A population growth is modelled by the differential equation 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 - 2006 - Paper 6

Setting P = 2P0 in the equation we found:

$2P_0 = P_0 e^{kt}.$

Dividing both sides by P0 gives:

$2 = e^{kt}.$

Taking the natural logarithm of both sides:

$\ln(2) = kt.$

Substituting k = 2.5, we find:

$t = \frac{\ln(2)}{2.5} \approx 0.2778282827\text{,...}$

This evaluates to approximately 399 minutes, rounding to the nearest minute yields: 399 minutes.

We start with the given differential equation:

dP/dt = λP cos(λt).

Separating the variables, we have:

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

Now integrating both sides:

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

This results in:

$\ln(P) = \frac{λ}{\lambda} sin(\lambda t) + C$

Next, applying the initial condition once more:

When t = 0, P = P0 implies:

$\ln(P_0) = C.$

Thus, we can express it as:

$$\ln(P) = sin(\lambda t) + \ln(P_0)$$

Exponentiating gives:

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

Setting P = 2P0 in our derived equation:

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

Dividing both sides by P0:

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

Taking the natural logarithm:

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

Recalling that λ = 2.5, we can find t:

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

which leads to:

$t = \sin^{-1}(\ln(2)).$

When calculated, we receive approximate values, rounding gives us around 441 minutes.

Thus, to the nearest minute: 441 minutes.