The rate of increase of the number, N, of fish in a lake is modelled by the differential equation dN dt = (k - t)(5000 - N) t > 0, 0 < N < 5000 In the given equation, the time t is measured in years from the start of January 2000 and k is a positive constant - Edexcel - A-Level Maths Pure - Question 7 - 2014 - Paper 8

Using the conditions provided:

1. At t = 1 (January 2001), N = 1200:

   $1200 = 5000 - A e^{-k(1)} \\ A e^{-k} = 3800 \quad \text{(1)}$

2. At t = 2 (January 2002), N = 1800:

   $1800 = 5000 - A e^{-k(2)} \\ A e^{-2k} = 3200 \quad \text{(2)}$

Dividing equation (1) by equation (2):

$\frac{3800}{3200} = e^{k} \\ k = \ln\left(\frac{19}{8}\right)$

Substituting back to find A:

Using (1):

$A e^{-k} = 3800 \\ A = 9025$

Thus, the values are:

• A = 9025
• k = \ln\left(\frac{19}{8}\right)

Using the formula found:

$N = 5000 - 9025 e^{-5k}$

Substituting k:

$N = 5000 - 9025 e^{-5\ln\left(\frac{19}{8}\right)}$

Calculating, we find:

$N = 4402.828401... \approx 4400 \text{ (to the nearest hundred)}$