,

x = \frac{1}{2} (1 + 4 t) e^{2 t} - \frac{1}{2} e^{- 2 t}, y = \frac{1}{2} (1 + 4 t) e^{2 t} + \frac{1}{2} e^{- 2 t}, z = \frac{1}{2} e^{- 2 t} - \frac{1}{2} (1 + 4 t) e^{2 t} .

These expressions have the required initial values and solve all three linear ordinary differential equations.

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