Differential Equation Modeling Applications | 微分方程建模应用

📚 Differential Equation Modeling Applications | 微分方程建模应用

Differential equations are among the most powerful tools in applied mathematics. They describe how quantities change continuously over time or space, forming the backbone of models in physics, biology, chemistry, economics, and engineering. In this revision guide, we will explore the core concepts of building and solving differential equations from real-world scenarios, with a focus on the A-Level mathematics syllabus.

微分方程是应用数学中最强大的工具之一。它们描述量如何随时间和空间连续变化，是物理、生物、化学、经济以及工程等领域模型的基石。在本复习指南中，我们将聚焦于A-Level数学考纲，探讨如何从现实情境中建立并求解微分方程。

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1. Why Model with Differential Equations? | 为什么用微分方程建模？

Many real-world processes involve rates of change. A statement such as “the rate of growth of a population is proportional to the current population” is naturally expressed as dy/dt = k y. This is a first-order differential equation. Modeling involves translating a verbal or physical description into a mathematical equation involving derivatives.

许多现实过程都涉及变化率。诸如”人口增长率与当前人口成正比”这样的表述，自然可以写成 dy/dt = k y。这是一个一阶微分方程。建模的过程就是把文字或物理描述翻译成包含导数的数学方程。

Key steps in building a model:

• Identify the independent variable (often time t) and the dependent variable (such as population N or temperature T).
• 独立变量（通常是时间 t）和依赖变量（如种群数量 N 或温度 T）的识别。
• Translate the given rate statement into an equation using derivatives, such as dN/dt.
• 将给定的变化率叙述用导数（如 dN/dt）写成方程。
• Include initial conditions or boundary conditions to make the solution unique.
• 加入初始条件或边界条件，使解具有唯一性。

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2. Separable First-Order Equations | 可分离变量的一阶方程

The simplest type of differential equation in modeling is the separable equation, which can be written in the form dy/dx = f(x) g(y). To solve it, we separate the variables: (1/g(y)) dy = f(x) dx, and then integrate both sides.

建模中最简单的微分方程类型是可分离变量方程，可写成 dy/dx = f(x) g(y) 的形式。解法是把变量分离： (1/g(y)) dy = f(x) dx，然后两边积分。

For example, consider the differential equation dy/dx = x y². Separating variables gives (1/y²) dy = x dx. Integrating both sides yields -1/y = x²/2 + C, so y = -2/(x² + 2C). This technique is essential for solving exponential growth and decay problems.

例如，考虑微分方程 dy/dx = x y²。分离变量得到 (1/y²) dy = x dx。两边积分得到 -1/y = x²/2 + C，因此 y = -2/(x² + 2C)。该技巧对求解指数增长与衰减问题至关重要。

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3. Exponential Growth and Decay | 指数增长与衰减

The classic model for exponential growth or decay is dN/dt = k N, where k is a constant. When k > 0, the quantity N increases with time; when k < 0, it decreases. The general solution is N(t) = N₀ e^{kt}, where N₀ is the initial value at t = 0.

经典指数增长或衰减模型是 dN/dt = k N，其中 k 为常数。当 k > 0 时，量 N 随时间增加；当 k < 0 时，它随时间减小。通解为 N(t) = N₀ e^{kt}，其中 N₀ 是 t = 0 时的初始值。

Applications include:

• Population growth of bacteria or animals under unlimited resources.
• 资源无限条件下细菌或动物种群的增长。
• Radioactive decay, where the half-life is given by t₁/₂ = ln 2 / λ.
• 放射性衰变，其半衰期为 t₁/₂ = ln 2 / λ。
• Continuous compound interest in finance: A = P e^{rt}.
• 金融中的连续复利： A = P e^{rt}。

dN/dt = k N → N(t) = N₀ e^{kt}

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4. Newton’s Law of Cooling | 牛顿冷却定律

Newton’s law of cooling states that the rate of change of temperature T of an object is proportional to the difference between T and the ambient temperature Tₐ. That is, dT/dt = -k (T – Tₐ), where k > 0. The negative sign indicates that hot objects cool down and cold objects warm up.

牛顿冷却定律指出，物体温度 T 的变化率与 T 和环境温度 Tₐ 之差成正比，即 dT/dt = -k (T – Tₐ)，其中 k > 0。负号表示热的物体会冷却，冷的物体会升温。

This equation is separable. Rewriting as dT/(T – Tₐ) = -k dt and integrating gives T – Tₐ = C e^{-kt}. Applying an initial condition T(0) = T₀ yields:

该方程是可分离的。改写为 dT/(T – Tₐ) = -k dt 并积分，得到 T – Tₐ = C e^{-kt}。利用初始条件 T(0) = T₀ 可得：

T(t) = Tₐ + (T₀ – Tₐ) e^{-kt}

This model is used to estimate the time of death in forensics, to design cooling systems, and to determine the optimal temperature for food storage.

该模型用于法医学中推测死亡时间、设计冷却系统，以及确定食品储存的最佳温度。

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5. Logistic Growth Model | 逻辑斯蒂增长模型

In reality, resources are limited. A population cannot grow exponentially forever. The logistic growth model modifies the exponential model by introducing a carrying capacity K, the maximum population the environment can sustain. The differential equation is:

在现实中资源是有限的。种群不可能永远指数增长。逻辑斯蒂增长模型通过引入环境容纳量 K（即环境能维持的最大种群数量）来修正指数模型。微分方程为：

dP/dt = r P (1 – P/K)

Here r is the intrinsic growth rate. When P is small, the factor (1 – P/K) is close to 1, so growth is nearly exponential. As P approaches K, the growth rate slows down to zero.

其中 r 为内禀增长率。当 P 很小时，(1 – P/K) 因子接近 1，因此增长近似指数。当 P 接近 K 时，增长率减慢并趋于零。

Solving this separable equation involves partial fractions. The solution is often written in the form:

求解这个可分离方程需要用到部分分式。解通常写成如下形式：

P(t) = K / (1 + A e^{-rt})

where A = (K – P₀)/P₀, with P₀ being the initial population. This S-shaped curve is typical for populations in nature, such as fish in a lake or bacteria in a petri dish.

其中 A = (K – P₀)/P₀，P₀ 是初始种群数量。这条S形曲线是自然界种群（如湖中鱼类或培养皿中的细菌）的典型特征。

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6. Mixing Problems | 混合问题

Mixing problems are a classic type of differential equation modeling. A tank contains a solution of salt and water. A mixture of a certain concentration flows in at a fixed rate, while the well-stirred mixture flows out at a fixed rate. We track the amount of salt Q(t) in the tank.

混合问题是微分方程建模中的经典类型。一个水箱装有盐和水的溶液。一定浓度的混合液以固定速率流入，同时充分搅拌的混合液以固定速率流出。我们跟踪水箱中盐的量 Q(t)。

The rate of change of salt is:

盐量的变化率为：

dQ/dt = (rate of salt in) – (rate of salt out)

Suppose pure water flows in at 2 L/min and the mixture flows out at 2 L/min. If the tank initially has 50 L of water with 10 kg of salt, then the concentration at time t is Q/50 kg/L. The outflow rate of salt is (2)(Q/50) = Q/25 kg/min. Hence dQ/dt = -Q/25.

假设纯水以2升/分钟流入，混合液以2升/分钟流出。若水箱初始有50升水和10千克盐，则 t 时刻的浓度为 Q/50 千克/升。盐的流出速率为 (2)(Q/50) = Q/25 千克/分钟。因此 dQ/dt = -Q/25。

The solution is Q(t) = 10 e^{-t/25}. After 25 minutes, half of the salt remains. This model is used in chemistry, environmental engineering, and pharmacology for drug dosage calculations.

解为 Q(t) = 10 e^{-t/25}。25分钟后，剩余的盐为原来的一半。该模型用于化学、环境工程以及药理学中的药物剂量计算。

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7. Motion with Resistance | 含阻力的运动

In mechanics, differential equations model motion under resistive forces. A classic example is a falling object subject to air resistance proportional to velocity: m dv/dt = mg – k v, where m is mass, g is acceleration due to gravity, and k is the drag coefficient.

在力学中，微分方程模拟物体在阻力作用下的运动。一个经典例子是物体在正比于速度的空气阻力下下落： m dv/dt = mg – k v，其中 m 是质量，g 是重力加速度，k 是阻力系数。

This equation is separable. Writing dv/(mg – k v) = dt/m and integrating yields the velocity as a function of time. As t → ∞, the velocity approaches the terminal velocity v_T = mg/k, where the gravitational force balances the drag force.

该方程是可分离的。将方程写为 dv/(mg – k v) = dt/m 并积分，得到速度作为时间的函数。当 t → ∞ 时，速度趋近于收尾速度 v_T = mg/k，此时重力与阻力平衡。

v(t) = (mg/k) (1 – e^{-kt/m})

This model is applied to skydiving, vehicle braking, and the motion of particles in viscous fluids.

该模型应用于跳伞、车辆制动以及粘性流体中微粒的运动。

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8. Electrical Circuits | 电路模型

In an RL circuit (a resistor and inductor in series), Kirchhoff’s voltage law gives the differential equation: L di/dt + R i = V, where L is inductance, R is resistance, V is the applied voltage, and i is the current.

在RL电路（电阻和电感串联）中，基尔霍夫电压定律给出微分方程： L di/dt + R i = V，其中 L 是电感，R 是电阻，V 是外加电压，i 是电流。

For a constant voltage V, the solution is:

对于恒定电压 V，解为：

i(t) = (V/R)(1 – e^{-Rt/L})

The current rises from 0 to its steady-state value V/R. The time constant L/R determines how quickly the current reaches the steady state. A similar equation governs the charging of a capacitor in an RC circuit.

电流从0上升到稳态值 V/R。时间常数 L/R 决定了电流达到稳态的速度。类似的方程也适用于RC电路中电容器的充电过程。

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9. Solving Technique: Integrating Factor | 求解技巧：积分因子

Some first-order differential equations are not separable but can be solved using an integrating factor. For an equation in the form dy/dx + P(x) y = Q(x), the integrating factor is:

某些一阶微分方程不可分离变量，但可以用积分因子求解。对于形如 dy/dx + P(x) y = Q(x) 的方程，积分因子为：

I(x) = e^{∫ P(x) dx}

Multiplying both sides by I(x) transforms the left-hand side into the derivative of I(x) y. Then we integrate both sides to find y.

将方程两边乘以 I(x)，左端就变成 I(x) y 的导数。然后两边积分即可求出 y。

For example, the equation dy/dx + 2x y = x has P(x) = 2x. The integrating factor is I = e^{x²}. Multiplying gives e^{x²} dy/dx + 2x e^{x²} y = x e^{x²}, which is d/dx (y e^{x²}) = x e^{x²}. Integration yields y = 1/2 + C e^{-x²}.

例如，方程 dy/dx + 2x y = x 中 P(x) = 2x。积分因子为 I = e^{x²}。两边乘以该因子得到 e^{x²} dy/dx + 2x e^{x²} y = x e^{x²}，即 d/dx (y e^{x²}) = x e^{x²}。积分得 y = 1/2 + C e^{-x²}。

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10. Numerical Methods: Euler’s Method | 数值方法：欧拉法

When a differential equation cannot be solved analytically, we can approximate the solution numerically. Euler’s method uses the idea that the derivative gives the slope of the tangent line. Starting from an initial point (x₀, y₀), we step forward by a small step size h:

当微分方程无法解析求解时，我们可以用数值方法逼近解。欧拉法利用导数是切线斜率的思想。从初始点 (x₀, y₀) 出发，以较小的步长 h 向前推进：

y_{n+1} = y_n + h f(x_n,

Published by TutorHao | Mathematics Revision Series | aleveler.com

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