Differential Equations for GCSE OCR Maths | GCSE OCR 数学：微分方程考点精讲

📚 Differential Equations for GCSE OCR Maths | GCSE OCR 数学：微分方程考点精讲

Differential equations may sound like advanced university material, but at GCSE OCR Higher tier they are simply equations that link a function to its rate of change. This guide walks you through the essentials: recognising a differential equation, reversing differentiation, handling constants of integration, and applying these ideas to motion and exponential models. By the end, you will be confident with every question style OCR can throw at you.

微分方程听起来像是大学进阶内容，但在 GCSE OCR 高等层级中，它们只是将函数与其变化率联系起来的方程。本指南带你掌握所有要点：识别微分方程、反向求导、处理积分常数，并将这些思路应用于运动学和指数模型。学完本文，你将能自信应对 OCR 可能出现的每一种题型。

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

A differential equation involves an unknown function and at least one of its derivatives. The simplest form you will see at GCSE is dy/dx = some expression in x, or occasionally dy/dx = ky, where k is a constant. The goal is to find the original function y.

微分方程包含一个未知函数及其至少一个导数。你在 GCSE 中会见到的最简形式是 dy/dx = 某个含 x 的表达式，或者偶尔见到 dy/dx = ky（k 为常数）。目的是求出原来的函数 y。

Think of it as a puzzle: you are told how y changes with respect to x, and you must reconstruct the curve itself. This is the reverse of taking a gradient function.

可以把它看作一个谜题：你被告知 y 如何随 x 变化，你必须重建曲线本身。这与求梯度函数的过程正好相反。

2. Revisiting Differentiation | 回顾微分

To solve differential equations, you must be fluent with differentiation. For a power function y = xⁿ, the derivative is dy/dx = nxⁿ⁻¹. You also need to know that the derivative of eᐟ is eᐟ, and the derivative of a constant is zero.

要解微分方程，你必须熟练掌握微分。对于幂函数 y = xⁿ，导数为 dy/dx = nxⁿ⁻¹。你还需要知道 eᐟ 的导数是 eᐟ，常数的导数为 0。

If you are given y = x³ + 5x, then dy/dx = 3x² + 5. This direction—from function to gradient—is everyday differentiation. Differential equations give you the derivative and ask you to go backwards.

若给定 y = x³ + 5x，则 dy/dx = 3x² + 5。这个方向——从函数到梯度——是常规微分。而微分方程给你导数，让你反方向求解。

3. From Derivative to Function: Integration | 从导数到函数：积分

Reversing differentiation is called integration. The integral of xⁿ (where n ≠ -1) is (xⁿ⁺¹)/(n+1). For example, integrating 4x³ gives x⁴, because the derivative of x⁴ is 4x³.

微分的逆运算称为积分。xⁿ 的积分（n ≠ -1）为 (xⁿ⁺¹)/(n+1)。例如，对 4x³ 积分得到 x⁴，因为 x⁴ 的导数是 4x³。

Integration always introduces an unknown constant, since any constant differentiates to zero. This is why the solution to a differential equation contains a ‘+ C’.

积分总会引入一个未知常数，因为任何常数求导后为 0。这就是微分方程的解中会出现 ‘+ C’ 的原因。

4. Solving dy/dx = f(x) | 求解 dy/dx = f(x)

When the derivative is given purely as a function of x, the solution is simply the indefinite integral. For instance, if dy/dx = 6x² – 2, then y = 2x³ – 2x + C. You integrate each term individually and add the constant.

当导数仅作为 x 的函数给定时，解就是简单的不定积分。例如，若 dy/dx = 6x² – 2，则 y = 2x³ – 2x + C。逐项积分并加上常数。

Learn to write your solution with an integral sign as an intermediate step: y = ∫ (6x² – 2) dx. This helps prevent errors when multiple terms are present.

学会将积分符号作为中间步骤写出：y = ∫ (6x² – 2) dx。当存在多个项时，这有助于避免出错。

Always check your answer by differentiating back; if you recover the original dy/dx, your integration is correct.

务必通过再微分来检验答案；若能还原出原 dy/dx，则积分正确。

5. The Constant of Integration | 积分常数

The +C represents an infinite family of curves, all having the same gradient function. Geometrically, these are vertical translations of one another. For a given dy/dx, every y = F(x) + C is a solution.

+C 表示一族无穷多的曲线，它们具有相同的梯度函数。从几何上看，它们互相是垂直平移。对给定的 dy/dx，每一个 y = F(x) + C 都是一个解。

Without extra information, you cannot determine C. This is why differential equations alone give a ‘general solution’. A particular solution requires a known point on the curve.

没有额外信息，你无法确定 C。这就是为什么仅凭微分方程只能给出“通解”。特解需要曲线上一个已知点。

6. Using Initial Conditions | 使用初始条件

To find the specific curve that passes through a point, substitute the coordinates into the general solution to solve for C. For example, given dy/dx = 2x and the point (1,4), integrate to get y = x² + C, then substitute x = 1, y = 4 to obtain 4 = 1 + C → C = 3, so y = x² + 3.

要找出经过某点的特定曲线，将坐标代入通解中解出 C。例如，给定 dy/dx = 2x 和点 (1,4)，积分得 y = x² + C，代入 x=1, y=4 得 4=1+C → C=3，故 y = x² + 3。

OCR exam questions frequently give a boundary condition or a point that the solution curve must satisfy. Always label your working clearly and state the particular solution.

OCR 考题经常给出一个边界条件或解曲线必须满足的点。清晰标出你的解题步骤并写出特解。

7. Exponential Growth and Decay | 指数增长与衰减 (dy/dx = ky)

One special type of differential equation at GCSE is dy/dx = ky. The general solution is y = Aeᵏᐟ, where A is a constant determined by initial conditions. This models exponential growth if k > 0, and exponential decay if k < 0.

GCSE 中一种特殊的微分方程是 dy/dx = ky。通解为 y = Aeᵏᐟ，其中 A 是由初始条件确定的常数。若 k > 0，建模指数增长；若 k < 0，建模指数衰减。

You can verify this by differentiating y = Aeᵏᐟ: dy/dx = kAeᵏᐟ = ky. Often OCR questions ask you to recognise that the rate of change is proportional to the quantity itself.

你只需对 y = Aeᵏᐟ 求导即可验证：dy/dx = kAeᵏᐟ = ky。OCR 考题常要求你识别出变化率与量本身成正比。

To find A, use an initial value. For example, if the population P satisfies dP/dt = 0.05P and P = 200 when t = 0, then P = 200e⁰·⁰⁵ᵗ.

要找到 A，利用初值。例如，若种群 P 满足 dP/dt = 0.05P 且 t=0 时 P=200，则 P = 200e⁰·⁰⁵ᵗ。

8. Motion in a Straight Line | 直线运动

In kinematics, velocity v is the derivative of displacement s with respect to time t: v = ds/dt. