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 3 - 2007 - Paper 8

Setting $P = 2P_0$ gives:

$2P_0 = P_0 e^{kt}$

Dividing both sides by $P_0$:

$2 = e^{kt}$

Taking the natural logarithm of both sides:

$\ln(2) = kt$

Now substituting $k = 2.5$:

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

Calculating this gives approximately:

$t \approx 0.277828822.$

Converting this into minutes:

$t \approx 0.277828822 \times 60 \approx 16.67 \text{ minutes}$

Thus, rounding to the nearest minute, the time taken is approximately 17 minutes.

To solve the second differential equation $\frac{dP}{dt} = \lambda P \cos(\lambda t)$, we separate variables:

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

Integrating both sides:

$\int \frac{dP}{P} = \int \lambda \cos(\lambda t) \, dt$

This results in:

$\ln(P) = \sin(\lambda t) + C$

Now, using the initial condition $P(0) = P_0$:

$\ln(P_0) = C.$

Thus, we can express the solution as:

$$P = P_0 e^{\sin(\lambda t)}.$

Setting $P = 2P_0$ gives:

$2P_0 = P_0 e^{\sin(\lambda t)}$.

Dividing both sides by $P_0$:

$2 = e^{\sin(\lambda t)}$.

Taking the natural logarithm of both sides:

$\ln(2) = \sin(\lambda t)$.

Given $\lambda = 2.5$, we have:

$\sin(\lambda t) = \sin(2.5t).$

To find $t$, we rearrange:

$t = \arcsin(\ln(2)) / 2.5$.

Calculating gives:

$t \approx 0.303638374 \times 60 \approx 18.21 ext{ minutes}.$

Thus, rounding to the nearest minute, the time taken is approximately 18 minutes.