A certain population can be modelled by the differential equation \(\frac{{{\text{d}}y}}{{{\text{d}}t}} = k\,y\cos kt\) , where y is the population at time t hours and k is a positive constant.

(a)     Given that \(y = {y_0}\) when t = 0 , express y in terms of k , t and \({y_0}\) .

(b)     Find the ratio of the minimum size of the population to the maximum size of the population.

(a)     \(\frac{{{\text{d}}y}}{{{\text{d}}t}} = ky\cos (kt)\)

\(\frac{{{\text{d}}y}}{y} = k\cos (kt){\text{d}}t\)     (M1)

\(\int {\frac{{{\text{d}}y}}{y} = \int {k\cos (kt){\text{d}}t} } \)     M1

\(\ln y = \sin (kt) + c\)     A1

\(y = A{{\text{e}}^{\sin (kt)}}\)

\(t = 0 \Rightarrow {y_0} = A\)     (M1)

\( \Rightarrow y = {y_0}{{\text{e}}^{\sin kt}}\)     A1

(b)     \( - 1 \leqslant \sin kt \leqslant 1\)     (M1)

\({y_0}{{\text{e}}^{ - 1}} \leqslant y \leqslant {y_0}{{\text{e}}^1}\)

so the ratio is \(\frac{1}{{\text{e}}}:{\text{e}}\,\,\,\,\,{\text{or }}1:{{\text{e}}^2}\)     A1

[7 marks]

Part (a) was done successfully by many candidates. However, very few attempted part (b).