已知 $z = e^{2x}\frac{dy}{dx}$，用 $x$、$\frac{dy}{dx}$ 和 $\frac{d^{2}y}{dx^{2}}$ 表示 $\frac{dz}{dx}$。 由此证明微分方程 $\displaystyle \frac{d^{2}y}{dx^{2}} + 2\frac{dy}{dx} = e^{-1-4x}$ 可以化简为 $\displaystyle \frac{dz}{dx} = e^{-1-2x}$。 求解该微分方程，用 $x$ 表示 $y$。 一个大鸟笼的饲养员在控制环境中启动了一个昆虫繁殖计划，初始昆虫数量为 5000 只，用于喂养鸟笼里的鸟。在繁殖计划开始后 $t$ 天，昆虫的出生率为昆虫数量 $I$（以千只为单位）的 $\tfrac{1}{25}$。同时，这些昆虫也在以每天 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微分方程 $\displaystyle \frac{dI}{dt} = 0.04(I - 6.25).$ 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，证明昆虫繁殖计划中的昆虫在一段时间后可能会被消耗殆尽。这将在经过多少个完整的整天后发生？

1. Differentiate $z$ with respect to $x$ using the product rule.

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

2. Show the reduction of the given differential equation.

Given the differential equation:
$\frac{d^2y}{dx^2} + 2\frac{dy}{dx} = e^{-1-4x}$
Multiply both sides by $e^{2x}$:
$e^{2x}\left( \frac{d^2y}{dx^2} + 2\frac{dy}{dx} \right) = e^{2x} e^{-1-4x}$
Substitute the expression for $\frac{dz}{dx}$ derived in the previous step:
$\frac{dz}{dx} = e^{-1-2x} \text{ (Shown)}$

3. Integrate to find $z$, remembering the constant of integration.

$z = \int e^{-1-2x} \, dx$
$z = -\frac{1}{2}e^{-1-2x} + C$

4. Substitute back $z = e^{2x}\frac{dy}{dx}$ and make $\frac{dy}{dx}$ the subject.

$e^{2x}\frac{dy}{dx} = -\frac{1}{2}e^{-1-2x} + C$
Divide by $e^{2x}$ (or multiply by $e^{-2x}$):
$\frac{dy}{dx} = -\frac{1}{2}e^{-1-2x} \cdot e^{-2x} + C e^{-2x}$
$\frac{dy}{dx} = -\frac{1}{2}e^{-1-4x} + Ce^{-2x}$

5. Integrate $\frac{dy}{dx}$ to find the general solution for $y$.

$y = \int \left( -\frac{1}{2}e^{-1-4x} + Ce^{-2x} \right) dx$
$y = -\frac{1}{2} \left( \frac{e^{-1-4x}}{-4} \right) + C \left( \frac{e^{-2x}}{-2} \right) + D$
$y = \frac{1}{8}e^{-1-4x} - \frac{C}{2}e^{-2x} + D$
Letting $A = -\frac{C}{2}$ and $B = D$, we get:
$y = \frac{1}{8}e^{-1-4x} + Ae^{-2x} + B$

1. Separate the variables and integrate both sides.

$\int \frac{1}{I - 6.25} \, \mathrm{d}I = \int 0.04 \, \mathrm{d}t$
$\ln |I - 6.25| = 0.04t + C$

2. Convert the logarithmic equation to exponential form to simplify finding the constant.

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

3. Apply the initial condition to find the constant.

Given the initial number of insects is 5000, and $I$ is in thousands, we have $I = 5$ when $t = 0$.
$5 - 6.25 = A e^{0.04(0)}$
$A = -1.25$

4. Express $I$ explicitly in terms of $t$.

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

1. Identify the particular solution from part (b)(i) and sketch its curve to show the possibility of depletion.

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

A sketch of $I$ against $t$ would show:

• A starting $I$-intercept at $(0, 5)$.
• A strictly decreasing, concave-downwards curve.
• The curve crossing the horizontal $t$-axis at some $t > 0$.

Because the sketch shows the solution curve intersecting the $t$-axis (where $I = 0$), it demonstrates that the insect population will drop to zero, meaning depletion is possible.

2. Calculate the exact time when the population reaches zero.

Set $I = 0$ to find the time of depletion:
$6.25 - 1.25e^{0.04t} = 0$
$1.25e^{0.04t} = 6.25$
$e^{0.04t} = 5$
$0.04t = \ln 5$
$t = 25\ln 5 \approx 40.2359$

3. Determine the number of complete days before depletion happens.

Since depletion occurs at $t \approx 40.2$ days, the population drops to zero during the 41st day.
Therefore, this will happen after 40 complete days.