IGCSE OCR Additional Maths: Differential Equations – Exam Tips | IGCSE OCR 附加数学：微分方程考点精讲

📚 IGCSE OCR Additional Maths: Differential Equations – Exam Tips | IGCSE OCR 附加数学：微分方程考点精讲

Differential equations link derivatives to functions and are a key topic in the OCR IGCSE Additional Mathematics (FSMQ 6993) syllabus. You need to be able to form, solve, and interpret first‑order differential equations, especially in contexts like growth, decay, and Newton’s law of cooling. This guide walks you through every essential skill, from separation of variables to applying initial conditions.

微分方程将导数与函数联系起来，是OCR IGCSE附加数学（FSMQ 6993）考纲中的重点。你需要会建立、求解并解释一阶微分方程，尤其要掌握增长、衰减和牛顿冷却定律等背景。本文带你过一遍所有关键技能，从分离变量法到初始条件的应用全覆盖。

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

A differential equation (DE) contains a derivative, such as dy/dx, and relates a function to its rate of change. For example, dy/dx = 3x is a first‑order differential equation because the highest derivative is the first derivative.

微分方程包含导数（如 dy/dx），将一个函数与其变化率联系起来。例如 dy/dx = 3x 就是一阶微分方程，因为其中最高阶导数是一阶导数。

In the OCR Additional Maths exam, you will only work with first‑order ordinary differential equations. You must be able to recognise equations that can be solved by separating the variables.

在OCR附加数学考试中，你只会遇到一阶常微分方程。你必须能识别出可通过分离变量法求解的方程。

2. Standard Form and Classification | 标准形式与分类

A first‑order DE can often be written in the form dy/dx = f(x)g(y), meaning the right‑hand side factors into a function of x alone and a function of y alone. This is the key requirement for the method of separation of variables.

一阶微分方程通常可写为 dy/dx = f(x)g(y)，即右边可分解为一个仅含 x 的函数与一个仅含 y 的函数的乘积。这是分离变量法的关键前提。

If the equation cannot be factorised in this way, it is not directly soluble by the separation method at this level. Always check for factorisation before proceeding.

如果方程不能这样分解，在这个阶段就不能直接用分离变量法求解。解题前务必先检查可否因式分解。

Equation	Can separate?
dy/dx = 2xy	Yes (2x · y)
dy/dx = x + y	No
dy/dx = (x²+1)/y	Yes ((x²+1) · y⁻¹)

Recognising the correct form saves time and prevents you from using an invalid technique in the exam.

识别正确形式能节省时间，避免在考试中使用不合法的方法。

3. Separation of Variables | 分离变量法

Once you have dy/dx = f(x)g(y), rewrite the equation so that all y‑terms are on the same side as dy and all x‑terms on the same side as dx:

(1/g(y)) dy = f(x) dx

一旦得到 dy/dx = f(x)g(y) 的形式，就把所有含 y 的项移到 dy 同一侧，所有含 x 的项移到 dx 同一侧。

This step must be done carefully. Do not forget to move the g(y) to the denominator on the y‑side. A common error is to write dy = f(x)g(y) dx, which is not fully separated.

这一步要仔细，不要忘记把 g(y) 移到 y 一侧的分母上。常见错误是写成 dy = f(x)g(y) dx，这并没有彻底分离变量。

After separation, both sides are ready for integration. No dx or dy should appear on the wrong side.

分离后两边就可以积分了，任何 dx 或 dy 都不应出现在错误的一侧。

4. Integrating to Find the General Solution | 积分求通解

With variables separated, integrate both sides independently:

∫ (1/g(y)) dy = ∫ f(x) dx

变量分离后，分别对两边积分。

Always include the constant of integration ‘+C’ on one side only (conventionally the x‑side). The constant is essential because it represents the family of all solution curves.

始终只在一侧加上积分常数 ‘+C’（习惯上加在 x 侧）。这个常数必不可少，因为它代表一族解曲线。

For example, solve dy/dx = 2xy. Separating gives (1/y) dy = 2x dx. Integrating yields ln|y| = x² + C, which gives the general solution y = A e^(x²), where A = ±e^C.

