证明通过代换 $y = ux^2$，微分方程 $\frac{\text{d}y}{\text{d}x} - \frac{2y}{x} = x^3$ 可以化简为 $\frac{\text{d}u}{\text{d}x} = x$。由此求出 $y$ 关于 $x$ 的通解。 科学家们正在通过监测实验开始后 $t$ 天时的细菌数量 $N$ 来研究某种细菌的生长情况。 在其中一个模型中，可以假设在时刻 $t$，生长速率与当时的细菌数量成正比。在 $t = 0$ 时，细菌数量为 5000，且初始生长速率为每天 200 个。写出该模型的微分方程，并求解以预测实验开始 50 天后的细菌数量。 经过进一步考虑，科学家们决定使用一种 冈珀茨模型（Gompertz model）来模拟细菌种群的增长，该模型表示为：
$\frac{\text{d}N}{\text{d}t} = kN(\ln M - \ln N),$
其中 $k$ 和 $M$ 是正常数。 求该微分方程的通解，并结合题目背景给出 $M$ 的实际意义。

1. Differentiate the substitution $y = ux^2$ with respect to $x$.

Using the product rule:
$\frac{\text{d}y}{\text{d}x} = x^2 \frac{\text{d}u}{\text{d}x} + u(2x)$

2. Substitute $y$ and $\frac{\text{d}y}{\text{d}x}$ into the original differential equation and simplify to show the required form.

Substitute into $\frac{\text{d}y}{\text{d}x} - \frac{2y}{x} = x^3$:
$x^2 \frac{\text{d}u}{\text{d}x} + 2ux - \frac{2(ux^2)}{x} = x^3$
$x^2 \frac{\text{d}u}{\text{d}x} + 2ux - 2ux = x^3$
$x^2 \frac{\text{d}u}{\text{d}x} = x^3$
Dividing both sides by $x^2$ (assuming $x \neq 0$):
$\frac{\text{d}u}{\text{d}x} = x \text{ (shown)}$

3. Solve the reduced differential equation to find $u$.

Integrate both sides with respect to $x$:
$\int \text{d}u = \int x \, \text{d}x$
$u = \frac{1}{2}x^2 + C$
where $C$ is an arbitrary constant.

4. Substitute back to find the general solution for $y$ in terms of $x$.

Since $y = ux^2$, substitute the expression for $u$:
$y = x^2 \left( \frac{1}{2}x^2 + C \right)$
$y = \frac{1}{2}x^4 + Cx^2$

1. Formulate the differential equation using the proportional relationship and initial conditions.

Since the growth rate is proportional to the number of bacteria present, we can write:
$\frac{\text{d}N}{\text{d}t} = kN$
At $t = 0$, $N = 5000$ and $\frac{\text{d}N}{\text{d}t} = 200$.
$200 = k(5000)$
$k = \frac{200}{5000} = 0.04$
Thus, the differential equation is $\frac{\text{d}N}{\text{d}t} = 0.04N$.

2. Solve the differential equation by separating the variables.

$\int \frac{1}{N} \,\text{d}N = \int 0.04 \,\text{d}t$
$\ln |N| = 0.04t + C$
Since $N > 0$ in this context, $\ln N = 0.04t + C$.

3. Apply the initial condition to find the particular solution.

Substitute $t = 0$ and $N = 5000$:
$\ln 5000 = 0.04(0) + C$
$C = \ln 5000$
Therefore, $\ln N = 0.04t + \ln 5000$
$\ln N - \ln 5000 = 0.04t$
$\ln\left(\frac{N}{5000}\right) = 0.04t$
$N = 5000e^{0.04t}$

4. Evaluate the model at $t = 50$ to obtain the projection.

When $t = 50$:
$N = 5000e^{0.04(50)}$
$N = 5000e^2$
$N \approx 36945.28$
The projected number of bacteria is $36945$ (to the nearest whole number).

1. Separate the variables to set up the differential equation for integration.

$\frac{\text{d}N}{\text{d}t} = kN(\ln M - \ln N)$
$\int \frac{1}{N(\ln M - \ln N)} \, \text{d}N = \int k \, \text{d}t$

2. Integrate both sides using the standard logarithmic form.

Notice that the numerator is related to the derivative of the denominator:
$-\int \frac{-\frac{1}{N}}{\ln M - \ln N} \, \text{d}N = \int k \, \text{d}t$
$-\ln|\ln M - \ln N| = kt + C$

3. Rearrange to make $N$ the subject to obtain the general solution.

$|\ln M - \ln N| = \text{e}^{-kt - C}$
$\ln M - \ln N = \pm \text{e}^{-C} \text{e}^{-kt}$
Let $A = \pm \text{e}^{-C}$, which is an arbitrary constant.
$\ln M - \ln N = A\text{e}^{-kt}$
$\ln N = \ln M - A\text{e}^{-kt}$
$N = \text{e}^{\ln M - A\text{e}^{-kt}} = M\text{e}^{-A\text{e}^{-kt}}$

4. Interpret the constant $M$ in the context of the bacteria population.

As $t \to \infty$, since $k$ is a positive constant, $\text{e}^{-kt} \to 0$.
$\ln N \to \ln M - 0 \implies N \to M$
Therefore, $M$ represents the limiting maximum population (or carrying capacity) of the bacteria as time approaches infinity.