Oil spill from a boat has a significant impact on the population of fish in a lake. The population (in thousands) $t$ days after an oil spill is denoted by $x$. Statistics show that the population increases at a rate inversely proportional to the square of the population and that fish die, due to the oil spill, at a rate proportional to the population. When $x = 0.5$, the number of fish remains constant.

1. Formulate an initial differential equation using two proportional constants.

Let the population increase rate be $\frac{A}{x^2}$ and the death rate be $Bx$, where $A$ and $B$ are positive constants.

The overall rate of change of the population is:
$\frac{dx}{dt} = \frac{A}{x^2} - Bx$

2. Use the condition for constant population to find the relationship between the two constants.

When $x = 0.5$, the population is constant, which means $\frac{dx}{dt} = 0$.
Substitute $x = 0.5$ and $\frac{dx}{dt} = 0$ into the equation:
$0 = \frac{A}{(0.5)^2} - B(0.5)$
$0 = \frac{A}{0.25} - 0.5B$
$0 = 4A - 0.5B$
$0.5B = 4A$
$B = 8A$

3. Substitute the relationship back to obtain the final equation.

Substitute $B = 8A$ back into the differential equation:
$\frac{dx}{dt} = \frac{A}{x^2} - 8Ax$
Combine the terms over a common denominator:
$\frac{dx}{dt} = \frac{A - 8Ax^3}{x^2}$
$\frac{dx}{dt} = \frac{A(1 - 8x^3)}{x^2}$
By letting $A = k$, where $k$ is a positive constant, we get:
$\frac{dx}{dt} = \frac{k(1 - 8x^3)}{x^2} \text{ (shown)}$

1. Separate the variables and integrate both sides of the differential equation.

$\frac{dx}{dt} = \frac{k(1-8x^3)}{x^2}$
$\int \frac{x^2}{1-8x^3} \, dx = \int k \, dt$
$-\frac{1}{24}\ln|1-8x^3| = kt + C$

2. Rearrange the equation into an exponential form.

$\ln|1-8x^3| = -24kt - 24C$
$1-8x^3 = \pm e^{-24kt - 24C}$
$1-8x^3 = A(e^{-24k})^t$
(where $A = \pm e^{-24C}$)

3. Use the given conditions to find the constants and show the target equation.

When the oil spill occurred ($t = 0$), population is 3000 ($x = 3$):
$1 - 8(3)^3 = A(e^{-24k})^0$
$1 - 216 = A \implies A = -215$

One day after ($t = 1$), population is 2000 ($x = 2$):
$1 - 8(2)^3 = -215(e^{-24k})^1$
$-63 = -215(e^{-24k}) \implies e^{-24k} = \frac{63}{215}$

Substitute $A$ and $e^{-24k}$ into the general equation:
$1 - 8x^3 = -215\left(\frac{63}{215}\right)^t \text{ (Shown)}$

4. Calculate the number of fish on the third day.

Substitute $t = 3$ into the equation:
$1 - 8x^3 = -215\left(\frac{63}{215}\right)^3$
$1 - 8x^3 = -215\left(\frac{250047}{9938375}\right) = -\frac{250047}{46225} \approx -5.40935$
$8x^3 \approx 6.40935$
$x^3 \approx 0.80117$
$x \approx 0.92877$

Since $x$ is in thousands, the number of fish is:
$0.92877 \times 1000 = 928.77$
To the nearest integer, there are $929$ fish.

1. Evaluate the behaviour of the exponential term as $t$ approaches infinity.

As $t \to \infty$, since $0 < \frac{63}{215} < 1$, the term $\left(\frac{63}{215}\right)^t \to 0$.

2. Solve for the limiting value of $x$ and interpret it in the context of the population.

Substitute the limit into the equation from (ii):
$1 - 8x^3 = 0$
$8x^3 = 1 \implies x^3 = \frac{1}{8} \implies x = 0.5$
Since $x$ is measured in thousands, the population of fish will approach $0.5 \times 1000 = 500$.