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Past exam of the mathematics course of the University of Cambridge / 2023 / ib / Paper 4 / 12B / c / Solution

The coefficient matrix has eigenvector

u_{+} = (1, 1)^{T}

with eigenvalue

3

and eigenvector

u_{-} = (1, - 1)^{T}

with eigenvalue

- 1

. Write

y (t) = p (t) u_{+} + q (t) u_{-} .

Since

(e^{2 t} - 2 t) = \frac{e ^{2 t} - 2 t}{2} u_{+} + \frac{e ^{2 t} + 2 t}{2} u_{-},

and

y (0) = (1, - 2)^{T} = - \frac{1}{2} u_{+} + \frac{3}{2} u_{-}

, the system diagonalises to

p^{'} = 3 p + \frac{e ^{2 t} - 2 t}{2}, qq u a d p (0) = - \frac{1}{2},

q^{'} = - q + \frac{e ^{2 t} + 2 t}{2}, qq u a d q (0) = \frac{3}{2} .

Using an integrating factor gives

p (t) = - \frac{1}{9} e^{3 t} - \frac{1}{2} e^{2 t} + \frac{t}{3} + \frac{1}{9},

q (t) = \frac{1}{6} e^{2 t} + t - 1 + \frac{7}{3} e^{- t} .

The unique fastest term is therefore

y (t) = - \frac{1}{9} e^{3 t} u_{+} + O (e^{2 t}) .

Consequently

e^{- n t} y (t)

diverges for

n < 3

, tends to zero for

n > 3

, and for

n = 3

tends to the finite nonzero vector

t \to \infty lim e^{- 3 t} y (t) = - \frac{1}{9} (11) .

Thus the only requested integer is

n = 3 .

This is an instance of dominant eigenmode in a forced linear system.

Solved by gpt-5.6-sol high.

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