例如解 dy/dx = 2xy。分离得 (1/y) dy = 2x dx。积分得 ln|y| = x² + C，通解为 y = A e^(x²)，其中 A = ±e^C。

Here the result is written using a single arbitrary constant A to simplify the expression.

这里用一个任意常数 A 来简化表达式。

5. Initial Conditions and Particular Solutions | 初始条件与特解

An initial condition gives a specific point (x₀, y₀) that the solution must satisfy. Substitute these values into the general solution to find the value of the constant A (or C).

初始条件会给出一个特定点 (x₀, y₀)，解必须满足该点。将这些值代入通解，求出常数 A（或 C）的具体值。

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

例如 y = A e^(x²)，已知 x = 0 时 y = 3，则 3 = A e⁰ ⇒ A = 3，特解为 y = 3 e^(x²)。

In word problems, the initial condition often corresponds to ‘at time t = 0’. Read the context carefully to extract the correct pair of values.

在应用题中，初始条件常对应“在 t = 0 时刻”。仔细阅读语境，提取正确的值对。

Without the initial condition you can only leave the answer in general form. The mark scheme often awards separate marks for finding the constant.

没有初始条件就只能保留通解形式。评分方案通常对求出常数的步骤单独给分。

6. Exponential Growth and Decay Models | 指数增长与衰减模型

A large class of exam questions involves the DE dy/dt = ky, where k is a constant. If k > 0, it models exponential growth; if k < 0, exponential decay.

一大类考题涉及微分方程 dy/dt = ky，其中 k 为常数。若 k > 0，模拟指数增长；若 k < 0，模拟指数衰减。

Separation gives (1/y) dy = k dt, and integration yields ln|y| = kt + C, or equivalently y = y₀ e^(kt), where y₀ is the initial amount.

分离得 (1/y) dy = k dt，积分得 ln|y| = kt + C，或等价地 y = y₀ e^(kt)，y₀ 为初始量。

You must be comfortable interpreting k from data: if a population doubles in 5 years, use the equation 2y₀ = y₀ e^(5k) to find k = (1/5)ln 2.

你必须能从数据中解读 k：如果种群数量在5年内翻倍，利用 2y₀ = y₀ e^(5k) 求出 k = (1/5)ln 2。

Examiners often ask for the value at a given time, the time to reach a certain size, or the rate of change at an instant.

考官常要求计算某一时刻的值、达到某一规模所需的时间或某一瞬间的变化率。

7. Newton’s Law of Cooling | 牛顿冷却定律

Newton’s law states that the rate of change of temperature of an object is proportional to the difference between its temperature T and the ambient temperature Tₐ:

dT/dt = -k (T – Tₐ)

牛顿定律指出，物体温度的变化率与其自身温度 T 和环境温度 Tₐ 的差成正比。

Here the negative sign ensures that if T > Tₐ, the object cools down. During solving, treat T – Tₐ as a single variable: let u = T – Tₐ, then du/dt = dT/dt, and the equation becomes du/dt = -k u, which is the decay model.

这里的负号保证了当 T > Tₐ 时物体降温。求解时把 T – Tₐ 视为一个整体变量：令 u = T – Tₐ，则 du/dt = dT/dt，方程化为 du/dt = -k u，即衰减模型。

Separation and integration give u = u₀ e^(-kt), so T – Tₐ = (T₀ – Tₐ) e^(-kt). Rearrange to T = Tₐ + (T₀ – Tₐ) e^(-kt).

分离积分得 u = u₀ e^(-kt)，从而 T – Tₐ = (T₀ – Tₐ) e^(-kt)，整理得 T = Tₐ + (T₀ – Tₐ) e^(-kt)。

In exams, always start from the given DE, then separate variables. Do not simply quote the final formula unless instructed.

考试中务必从给出的微分方程出发，再分离变量。除非有明确指示，否则不要直接套用最终公式。

8. Rate of Change Application Problems | 变化率应用题

Words like ‘rate’, ‘speed of increase’, or ‘proportional to’ signal that you need a DE. Translate the sentence directly into symbols: e.g., ‘the rate of increase of population P is proportional to P’ becomes dP/dt = kP.

题目中若出现“速率”、“增长速度”、“与…成正比”之类词语，就提示你需要建立微分方程。直接把句子转译成符号：例如“人口 P 的增长率与 P 成正比”写作 dP/dt = kP。

Some problems state the rate is proportional to the square root of the quantity, leading to dy/dt = k√y, which is also separable.

有些题目会说变化率与量的平方根成正比，会得到 dy/dt = k√y，同样可分离。

Always define your variables clearly and note the initial condition provided. A sketch of the situation helps you verify whether k should be positive or negative.

始终清晰定义变量，并记下给出的初始条件。一个情境草图能帮你检验 k 应为正还是负。

A typical question: ‘The rate of decrease of radioactive substance is proportional to the amount remaining.’ This gives dM/dt = -kM.

典型问题：“放射性物质的衰减速率与剩余量成正比。” 这得到 dM/dt = -kM。

9. Common Mistakes and Exam Strategies | 常见错误与解题策略

Forgetting the constant of integration. Always write ‘+C’ immediately after integrating. Losing this constant can cost you two or more marks.

忘记加积分常数。积分后立即写上 ‘+C’。遗漏这个常数会让你至少丢掉两分。
Misplacing dx and dy. Be systematic: separate all y terms with dy, all x terms with dx on the other side.

dx 与 dy 放错位置。要有条理：把含 y 的项与 dy 放在一侧，含 x 的项与 dx 放在另一侧。
Applying the initial condition at the wrong stage. Substitute only after you have the general solution, not before integration.

在错误阶段代入初始条件。只有在得到通解后再代入，不要在积分前代入。
Ignoring absolute values in logarithms. In growth/decay, the quantity is positive, so you can safely drop the absolute value after explaining.

忽略对数中的绝对值。在增长/衰减模型中，量为正，因此你可以在解释后安全地去掉绝对值符号。

Examiners look for method marks: showing the separation, correct integration, and handling of the constant. Even if the final answer is wrong, clear working can secure most of the marks.

考官看重步骤分：展示变量分离、正确积分、处理常数。即便最终答案有误，清晰的步骤也能保住大部分分数。

10. Worked Example | 典型例题分析

Question: A population P satisfies dP/dt = 0.2P, with P = 500 when t = 0. Find P after 10 units of time.

题目： 某种群 P 满足 dP/dt = 0.2P，t = 0 时 P = 500。求 10 单位时间后的 P。

Solution: Separate: (1/P) dP = 0.2 dt. Integrate: ln P = 0.2t + C. Use initial condition: ln 500 = 0 + C ⇒ C = ln 500. So ln P = 0.2t + ln 500. Combine: ln(P/500) = 0.2t ⇒ P/500 = e^(0.2t) ⇒ P = 500 e^(0.2t). At t = 10: P = 500 e² ≈ 500 × 7.389 = 3694.5.

解：分离变量：(1/P) dP = 0.2 dt。积分： ln P = 0.2t + C。代入初始条件： ln 500 = 0 + C ⇒ C = ln 500。因此 ln P = 0.2t + ln 500。合并： ln(P/500) = 0.2t ⇒ P/500 = e^(0.2t) ⇒ P = 500 e^(0.2t)。t = 10 时： P = 500 e² ≈ 500 × 7.389 = 3694.5。

Always present the final answer in exact form if the question asks, but a decimal approximation is often acceptable. Show the substitution step clearly.

如果题目要求，保留精确形式；否则十进制近似值通常也可接受。清晰展示代入步骤。

11. Revision Checklist | 复习检查清单

Can I recognise a differential equation that can be solved by separation?
我能识别出能用分离法求解的微分方程吗？
Do I always include the constant of integration?
我是否总是包含积分常数？
Can I use an initial condition to find a particular solution?
我能利用初始条件求出特解吗？
Can I model exponential growth/decay and Newton’s cooling using DEs?
我能用微分方程建模指数增长/衰减和牛顿冷却吗？
Do I check my final formula by differentiating back?
我会通过求导回代来检验最终公式吗？

Mastering these points will give you confidence to tackle any differential equation question in the OCR IGCSE Additional Maths exam.

掌握这些要点，你就能从容应对OCR IGCSE附加数学考试中任意一道微分方程题。

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IGCSE OCR Additional Maths: Differential Equations – Exam Tips | IGCSE OCR 附加数学：微分方程考点精讲