Modelling with Differential Equations | 微分方程建模

📚 Modelling with Differential Equations | 微分方程建模

In the real world, quantities rarely stay constant. Populations grow, radioactive substances decay, objects cool down, and pollutants mix into reservoirs. All these changes can be captured using differential equations – equations that involve a function and its derivatives. This topic explores how to construct such equations from worded contexts, how to solve them using separation of variables, and how to interpret the solution within the original scenario. Mastering modelling with differential equations gives you a powerful toolkit for linking pure mathematics to physics, biology, chemistry, and engineering.

在现实世界中，量很少保持不变。人口增长、放射性物质衰变、物体冷却、污染物融入水库——所有这些变化都可以用微分方程来描述，即包含函数及其导数的方程。本主题探讨如何从文字语境中构建这样的方程，如何使用分离变量法求解它们，以及如何在原始情境中解释解。掌握微分方程建模将为你提供一个强大的工具包，将纯数学与物理、生物、化学和工程学联系起来。

1. What is a Differential Equation? | 什么是微分方程？

A differential equation is an equation that relates a function to its derivatives. For example, dy/dx = 3x² is a simple first-order differential equation because it involves the first derivative. The solution to a differential equation is not just a number but a family of functions. When we solve dy/dx = 3x², we obtain y = x³ + C, where C is an arbitrary constant. The order of a differential equation refers to the highest derivative present: dy/dx is first-order, d²y/dx² is second-order. In A-Level modelling, we focus almost exclusively on first-order ordinary differential equations that can be solved by separation of variables.

微分方程是将函数与其导数联系起来的方程。例如，dy/dx = 3x² 是一个简单的一阶微分方程，因为它包含一阶导数。微分方程的解不是一个数值，而是一族函数。当我们求解 dy/dx = 3x² 时，得到 y = x³ + C，其中 C 是任意常数。微分方程的阶数指的是所出现的最高阶导数：dy/dx 是一阶，d²y/dx² 是二阶。在A-Level建模中，我们几乎只关注可以通过分离变量法求解的一阶常微分方程。

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2. Modelling with Rates of Change | 用变化率建模

Many modelling scenarios start with a statement about how quickly something changes. If the rate of change of a quantity P with respect to time t is proportional to the current amount, we write dP/dt ∝ P. Introducing a constant of proportionality k transforms this into the differential equation dP/dt = kP. If P represents population, k > 0 gives growth; if it represents a decaying substance, k < 0 gives decay. The key to modelling is translating phrases like 'the rate of increase is directly proportional to …' into mathematical language using derivatives.

许多建模场景都始于关于某量变化有多快的陈述。如果量 P 关于时间 t 的变化率与当前数量成正比，我们写成 dP/dt ∝ P。引入比例常数 k，将其转化为微分方程 dP/dt = kP。如果 P 代表人口，k > 0 表示增长；如果代表衰变物质，k < 0 表示衰减。建模的关键在于将诸如“增长率与……成正比”的语句用导数翻译成数学语言。

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3. Setting Up Differential Equations | 建立微分方程

To build a differential equation from a real-world problem, follow these steps: (1) Identify the variable and its rate of change, clearly defining all symbols. (2) Express the given proportionality or balance law using derivatives. For instance, the rate of temperature loss of a body is proportional to the difference between its temperature T and the ambient temperature Tₐ, giving dT/dt = −k(T − Tₐ). (3) Include any constant of proportionality, noting whether it is positive or negative. (4) Attach initial conditions such as ‘at t = 0, T = 100°C’. These will be needed later to find a particular solution.

从现实世界问题建立微分方程，请遵循以下步骤：(1) 确定变量及其变化率，明确定义所有符号。(2) 用导数表达给定的比例关系或平衡定律。例如，物体温度下降的速率与自身温度 T 和环境温度 Tₐ 之差成正比，得到 dT/dt = −k(T − Tₐ)。(3) 引入比例常数，注意正负号。(4) 附加初始条件，如“当 t = 0 时，T = 100°C”。这些条件稍后用于求特解。

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4. Separation of Variables | 分离变量法

Once we have a differential equation of the form dy/dx = f(x)g(y), we can solve it by separating the variables. Rewrite the equation so that all y‑terms, including dy, appear on one side and all x‑terms, including dx, on the other:

∫ (1/g(y)) dy = ∫ f(x) dx

Then integrate both sides. Do not forget the constant of integration, usually written as +C on one side. After integration, rearrange to make y the subject if possible. Even if an explicit function cannot be found, the implicit form is acceptable. This method works because the derivative dy/dx can be treated as a ratio of differentials, allowing algebraic manipulation.

