Modelling with rate proportional to difference

9 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 $$\frac { \mathrm { d } \theta } { \mathrm {~d} t } = - k ( \theta - A )$$ where \(k\) is a positive constant.

1. Find, in any form, the solution of this differential equation, given that \(\theta = 4 A\) when \(t = 0\).
2. Given also that \(\theta = 3 A\) when \(t = 1\), show that \(k = \ln \frac { 3 } { 2 }\).
3. Find \(\theta\) in terms of \(A\) when \(t = 2\), expressing your answer in its simplest form.

The current, \(x\) amps, at time \(t\) seconds after a switch is closed in a particular electric circuit is modelled by the equation $$\frac{dx}{dt} = k - 3x$$ where \(k\) is a constant. Initially there is zero current in the circuit.

1. Solve the differential equation to find an equation, in terms of \(k\), for the current in the circuit at time \(t\) seconds. Give your answer in the form \(x = f(t)\). [6]

Given that in the long term the current in the circuit approaches \(7\) amps,

1. find the value of \(k\). [2]
2. Hence find the time in seconds it takes for the current to reach \(5\) amps, giving your answer to \(2\) significant figures. [3]