Differential Equations | 微分方程考点精讲

📚 Differential Equations | 微分方程考点精讲

Differential equations are a vital part of calculus, linking a function with its derivatives. They model countless real-world phenomena such as population growth, radioactive decay, cooling rates and motion. In GCSE CIE Mathematics, you are expected to understand how to form, classify, solve and apply simple first‑order differential equations, especially using the method of separation of variables. This guide covers all essential concepts, techniques and common pitfalls to help you master differential equations with confidence.

微分方程是微积分的重要组成部分，它将一个函数与其导数联系起来。微分方程可以模拟众多现实世界现象，如人口增长、放射性衰变、冷却速率和运动。在GCSE CIE数学中，你需要理解如何建立、分类、求解并应用简单的一阶微分方程，特别是使用分离变量法。本指南涵盖所有核心概念、解题技巧和常见误区，帮助你自信掌握微分方程。

1. Introduction to Differential Equations | 微分方程简介

A differential equation is any equation that contains a derivative, such as dy/dx or d²y/dx². It relates an unknown function y(x) to its rates of change. For example, the equation dy/dx = 3x² is a differential equation whose solution is y = x³ + C.

微分方程是任何包含导数（如 dy/dx 或 d²y/dx²）的方程。它将未知函数 y(x) 与其变化率联系起来。例如，方程 dy/dx = 3x² 就是一个微分方程，其解为 y = x³ + C。

At GCSE level, we mainly work with first‑order ordinary differential equations – those containing only the first derivative dy/dx. The goal is to find the original function y = f(x) from information about its derivative.

在GCSE阶段，我们主要研究一阶常微分方程——即仅含一阶导数 dy/dx 的方程。目标是通过导数的信息来找出原函数 y = f(x)。

2. Order and Degree of a Differential Equation | 微分方程的阶与次

The order of a differential equation is the highest derivative that appears in it. For instance, dy/dx = 2x has order 1, while d²y/dx² + 3dy/dx = 0 has order 2. The degree is the power of the highest derivative, provided the equation is a polynomial in derivatives. In GCSE CIE, we only deal with first‑order, first‑degree equations.

微分方程的阶是指方程中出现的最高阶导数的阶数。例如，dy/dx = 2x 的阶为1，而 d²y/dx² + 3dy/dx = 0 的阶为2。次是指最高阶导数的幂次，前提是方程关于导数是多项式。在GCSE CIE中，我们只处理一阶一次方程。

Understanding order and degree helps you choose the correct solving strategy. All equations you encounter will be of the form dy/dx = f(x, y), ready for separation of variables.

理解阶与次有助于选择正确的求解策略。你遇到的所有方程都将是 dy/dx = f(x, y) 的形式，可以直接使用分离变量法。

3. Forming Differential Equations | 建立微分方程

Many exam questions ask you to form a differential equation from a word problem. Look for phrases describing rates of change – for example, ‘the rate of increase of a population is proportional to its current size’ translates to dP/dt = kP. Similarly, ‘the temperature of an object decreases at a rate proportional to the difference between its temperature and the room temperature’ becomes dT/dt = -k(T – Tₐ).

许多考题要求你根据文字问题建立微分方程。关注描述变化率的短语——例如，“一种群的增长速率与其当前规模成正比”转化为 dP/dt = kP。同样，“物体温度下降的速率与其温度和室温之差成正比”变成 dT/dt = -k(T – Tₐ)。

A minus sign is crucial when a quantity is decreasing. Always define your variables clearly: let y be the quantity changing, x be the independent variable (often time t), and k a positive constant of proportionality.

当某个量在减少时，负号至关重要。始终清晰地定义变量：设 y 为变化中的量，x 为自变量（通常是时间 t），k 为正比例常数。

4. Solving Differential Equations: Separation of Variables | 解微分方程：分离变量法

The main technique at GCSE is separating the variables. If a differential equation can be written as dy/dx = g(x)h(y), we rearrange it so that all y‑terms are on one side with dy and all x‑terms on the other with dx. This gives (1/h(y)) dy = g(x) dx. Then we integrate both sides.

GCSE阶段的主要方法是分离变量法。如果一个微分方程可以写成 dy/dx = g(x)h(y) 的形式，我们重新排列，将所有含 y 的项与 dy 放在一边，所有含 x 的项与 dx 放在另一边。得到 (1/h(y)) dy = g(x) dx 后，再对两边积分。

For example, solve dy/dx = xy. We rewrite as (1/y) dy = x dx. Integrating gives ln|y| = ½x² + C. Finally, exponentiate to obtain y = A e^(½x²), where A = ±e^C.