一旦我们得到了形如 dy/dx = f(x)g(y) 的微分方程，就可以通过分离变量来求解。重写方程，使所有含 y 的项（包括 dy）位于一边，所有含 x 的项（包括 dx）位于另一边：

∫ (1/g(y)) dy = ∫ f(x) dx

然后两边积分。切勿忘记积分常数，通常在一侧写为 +C。积分后，尽可能解出 y。即使无法求得显函数，隐式形式也是可以接受的。该方法之所以有效，是因为导数 dy/dx 可视为微分之比，从而允许代数操作。

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5. General Solutions and Particular Solutions | 通解与特解

The general solution of a first-order differential equation contains one arbitrary constant, C. If an initial condition is given, say y(x₀) = y₀, we can substitute these values to find C and obtain a particular solution. For example, solving dy/dx = 2xy with y(0)=3 leads first to ln|y| = x² + C, giving y = Ae^(x²) (where A = e^C). Using y(0)=3 gives A = 3, so the particular solution is y = 3e^(x²). Always check that the particular solution satisfies the original differential equation and the initial condition.

一阶微分方程的通解包含一个任意常数 C。如果给定了初始条件，例如 y(x₀) = y₀，我们可以代入这些值求出 C 并获得特解。例如，求解 dy/dx = 2xy 且 y(0)=3，首先得到 ln|y| = x² + C，进而得 y = Ae^(x²)（其中 A = e^C）。利用 y(0)=3 得 A = 3，所以特解为 y = 3e^(x²)。务必检查特解是否满足原微分方程和初始条件。

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6. Exponential Growth Model | 指数增长模型

A fundamental model is exponential growth, governed by dP/dt = kP with k > 0. The solution, using separation of variables, is P = P₀ e^(kt) where P₀ is the initial amount at t = 0. This model applies well to unrestricted population growth, compound interest, and certain biological processes. The doubling time, T_d, is the time taken for P to double, found from e^(kT_d) = 2 ⇒ T_d = ln(2)/k. In exam questions, you may be asked to find k from a given doubling time or to predict future population given a current value and a growth rate.

一个基础模型是指数增长，由 dP/dt = kP（k > 0）控制。利用分离变量法，解为 P = P₀ e^(kt)，其中 P₀ 是 t = 0 时的初始数量。该模型适用于无限制的人口增长、复利和某些生物过程。倍增时间 T_d 是 P 翻倍所需的时间，由 e^(kT_d) = 2 ⇒ T_d = ln(2)/k 求得。在考题中，你可能需要根据给定的倍增时间求出 k，或根据当前值和增长率预测未来人口。

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7. Exponential Decay and Half-Life | 指数衰减与半衰期

When a quantity decreases at a rate proportional to its current value, we have dN/dt = −λN, where λ > 0 is the decay constant. The solution is N = N₀ e^(−λt). This model describes radioactive decay, drug elimination from the bloodstream, and the depreciation of assets. A key concept is half-life, t₁/₂, the time required for N to halve: e^(−λ t₁/₂) = 1/2 ⇒ t₁/₂ = ln(2)/λ. Note that half-life is independent of the initial amount. Typical exam problems involve finding λ from a given half-life or calculating the amount remaining after a certain time.

当某量以与其当前值成正比的速率减少时，我们有 dN/dt = −λN，其中 λ > 0 为衰变常数。其解为 N = N₀ e^(−λt)。该模型描述放射性衰变、血液中药物消除和资产折旧。关键概念是半衰期 t₁/₂，即 N 减半所需的时间：e^(−λ t₁/₂) = 1/2 ⇒ t₁/₂ = ln(2)/λ。注意半衰期与初始量无关。典型考题涉及根据给定半衰期求 λ 或计算一时间后剩余的量。

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8. Newton’s Law of Cooling | 牛顿冷却定律

Newton’s Law of Cooling states that the rate of change of the temperature T of an object is proportional to the difference between its own temperature and the ambient temperature Tₐ (assumed constant). The differential equation is dT/dt = −k(T − Tₐ), where k > 0. Using separation of variables and integrating gives ln|T − Tₐ| = −kt + C, so T = Tₐ + (T₀ − Tₐ)e^(−kt), where T₀ is the initial temperature at t = 0. This model yields an exponential decay of the temperature difference. Questions often provide two temperature readings at different times to determine k and Tₐ, or to forecast the temperature at a later time.

