已知 $z = e^{2x}\frac{dy}{dx}$，用 $x$、$\frac{dy}{dx}$ 和 $\frac{d^2y}{dx^2}$ 表示 $\frac{dz}{dx}$。由此证明微分方程 $\frac{d^2y}{dx^2} + 2\frac{dy}{dx} = e^{-1-4x}$ 可以简化为 $\frac{dz}{dx} = e^{-1-2x}.$ 解此微分方程，用 $x$ 表示 $y$。 一位鸟舍饲养员在受控环境中启动了一项昆虫繁殖计划，初始昆虫数量为 5000 只，用于喂养鸟舍里的鸟类。在繁殖计划开始后的 $t$ 天，昆虫的出生率是昆虫数量 $I$（以千只为单位）的 $\tfrac{1}{25}$。同时，昆虫也以每天 0.25（以千只为单位）的恒定速率被喂给鸟类。 证明昆虫的数量可以用以下微分方程建模 $\frac{dI}{dt} = 0.04(I - 6.25).$ 由此求解该微分方程，得到 $I$ 关于 $t$ 的特解。 通过绘制解曲线的草图，证明该昆虫繁殖计划中的昆虫在一段时间后有可能会被消耗殆尽。这将在经过多少个完整的整天后发生？

1. Find an expression for $\frac{dz}{dx}$ by differentiating the given substitution with respect to $x$.

$z = e^{2x}\frac{dy}{dx}$
Using the product rule,
$\frac{dz}{dx} = e^{2x} \frac{d}{dx}\left(\frac{dy}{dx}\right) + \frac{dy}{dx} \frac{d}{dx}\left(e^{2x}\right)$
$\frac{dz}{dx} = e^{2x}\frac{d^2y}{dx^2} + 2e^{2x}\frac{dy}{dx}$

2. Show that the differential equation reduces to the required form.

Given the differential equation:
$\frac{d^2y}{dx^2} + 2\frac{dy}{dx} = e^{-1-4x}$
Multiply both sides by $e^{2x}$:
$e^{2x}\left(\frac{d^2y}{dx^2} + 2\frac{dy}{dx}\right) = e^{2x}(e^{-1-4x})$
$e^{2x}\frac{d^2y}{dx^2} + 2e^{2x}\frac{dy}{dx} = e^{-1-2x}$
Since the left-hand side is exactly $\frac{dz}{dx}$ from Step 1:
$\frac{dz}{dx} = e^{-1-2x} \quad \text{(Shown)}$

3. Solve for $z$ by integrating.

$\frac{dz}{dx} = e^{-1-2x}$
Integrating both sides with respect to $x$:
$z = \int e^{-1-2x} \, dx$
$z = -\frac{1}{2}e^{-1-2x} + A$
where $A$ is an arbitrary constant.

4. Substitute $z$ back to express $\frac{dy}{dx}$ in terms of $x$.

Substitute $z = e^{2x}\frac{dy}{dx}$ back into the equation:
$e^{2x}\frac{dy}{dx} = -\frac{1}{2}e^{-1-2x} + A$
Divide both sides by $e^{2x}$ (or multiply by $e^{-2x}$):
$\frac{dy}{dx} = -\frac{1}{2}e^{-1-4x} + Ae^{-2x}$

5. Solve for $y$ by integrating $\frac{dy}{dx}$.

$y = \int \left( -\frac{1}{2}e^{-1-4x} + Ae^{-2x} \right) \, dx$
$y = -\frac{1}{2}\left(\frac{e^{-1-4x}}{-4}\right) + A\left(\frac{e^{-2x}}{-2}\right) + B$
$y = \frac{1}{8}e^{-1-4x} - \frac{A}{2}e^{-2x} + B$
Letting $C_1 = -\frac{A}{2}$ and $C_2 = B$, we obtain the general solution:
$y = \frac{1}{8}e^{-1-4x} + C_1e^{-2x} + C_2$

1. Express the net rate of change of the insect population as the birth rate minus the consumption rate.

$\frac{dI}{dt} = \text{Rate of birth} - \text{Rate of feeding}$
$\frac{dI}{dt} = \frac{1}{25}I - 0.25$

2. Convert the fractional rate to a decimal and factorise to obtain the required form.

$\frac{dI}{dt} = 0.04I - 0.25$
$\frac{dI}{dt} = 0.04\left(I - \frac{0.25}{0.04}\right)$
$\frac{dI}{dt} = 0.04(I - 6.25) \quad \text{(Shown)}$

1. Separate the variables and integrate both sides.

$\int \frac{1}{I - 6.25} \, dI = \int 0.04 \, dt$
$\ln |I - 6.25| = 0.04t + c$

2. Rewrite the equation in exponential form to remove the logarithm.

$I - 6.25 = \pm e^{0.04t + c}$
$I - 6.25 = A e^{0.04t}$ (where $A = \pm e^c$)

3. Use the initial condition to determine the arbitrary constant $A$.

The initial number of insects is $5000$, so $I = 5$ when $t = 0$.
$5 - 6.25 = A e^{0.04(0)}$
$A = -1.25$

4. Express $I$ as the subject of the formula.

$I - 6.25 = -1.25e^{0.04t}$
$I = 6.25 - 1.25e^{0.04t}$

1. State the particular solution from (b)(ii) and describe its sketch to show depletion.

From (b)(ii), the particular solution is $I = 6.25 - 1.25e^{0.04t}$.

A sketch of $I$ against $t$ for $t \ge 0$ would show the $I$-intercept at $(0, 5)$. Since $e^{0.04t}$ increases as $t$ increases, the curve is strictly decreasing. The curve will eventually intersect the $t$-axis ($I=0$) and become negative. This shows that the insect population will drop to $0$ and be depleted after some time.

2. Set $I = 0$ to find the exact time of depletion and deduce the number of complete days.

To find when the insects are depleted, set $I = 0$:
$6.25 - 1.25e^{0.04t} = 0$
$1.25e^{0.04t} = 6.25$
$e^{0.04t} = 5$
$0.04t = \ln 5$
$t = 25\ln 5 \approx 40.236 \text{ days}$

Since $I > 0$ at $t = 40$ and $I < 0$ at $t = 41$, the population will be fully depleted after $41$ complete days.