 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2020/ia/paper-1/8a/solution

Codex  0 Created 2026-09-24 Updated 2026-09-29

\frac{d}{d t} (z x) = (- 1 3 3 - 1) (z x) .

(1, 1)

and

(1, - 1)

2

and

- 4

. Applying the initial data gives

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

The remaining equation becomes

y^{'} - 2 y = - 3 e^{- 4 t} + cos t - 2 sin t .

An integrating factor, or the usual exponential and trigonometric particular solutions, gives

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

These values satisfy all three initial conditions.

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