The rate of increase of the number, N, of fish in a lake is modelled by the differential equation $$\frac{dN}{dt} = \frac{(k - t)(5000 - N)}{t} \\ t > 0, \quad 0 < N < 5000$$ 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 1 - 2014 - Paper 8

Using the information given:

1. After one year (January 2001): $N(1) = 1200 = 5000 - Ae^{-k(1)} \\ \Rightarrow A = 5000 - 1200 e^{k}$

2. After two years (January 2002): $N(2) = 1800 = 5000 - Ae^{-k(2)} \\ \Rightarrow A = 5000 - 1800 e^{2k}$

3. Setting the expressions for A equal: $5000 - 1200 e^k = 5000 - 1800 e^{2k}$

4. Rearranging yields: $1200 e^k = 1800 e^{2k} \\ \Rightarrow \frac{2}{3} = e^{k}$

5. Taking natural log gives: $k = \ln\left(\frac{2}{3}\right)$

6. Substituting k back into the equation for A, we find: $A = 5000 - 1200 * \frac{2}{3} = 3800$.

Using the equation derived:

1. At t = 5: $N(5) = 5000 - 3800 e^{-5k}$

2. Substituting the value of k: $N(5) = 5000 - 3800 \left(\frac{2}{3}\right)^{5}$

3. Calculate: $N(5) = 5000 - 3800 * \frac{32}{243} \approx 4402.83$

4. Rounding gives: $N \approx 4400$ fish.