例如，求解 dy/dx = xy。改写为 (1/y) dy = x dx。积分得 ln|y| = ½x² + C。最后取指数得到 y = A e^(½x²)，其中 A = ±e^C。

Method: ∫ (1/y) dy = ∫ x dx → ln|y| = ½x² + C

5. Integrating Both Sides | 两边积分

After separating, you must perform the integration correctly. Always include the constant of integration C on one side only – usually the x‑side – to keep the solution general. Be familiar with basic integrals: ∫ xⁿ dx = xⁿ⁺¹/(n+1) + C, ∫ (1/x) dx = ln|x| + C, ∫ eˣ dx = eˣ + C, and trigonometric integrals like ∫ cos x dx = sin x + C.

在分离变量后，你必须正确进行积分。始终只在一边加上积分常数 C——通常加在 x 一边，以保持解的一般性。熟悉基本积分公式：∫ xⁿ dx = xⁿ⁺¹/(n+1) + C，∫ (1/x) dx = ln|x| + C，∫ eˣ dx = eˣ + C，以及三角函数的积分如 ∫ cos x dx = sin x + C。

When integrating after separation, you may need to use substitution or linear expressions. For instance, ∫ 1/(2y+1) dy = (1/2) ln|2y+1|. Never forget to multiply by the coefficient reciprocal.

分离后积分时，你可能需要换元或处理线性表达式。例如，∫ 1/(2y+1) dy = (1/2) ln|2y+1|。永远不要忘记乘以系数的倒数。

6. General and Particular Solutions | 通解与特解

The general solution of a differential equation contains an arbitrary constant C. It represents a whole family of curves. When you are given an initial condition or boundary condition – for example, y = 2 when x = 0 – you can substitute these values into the general solution and solve for C. The resulting expression is called a particular solution.

微分方程的通解含有一个任意常数 C。它代表了整个曲线族。当你给定初始条件或边界条件——例如，当 x = 0 时 y = 2——你可以将这些值代入通解并解出 C。所得表达式称为特解。

For example, if the general solution is y = A e^(½x²) and we know y = 3 when x = 0, then 3 = A e⁰, so A = 3, giving the particular solution y = 3e^(½x²).

例如，若通解为 y = A e^(½x²)，且已知 x = 0 时 y = 3，则 3 = A e⁰，因此 A = 3，得到特解 y = 3e^(½x²)。

7. Verifying Solutions | 验证解

You can check whether a given function satisfies a differential equation by differentiating it and substituting back. If the original equation holds true for all x, the function is a valid solution. This skill is often tested in multiple‑choice sections.

你可以通过对给定函数求导并代回原方程来检验它是否满足微分方程。如果原方程对所有 x 均成立，则该函数是有效的解。这一技巧常在选择题部分考查。

For example, to verify that y = 4e^(3x) solves dy/dx = 3y, compute dy/dx = 12e^(3x). Substituting gives 12e^(3x) = 3(4e^(3x)) = 12e^(3x), so the identity holds. Always show your derivative and substitution clearly.

例如，要验证 y = 4e^(3x) 是 dy/dx = 3y 的解，计算 dy/dx = 12e^(3x)。代回得 12e^(3x) = 3(4e^(3x)) = 12e^(3x)，恒等式成立。始终清晰地展示你的求导和代换过程。

8. Applications: Growth and Decay | 应用：增长与衰减

Exponential growth and decay are modelled by dy/dt = ky, where k > 0 gives growth and k < 0 gives decay. The general solution is y = y₀ e^(kt), where y₀ is the initial amount at t = 0. This applies to populations, bacteria colonies, radioactive substances and compound interest.

指数增长与衰减由 dy/dt = ky 建模，其中 k > 0 表示增长，k < 0 表示衰减。通解为 y = y₀ e^(kt)，其中 y₀ 是 t = 0 时的初始量。这适用于种群、细菌菌落、放射性物质和复利等情景。

In exams, you may be asked to find the percentage increase after a certain time, or the time needed for a quantity to double (for growth) or halve (for decay). For doubling, set y = 2y₀ and solve e^(kt) = 2 to get t = (ln 2)/k.

