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 7

We set the equation from part (a) with the target population:

$2P_0 = P_0 e^{kt}$

Dividing through by $P_0$ yields:

$2 = e^{kt}$

Taking the natural logarithm of both sides leads to:

$\ln 2 = kt$

Now, substituting $k = 2.5$:

$t = \frac{\ln 2}{2.5} \approx 0.277258...\text{ days}$

Converting days to minutes (1 day = 1440 minutes):

$t \approx 0.277258 \times 1440 \approx 399.36... \text{ minutes}$

Thus, rounding to the nearest minute, we find:

$t \approx 399 \text{ minutes}$.

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

Again, we separate the variables: $\frac{dP}{P} = \lambda \cos(At) \, dt$

Next, we integrate both sides: $\int \frac{dP}{P} = \int \lambda \cos(At) \, dt$

This results in: $\ln P = \lambda \frac{\sin(At)}{A} + C$

Using the initial condition $P(0) = P_0$, we find: $\ln P_0 = C$

Thus, we can express our equation as: $\ln P = \lambda \frac{\sin(At)}{A} + \ln P_0$

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

Setting the population equation from part (c) to $2P_0$: $2P_0 = P_0 e^{\lambda \frac{\sin(At)}{A}}$

Dividing through by $P_0$ gives: $2 = e^{\lambda \frac{\sin(At)}{A}}$

Taking the logarithm of both sides leads to: $\ln 2 = \lambda \frac{\sin(At)}{A}$

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

To isolate $t$, we multiply both sides by $A/2.5$: $t = \frac{A \ln 2}{2.5 \sin(At)}$

This equation requires numerical methods to solve for $t$. Assuming specific values for $A$, we can find the corresponding time. After numerical evaluation and approximation, round to the nearest minute.