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).