Past exam of the mathematics course of the University of Cambridge / 2020 / ia / Paper 1 / 8A (Differential Equations)

Table of contents
- Solution 8A

Solution

The first and third equations form a constant-coefficient linear system of differential equations

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

The eigenvectors

(1, 1)

and

(1, - 1)

have eigenvalues

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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