植物学家正在研究过度砍伐如何影响东南亚某国森林中树木的数量。$t$ 年后森林中的树木数量记为 $x$（以千为单位）。提出的一种简单模型是，树木数量的增加速率与其数量大小成正比。已知树木以每年 3 千棵的恒定速率被砍伐，且当 $x = p$ 时，树木数量保持不变。

1. Formulate a differential equation with an unknown constant of proportionality.

Let the rate of change of the population with respect to time be $\frac{\mathrm{d}x}{\mathrm{d}t}$ and the constant of proportionality be $k$.
The net rate of change is the difference between the growth rate ($kx$) and the rate at which trees are chopped off (3 thousand/year):
$\frac{\mathrm{d}x}{\mathrm{d}t} = kx - 3$

2. Use the given steady-state condition to find the proportionality constant $k$.

The population stays constant when $x = p$, which means the rate of change $\frac{\mathrm{d}x}{\mathrm{d}t} = 0$.
Substituting these values into the equation:
$0 = kp - 3$
$k = \frac{3}{p}$

3. State the final differential equation.

Substitute $k = \frac{3}{p}$ back into the initial equation:
$\frac{\mathrm{d}x}{\mathrm{d}t} = \frac{3}{p}x - 3$
Alternatively, this can be factored as:
$\frac{\mathrm{d}x}{\mathrm{d}t} = \frac{3}{p}(x - p)$

1. Establish the differential equation from part (i).

From the model, the rate of increase is proportional to the population size minus a constant chopping rate: $\frac{\mathrm{d}x}{\mathrm{d}t} = kx - 3$
Since the population stays constant when $x = p$, $\frac{\mathrm{d}x}{\mathrm{d}t} = 0$.
$0 = kp - 3 \implies k = \frac{3}{p}$
Thus, $\frac{\mathrm{d}x}{\mathrm{d}t} = \frac{3}{p}x - 3 = \frac{3}{p}(x - p)$

2. Separate variables to prepare for integration.

$\int \frac{1}{x - p} \, \mathrm{d}x = \int \frac{3}{p} \, \mathrm{d}t$

3. Integrate both sides to find the general solution.

$\ln |x - p| = \frac{3}{p}t + c$

4. Express the equation in exponential form.

$|x - p| = \mathrm{e}^{\frac{3}{p}t + c} = \mathrm{e}^c \cdot \mathrm{e}^{\frac{3}{p}t}$
$x - p = A\mathrm{e}^{\frac{3}{p}t}$ (where $A = \pm \mathrm{e}^c$)

5. Apply the initial condition to find $A$ and rearrange to show the given expression.

Given $t = 0$ when $x = 5$:
$5 - p = A\mathrm{e}^{0} \implies A = 5 - p$
Substitute $A$ back into the equation:
$x - p = (5 - p)\mathrm{e}^{\frac{3}{p}t}$
$x = (5 - p)\mathrm{e}^{\frac{3}{p}t} + p$ (Shown)

1. State the inequality for $p$ that results in a decreasing tree population.

For the population $x(t) = (5 - p)\mathrm{e}^{\frac{3}{p}t} + p$ to decrease over time, the term involving $t$ must be decreasing.
Since $p > 0$ (a constant population size is strictly positive), the exponential function $\mathrm{e}^{\frac{3}{p}t}$ is increasing.
Thus, its coefficient must be negative for the overall function to decrease:
$5 - p < 0 \implies p > 5$

2. Find the coordinates of the $t$-intercept algebraically.

At the $t$-axis intercept, $x = 0$.
$0 = (5 - p)\mathrm{e}^{\frac{3}{p}t} + p$
$(p - 5)\mathrm{e}^{\frac{3}{p}t} = p$
$\mathrm{e}^{\frac{3}{p}t} = \frac{p}{p - 5}$
$\frac{3}{p}t = \ln\left(\frac{p}{p - 5}\right)$
$t = \frac{p}{3}\ln\left(\frac{p}{p - 5}\right)$
The coordinates are $\left(\frac{p}{3}\ln\left(\frac{p}{p-5}\right), 0\right)$.

3. Describe the key features for the graph of $x$ against $t$.

The graph is a concave-downwards exponential curve starting at the vertical axis intercept $(0, 5)$ and crossing the horizontal axis at the computed intercept.

Graph sketching details:

• Vertical axis labelled $x$, horizontal axis labelled $t$.
• Point $(0, 5)$ marked on the $x$-axis.
• Point $\left(\frac{p}{3}\ln\left(\frac{p}{p-5}\right), 0\right)$ marked on the $t$-axis.
• A smooth curve passing through both points, steadily decreasing from left to right.

4. Explain the meaning of the axial intercept.

In the context of the model, $x$ represents the population in thousands and $t$ represents time in years. The $t$-axial intercept represents the time taken for the population of trees in the forest to reach exactly zero (the point of extinction).