The temperature of a quantity of liquid at time \(t\) is \(\theta\). The liquid is cooling in an atmosphere whose temperature is constant and equal to \(A\). The rate of decrease of \(\theta\) is proportional to the temperature difference \((\theta - A)\). Thus \(\theta\) and \(t\) satisfy the differential equation

\(\frac{d\theta}{dt} = -k(\theta - A),\)

where \(k\) is a positive constant.

(i) Find, in any form, the solution of this differential equation, given that \(\theta = 4A\) when \(t = 0\). [5]

(ii) Given also that \(\theta = 3A\) when \(t = 1\), show that \(k = \ln \frac{3}{2}\). [1]

(iii) Find \(\theta\) in terms of \(A\) when \(t = 2\), expressing your answer in its simplest form. [3]