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 1 - 2007 - Paper 7

We know:

$P = 2P_0$

From the equation we derived earlier:

$2P_0 = P_0 e^{kt}$

Dividing both sides by $P_0$ (assuming $P_0 > 0$), we get:

$2 = e^{kt}$

Taking the natural logarithm of both sides results in:

$\ln(2) = kt.$

Substituting $k = 2.5$ gives:

$t = \frac{\ln(2)}{2.5}.$
Calculating:

$t \approx \frac{0.693147}{2.5} \approx 0.2772588... \text{ days} = 0.2772588 \times 1440, ext{ (since 1 day = 1440 min)}$

This yields approximately $399.8$ minutes, or about $6$ hours and $39$ minutes when rounded to the nearest minute.

The new equation is:

$\frac{dP}{dt} = \lambda P \cos At$

We again separate the variables:

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

Integrating both sides yields:

$\int \frac{1}{P} dP = \lambda \int \cos At \, dt$

This leads us to:

$\ln|P| = \frac{\lambda}{A} \sin At + C$

Exponentiating both sides gives:

$P = e^{\frac{\lambda}{A} \sin At + C} = e^C e^{\frac{\lambda}{A} \sin At}.$

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

$P_0 = e^C \Rightarrow C = \ln P_0,$
so we find:

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

Setting the population to double:

$P = 2P_0,$

yielding:

$2P_0 = P_0 e^{\frac{\lambda}{A} \sin At}$

Dividing by $P_0$ gives us:

$2 = e^{\frac{\lambda}{A} \sin At}$

Taking logarithm:

$\ln(2) = \frac{\lambda}{A} \sin At.$
Substituting $lambda = 2.5$ yields:

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

To solve this, we need the value of $A$. Rearranging gives:

$t = \frac{\arcsin \left( \frac{\ln(2) A}{2.5} \right)}{A}.$ Hence, we need a specific value of $A$ to compute the exact time.