在考试中，你可能被要求求出一段时间后的增长百分比，或是某个量翻倍（增长）或减半（衰减）所需的时间。对于翻倍，设 y = 2y₀，解 e^(kt) = 2 得 t = (ln 2)/k。

Doubling time: t = ln 2 / k

9. Applications: Cooling and Heating | 应用：冷却与加热

Newton’s Law of Cooling states that the rate of change of an object’s temperature is proportional to the difference between its temperature T and the ambient temperature Tₐ. This gives dT/dt = -k(T – Tₐ), with k > 0. The solution is T = Tₐ + (T₀ – Tₐ)e^(-kt), where T₀ is the initial temperature.

牛顿冷却定律指出，物体温度的变化速率与其温度 T 和环境温度 Tₐ 之差成正比。这给出 dT/dt = -k(T – Tₐ)，且 k > 0。解为 T = Tₐ + (T₀ – Tₐ)e^(-kt)，其中 T₀ 为初始温度。

Exam questions might ask you to find the temperature after a given time or the time at which the object reaches a certain temperature. Be careful with the sign: the difference T – Tₐ decays exponentially toward zero.

考题可能会要求你求出给定时间后的温度，或物体达到某个温度所需的时间。注意符号：差值 T – Tₐ 随时间指数衰减至零。

Variable	Meaning
T₀	Initial temperature
Tₐ	Ambient (surrounding) temperature
k	Positive cooling constant
t	Time

10. Using Boundary Conditions | 使用边界条件

Boundary conditions or initial conditions allow you to find the specific constant in a solution. Always use the condition right after integration, before you manipulate the expression too much. If you have ln|y| = 2x + C, and y = 5 at x = 0, plug in to get ln 5 = C, then write ln|y| = 2x + ln 5, which simplifies to y = 5 e^(2x).

边界条件或初始条件让你能够求出解中的具体常数。总是尽量在积分后立即使用条件，在过多操作表达式之前。如果你得到 ln|y| = 2x + C，且已知 x = 0 时 y = 5，代入得 ln 5 = C，然后写 ln|y| = 2x + ln 5，化简得 y = 5 e^(2x)。

When the condition involves y and x, substitution at an early stage often avoids messy algebra later. In population problems, t = 0 usually gives the initial population directly.

当条件涉及 y 和 x 时，早期代换通常能避免后期繁琐的代数运算。在种群问题中，t = 0 通常直接给出初始种群。

11. Common Mistakes | 常见错误

One of the most frequent errors is forgetting the constant of integration. Without it, you lose the family of solutions and cannot satisfy an initial condition. Another is mishandling the absolute value after integrating 1/y – always write ln|y| unless you know y is positive. Also, students often misplace the negative sign in decay or cooling equations, leading to unrealistic results.

最常见的错误之一是忘记积分常数。没有了它，你就会失去解族，无法满足初始条件。另一个错误是在积分 1/y 后没有正确使用绝对值——除非你知道 y 为正，否则始终写成 ln|y|。另外，学生常常在衰减或冷却方程中放错负号，导致不切实际的结果。

When separating variables, ensure every term involving y is multiplied by dy, and every term involving x by dx. Do not mix differentials. For instance, dy/dx = y + x cannot be separated directly; it requires a different method not in GCSE.

分离变量时，确保每个含 y 的项都乘以 dy，每个含 x 的项都乘以 dx。不要将微分混淆。例如，dy/dx = y + x 不能直接分离，需要其他方法，不在GCSE范围内。

Do: dy/dx = 3xy → (1/y) dy = 3x dx
Don’t: dy/dx = x + y → cannot be separated like (1/y) dy = x dx

12. Exam Tips | 考试技巧

Always show all steps of separation and integration clearly. Even if you know the final form, examiners award marks for method. When forming differential equations from words, write down the proportionality statement first, then introduce the constant k. State clearly: ‘Let y be … and x be …’ before building the equation.

始终清晰地展示分离和积分的所有步骤。即使你知道最终形式，考官也会根据方法给分。从文字中建立微分方程时，先写下比例关系陈述，再引入常数 k。在建立方程前清晰地说明：“设 y 为……，x 为……”。

Check your final answer against the given condition. If a particular solution is required, substitute back to verify. Pay attention to units when time or other quantities are involved. Practice past CIE questions to become familiar with the phrasing and mark schemes.

用所给条件检查你的最终答案。如果需要求特解，代回验证。当涉及时间或其他量时，注意单位。多练习以往的CIE真题，熟悉题目表述和评分方案。

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Differential Equations | 微分方程考点精讲