Past exam of the mathematics course of the University of Cambridge / 2018 / ia / Paper 2 / 5B (Differential Equations)

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- Solution 5B

Solution

After multiplying by the inverse of the derivative matrix, the system is

\overset{z}{˙} + A z = e^{- 4 t} (3 b /2 - 1 2) + e^{- t} (- c /2 - 1 c - 1), A = (2 - 2 - 1 3) .

Choose eigenvectors

(1, 1)^{T}

and

(- 1/2, 1)^{T}

1

and

4

, and write

(y x) = (11 - 1/2 1) (w _{2} w _{1}) .

Then

\overset{w}{˙}_{1} + w_{1} \overset{w}{˙}_{2} + 4 w_{2} = (b /2 + 1) e^{- 4 t} + (c /2 - 1) e^{- t}, = (b - 2) e^{- 4 t} - c e^{- t} .

The zero initial data give

w_{1} w_{2} x = \frac{b + 2}{6} (e^{- t} - e^{- 4 t}) + \frac{c - 2}{2} t e^{- t}, = (b - 2) t e^{- 4 t} - \frac{c}{3} (e^{- t} - e^{- 4 t}), = w_{1} - \frac{1}{2} w_{2}, y = w_{1} + w_{2} .

The

t e^{- 4 t}

and

t e^{- t}

terms are resonant responses. When

b = 2

or

c = 2

, respectively, the corresponding resonant forcing vanishes and so does that secular factor.

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