A team of naturalists is studying the change in population of wild boars on an island. It is suggested that the population of wild boars, x hundred, at time t years, can be modelled by the differential equation $\frac{dx}{dt}=\frac{1}{10}x(5-x).$

1. Separate the variables to set up the integrals.

$\int \frac{1}{x(5-x)} \, dx = \int \frac{1}{10} \, dt$

2. Express the rational function as partial fractions.

$\frac{1}{x(5-x)} = \frac{A}{x} + \frac{B}{5-x}$
$1 = A(5-x) + Bx$
Substituting $x = 0 \implies 5A = 1 \implies A = \frac{1}{5}$
Substituting $x = 5 \implies 5B = 1 \implies B = \frac{1}{5}$
$\int \left( \frac{1}{5x} + \frac{1}{5(5-x)} \right) \, dx = \int \frac{1}{10} \, dt$

3. Integrate both sides of the equation.

$\frac{1}{5} \ln|x| - \frac{1}{5} \ln|5-x| = \frac{1}{10}t + C$

4. Use the initial condition to find the constant of integration.

Given $x = 1$ when $t = 0$ (noting that $x>0$ and $5-x>0$, we can drop the absolute values):
$\frac{1}{5} \ln 1 - \frac{1}{5} \ln(5-1) = \frac{1}{10}(0) + C$
$C = -\frac{1}{5} \ln 4$

5. Rearrange the equation to make $x$ the subject.

$\frac{1}{5} \ln \left( \frac{x}{5-x} \right) = \frac{1}{10}t - \frac{1}{5} \ln 4$
Multiply by 5:
$\ln \left( \frac{x}{5-x} \right) = \frac{1}{2}t - \ln 4$
$\frac{x}{5-x} = e^{\frac{1}{2}t - \ln 4} = \frac{1}{4}e^{\frac{1}{2}t}$
$4x = 5e^{\frac{1}{2}t} - xe^{\frac{1}{2}t}$
$x(4 + e^{\frac{1}{2}t}) = 5e^{\frac{1}{2}t}$
$x = \frac{5e^{\frac{1}{2}t}}{4 + e^{\frac{1}{2}t}}$

1. Solve part (i) concisely to obtain the relationship between $x$ and $t$ (dependency).

Since part (i) is a prerequisite, we first find $x$ in terms of $t$:
$\int \frac{1}{x(5-x)} \,dx = \int \frac{1}{10} \,dt$
Using partial fractions:
$\frac{1}{5} \int \left( \frac{1}{x} + \frac{1}{5-x} \right) \,dx = \frac{1}{10} t + C$
$\ln|x| - \ln|5-x| = \frac{1}{2}t + 5C$
Given $x=1$ when $t=0$:
$\ln 1 - \ln 4 = 0 + 5C \implies 5C = -\ln 4$
$\ln\left(\frac{x}{5-x}\right) = \frac{1}{2}t - \ln 4$
$\frac{x}{5-x} = \frac{1}{4}e^{0.5t}$
$4x = 5e^{0.5t} - x e^{0.5t} \implies x = \frac{5e^{0.5t}}{4+e^{0.5t}}$

2. Interpret the population value and solve for exact $t$.

The population is $x$ hundred, so a population of 200 wild boars corresponds to $x = 2$.
Substitute $x = 2$ into the intermediate equation (or the final expression for $x$):
$\ln\left(\frac{2}{5-2}\right) = \frac{1}{2}t - \ln 4$
$\ln\left(\frac{2}{3}\right) + \ln 4 = \frac{1}{2}t$
$\frac{1}{2}t = \ln\left(\frac{8}{3}\right)$
$t = 2 \ln\left(\frac{8}{3}\right)$

1. Evaluate the limit of $x$ as $t \to \infty$ using the result from (i)

From part (i), the solution is typically expressed in the form $x = \frac{5}{1 + 4e^{-0.5t}}$.
As $t \to \infty$, the term $e^{-0.5t} \to 0$.
Therefore, $x \to \frac{5}{1 + 0} = 5$.

2. Interpret the mathematical limit in the context of the problem

Since $x$ represents the population in hundreds, a value of $x = 5$ corresponds to $5 \times 100 = 500$.
Hence, the population will approach 500 over a long time.