牛顿冷却定律指出，物体温度 T 的变化率与其自身温度和环境温度 Tₐ（假设恒定）之差成正比。微分方程为 dT/dt = −k(T − Tₐ)，其中 k > 0。利用分离变量法并积分，得 ln|T − Tₐ| = −kt + C，故 T = Tₐ + (T₀ − Tₐ)e^(−kt)，T₀ 为 t = 0 时的初始温度。该模型表现为温度差呈指数衰减。题目常提供两个不同时刻的温度读数，以确定 k 和 Tₐ，或预测后续时刻的温度。

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9. Mixing Problems | 混合问题

Mixing problems involve a tank containing a solution with a dissolved substance, where liquid with a certain concentration flows in and the well-stirred mixture flows out at the same rate. Let x(t) be the amount of substance in the tank at time t. The rate of change dx/dt equals the rate in minus the rate out. The rate in is (inflow concentration) × (inflow rate). The rate out is (x(t)/volume) × (outflow rate), since the mixture is uniform. If the volume V remains constant (inflow equals outflow), the differential equation takes the form dx/dt = R_in − (R_out / V) x. This is a linear first-order equation that can be solved by separation of variables or an integrating factor. After solving, you can find the steady-state amount as t → ∞.

混合问题涉及一个含有溶解物质溶液的容器，其中一定浓度的液体流入，同时搅拌均匀后的混合物以相同速率流出。设 x(t) 为 t 时刻容器内物质的量。变化率 dx/dt 等于流入速率减去流出速率。流入速率为（流入浓度）×（流入速率）。由于混合物是均匀的，流出速率为（x(t)/体积）×（流出速率）。如果体积 V 保持不变（流入等于流出），微分方程形式为 dx/dt = R_in − (R_out / V) x。这是一阶线性方程，可用分离变量法或积分因子求解。求解后，可求得 t → ∞ 时的稳态量。

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10. Motion with Velocity-Dependent Resistance | 含速度相关阻力的运动

In mechanics, differential equations arise when the resultant force depends on velocity. For example, a particle falling under gravity with air resistance proportional to its velocity v can be modelled by m (dv/dt) = mg − kv, where k is a positive constant. This can be rearranged to a separable equation: dv/dt = g − (k/m)v. Separation of variables yields the velocity function v(t). As t increases, velocity approaches the terminal velocity v_term = mg/k. Such models are common in A-Level mechanics and provide a bridge between pure mathematics and physics. Knowing how to set up the equation from Newton’s second law is essential.

在力学中，当合力依赖于速度时就会产生微分方程。例如，一个在重力作用下下落且受到与速度成正比的空气阻力的质点，可建模为 m (dv/dt) = mg − kv，其中 k 为正常数。这可以重排为可分离变量的方程：dv/dt = g − (k/m)v。分离变量法给出速度函数 v(t)。随着 t 增大，速度趋近于终端速度 v_term = mg/k。这类模型在A-Level力学中很常见，为纯数学和物理之间架起了桥梁。学会从牛顿第二定律建立方程至关重要。

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11. Interpreting the Solution and Model Limitations | 解的解释与模型局限性

After finding a particular solution, always relate it back to the real-world context. For population growth, an exponential model predicts unlimited growth – but in reality, resources are finite, so the model might only be valid for small populations over short time intervals. In cooling, the model assumes constant ambient temperature and perfect conduction, which may not hold. When a question asks ‘comment on the suitability of the model’, discuss whether the assumptions are realistic and what long-term behaviour is predicted. Understanding limitations helps you write better conclusions and earn evaluation marks.

求出特解后，一定要将其与真实世界语境联系起来。对于人口增长，指数模型预测无限制的增长——但现实中资源有限，因此模型可能仅在短时间内对小种群有效。在冷却问题中，模型假设环境恒定且热传导理想，这或许不成立。当题目要求“评论模型的适宜性”时，应讨论假设是否现实以及模型预测的长期行为。理解局限性有助于写出更好的结论并赢得评价分数。

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12. Exam Tips and Common Pitfalls | 考试技巧与常见错误

– Always begin by stating clearly what the variables represent (e.g. ‘let P be the population at time t hours’).
– When separating variables, be meticulous with algebra: ensure division by g(y) does not introduce division by zero; handle absolute values in logarithms properly if the context requires.
– Remember to include the constant of integration immediately after integrating, not at the end.
– Use initial conditions to find the constant; if two conditions are given, you may need to find two constants (e.g. k and C).
– Check that your final solution has the correct units and behaves as expected (e.g. temperature tends to ambient).
– If the equation is of the form dy/dx = f(x)g(y), always check if g(y) = 0 yields a constant solution that might be missed.

– 开始务必清晰说明变量含义（例如“设 P 为 t 小时后的人口数量”）。
– 分离变量时，在代数上要一丝不苟：确保除以 g(y) 不会引入除以零的情况；如果需要，正确对待对数中的绝对值。
– 积分后立即加上积分常数，而不要等到最后。
– 利用初始条件求出常数；如果给出了两个条件，可能需要求出两个常数（例如 k 和 C）。
– 检查最终解的单位是否正确，且行为是否符合预期（如温度趋近环境温度）。
– 如果方程形如 dy/dx = f(x)g(y)，务必检查 g(y) = 0 是否会给出可能被忽略的常数解。

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