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A-Level Mathematics: Differential Equation Modelling in Action | A-Level 数学：微分方程建模实战

2026-08-30 11:34 · 本文部分内容由AI辅助创作

📚 A-Level Mathematics: Differential Equation Modelling in Action | A-Level 数学：微分方程建模实战

Differential equations are among the most powerful tools in applied mathematics. In A-Level Mathematics, you are often asked to translate a real-world situation into a differential equation, solve it, and then interpret the solution. This article walks through the common modelling contexts and the solving techniques required in exams.

微分方程是应用数学中最强大的工具之一。在 A-Level 数学中，你经常需要将现实情境转化为微分方程，求解并解释结果。本文系统梳理常见建模情境与考试所需的求解技巧，帮助你快速掌握“建模实战”。

1. Expressing Rates in Mathematical Language | 用数学语言表达变化率

The phrase "the rate of change" is the clue to use a derivative. If a quantity y changes with time t , then the expression dydt represents its rate of change. The phrase "proportional to" translates to a constant multiple, so "y increases at a rate proportional to y" becomes dydt = ky . The phrase "inversely proportional" becomes dydt = ky .

遇到“变化率”就要联想到导数。若某个量 y 随时间 t 变化，那么 dy/dt 就表示 y 的变化率。“成正比”对应乘以一个常数，因此“y 的增长速率与 y 成正比”可写作 dy/dt = ky；“成反比”则写作 dy/dt = k/y。

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

The simplest and most important model is the exponential model. It describes populations, radioactive decay, cooling, and continuously compounded interest. If the rate of growth of a quantity P is proportional to P itself, then the differential equation is:

最重要且最基本的模型是指数模型，它适用于人口增长、放射性衰变、冷却以及连续复利等问题。若量 P 的增长速率与 P 本身成正比，则微分方程为：

dP/dt = kP

Separating variables and integrating gives the general solution:

分离变量并积分可得通解：

P(t) = P₀ ekt

k > 0, the quantity grows exponentially; if k < 0, it decays exponentially. In radioactive decay, applications include half-life: the time taken for P to fall to half of its original value.

其中 P₀ 是 t = 0 时的初始值。若 k > 0，该量呈指数增长；若 k < 0，则呈指数衰减。在放射性衰变中，还常涉及半衰期：即 P 下降到初始值一半所需的时间。

3. Limited Growth and Logistic Models | 受限增长与逻辑斯蒂模型

In real populations, resources are limited. A more realistic model is the logistic equation, where the growth rate is proportional both to the current population and to the remaining capacity:

现实种群中资源是有限的。更符合实际的模型是逻辑斯蒂方程，其增长率同时与当前种群数量和剩余容量成正比：

dP/dt = rP(1 - P/K)

The constant K is the carrying capacity. When P is near 0, the growth is almost exponential. When P approaches K, the growth slows down and eventually stops. The general solution is:

常数 K 称为环境容纳量。当 P 接近 0 时，增长近似指数；当 P 接近 K 时，增长放缓并最终停止。其通解为：

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

nd from the initial condition. If P₀ = K/2, then A = 1, giving the classic S-shaped curve.

其中 A 是由初始条件确定的常数。若 P₀ = K/2，则 A = 1，得到经典的 S 形曲线。

Model Differential Equation Solution Form
Exponential dP/dt = kP P = P₀ ekt
Limited Growth dP/dt = k(L - P) P = L - (L - P₀)e-kt
Logistic dP/dt = rP(1 - P/K) P = K / (1 + A e-rt)

4. Newton's Law of Cooling | 牛顿冷却定律

Newton's law of cooling states that the rate of change of an object's temperature T is proportional to the difference between T and the surrounding temperature Tₛ. This gives a linear first-order differential equation:

牛顿冷却定律指出：物体温度 T 的变化率与 T 和周围环境温度 Tₛ 之差成正比，于是得到一阶线性微分方程：

dT/dt = -k(T - Tₛ)

where k > 0. If the object is hotter than its surroundings, then dT/dt is negative, so the object cools. If the object is colder than its surroundings, then dT/dt is positive, so it warms up. The solution is:

其中 k > 0。若物体比周围环境热，则 dT/dt 为负，物体降温；若物体比周围环境冷，则 dT/dt 为正，物体升温。其解为：

T(t) = Tₛ + (T₀ - Tₛ)e-kt

This solution shows that the temperature difference T - Tₛ decays exponentially to zero, so the object gradually reaches thermal equilibrium with the environment.

该解表明温度差 T - Tₛ 按指数规律衰减到 0，因此物体最终与环境达到热平衡。

5. Motion with Resistance | 含阻力的运动问题

Modelling motion often produces differential equations in velocity v. A common situation is a falling object subject to gravity and air resistance proportional to velocity. The acceleration is dv/dt, so Newton's second law gives:

运动建模常会得到关于速度 v 的微分方程。常见情形是物体在下落过程中同时受到重力与正比于速度的阻力。由于加速度为 dv/dt，由牛顿第二定律可得：

dv/dt = g - kv

leration is g. As v increases, the resistance term kv grows. The terminal velocity vₜ occurs when the acceleration is zero:

其中 g 是重力加速度，k 是正的阻力常数。初始时 v = 0，加速度为 g；随着 v 增大，阻力项 kv 增大。当加速度为 0 时达到终端速度 vₜ：

vₜ = g/k

Solving the differential equation with v(0) = 0 gives:

结合 v(0) = 0 求解上述微分方程，可得：

v(t) = (g/k)(1 - e-kt)

This velocity increases toward the terminal velocity but never exceeds it, reflecting the balance between weight and resistance.

该速度逐渐趋近终端速度但不会超过它，体现了重力与阻力之间的平衡。

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6. Mixing and Concentration Problems | 混合与浓度问题

Another common modelling context is the well-mixed tank problem. If Q(t) is the amount of substance dissolved in a tank, the differential equation is built from the principle:

另一个常见建模背景是充分混合的水箱问题。若 Q(t) 表示水箱中溶解物质的总量，则微分方程建立在如下原则上：

dQ/dt = rate of substance entering - rate of substance leaving

2 × 2 = 0.4 kg/min. The leaving rate is the concentration Q/100 times the flow rate 2, so:

例如，设一个水箱中有 100 L 水和 10 kg 盐。浓度为 0.2 kg/L 的盐溶液以 2 L/min 的流速进入，同时充分混合的液体以 2 L/min 的流速离开。进入速率为 0.2 × 2 = 0.4 kg/min；离开速率为浓度 Q/100 乘以流速 2，所以：

dQ/dt = 0.4 - Q/50

variables or the integrating factor method gives the particular solution Q(t) = 20 - 10 e-t/50.

这是一阶线性微分方程。改写为 dQ/dt = (20 - Q)/50 后，使用分离变量法或积分因子法，可得特解 Q(t) = 20 - 10 e-t/50。

7. Separation of Variables: Solving the Equations | 分离变量法：求解微分方程

). The method of separation of variables rewrites the derivative as a fraction-like expression and integrates both sides:

A-Level 中的绝大多数微分方程模型都需要求解形如 dy/dx = f(x)g(y) 的一阶方程。分离变量法将导数改写为类似分式的形式，然后两边积分：

∫ (1/g(y)) dy = ∫ f(x) dx

tial model dP/dt = kP, separate variables to get ∫ (1/P) dP = ∫ k dt. Integrating gives ln|P| = kt + C. Because populations are usually positive, this can be rearranged to P = A ekt, where A = ±eC.

例如，对于指数模型 dP/dt = kP，分离变量得 ∫ (1/P) dP = ∫ k dt。积分得 ln|P| = kt + C。由于种群数量通常为正，可整理为 P = A ekt，其中 A = ±eC。

8. Using Initial Conditions and Particular Solutions | 利用初始条件求特解

Suppose dv/dt = g - kv with v(0) = 0. Separating variables gives ∫ (1/(g - kv)) dv = ∫ dt.

在找到任意常数之前，模型是不完整的。通解中含有一个任意常数，而特解满足给定的初始条件。设 dv/dt = g - kv 且 v(0) = 0。分离变量得 ∫ (1/(g - kv)) dv = ∫ dt。

ating and applying the initial condition yields the particular solution v(t) = (g/k)(1 - e-kt). This is the exact solution used in terminal velocity problems.

积分并代入初始条件后得到特解 v(t) = (g/k)(1 - e-kt)。这正是终端速度问题中所使用的精确解。

mixing example dQ/dt = 0.4 - Q/50 with Q(0) = 10, the general solution is Q(t) = 20 + A e-t/50. Substituting Q(0) = 10 gives A = -10, so Q(t) = 20 - 10 e-t/50.

类似地，对于混合问题 dQ/dt = 0.4 - Q/50，且 Q(0) = 10，通解为 Q(t) = 20 + A e-t/50。代入 Q(0) = 10 得 A = -10，因此 Q(t) = 20 - 10 e-t/50。

9. Equilibrium Points and Stability | 平衡点与稳定性

c equation gives two equilibrium populations: P = 0 and P = K. Setting dT/dt = 0 in Newton's law of cooling gives T = Tₛ, the surrounding temperature.

当变化率为零时对应平衡点。在逻辑斯蒂方程中令 dP/dt = 0，可得两个平衡种群：P = 0 和 P = K；在牛顿冷却定律中令 dT/dt = 0，则得 T = Tₛ，即环境温度。

ow K, the population increases toward K; if P is slightly above K, the population decreases toward K.

在逻辑斯蒂模型中，P = 0 是不稳定平衡点，因为任意微小的正种群都会增长；相反，P = K 是稳定平衡点：若 P 略低于 K，种群会增大并趋于 K；若 P 略高于 K，种群会减小并趋于 K。

In exam questions, the sign of dP/dt or dv/dt is often used to justify the long-term behaviour of a model. Always check whether the equilibrium is approached from above or below.

在考试题中，常利用 dP/dt 或 dv/dt 的符号来讨论模型的长期行为。务必判断平衡点是从上方还是从下方趋近。

10. The Modelling Cycle and Verification | 建模循环与验证

Modelling with differential equations is not just about solving equations. The full cycle involves creating, solving, interpreting, and refining the model. Examiners reward clear communication of this process.

微分方程建模不仅仅是求解方程。完整的循环包括建立模型、求解、解释以及改进模型。考官非常看重对这一过程的清晰表述。

11. Common Pitfalls and Exam Tips | 常见错误与考试技巧

Differential equation questions are a reliable source of marks if you follow a consistent routine. However, small errors can lose several marks quickly. Watch out for the following common pitfalls.

只要遵循固定步骤，微分方程题是稳定得分的来源。但一个小错误可能迅速导致失分。请特别注意以下常见陷阱。