A population growth is modelled by the differential equation $$\frac{dP}{dt} = kP,$$ where $P$ is the population, $t$ is the time measured in days and $k$ is a positive constant - Edexcel - A-Level Maths Pure - Question 2 - 2007 - Paper 8

Given $k = 2.5$, we set $P = 2P_0$ in our solution:

$2P_0 = P_0 e^{2.5t}.$

Dividing both sides by $P_0$ results in:

$2 = e^{2.5t}.$

Taking the natural logarithm of both sides gives:

$\ln 2 = 2.5t \Rightarrow t = \frac{\ln 2}{2.5}.$

Calculating this yields:

$t \approx 0.277258872.$

Converting this to minutes,

$$t \approx 0.277258872 \times 60 \approx 16.63663231 \text{ minutes} \Rightarrow 17 \text{ minutes (to the nearest minute)}.$

The second differential equation given is $\frac{dP}{dt} = \lambda P \cos(At).$

Separating the variables, we get:

$\frac{1}{P} dP = \lambda \cos(At) \, dt.$

Integrating both sides results in:

$\ln P = \lambda \int \cos(At) \, dt + C.$

The integral of $\, \cos(At)$ is $\, \frac{1}{A} \sin(At)$. Therefore:

$\ln P = \frac{\lambda}{A} \sin(At) + C.$

Using the initial condition $P(0) = P_0$:

$\ln P_0 = C.$

Hence, we have:

$$P = P_0 e^{\frac{\lambda}{A} \sin(At)}.$

Given that $\lambda = 2.5$, we set $P = 2P_0$ in our previous solution:

$2P_0 = P_0 e^{\frac{2.5}{A} \sin(At)}.$

Dividing both sides by $P_0$ yields:

$2 = e^{\frac{2.5}{A} \sin(At)}.$

Taking the natural logarithm gives:

$\ln 2 = \frac{2.5}{A} \sin(At).$

Rearranging, and isolating $\sin(At)$ provides:

$\sin(At) = \frac{A \ln 2}{2.5}.$

We need to solve for $t$ for the first instance when this equals 2; hence selecting appropriate values for $A$ will allow us to find the time $t$.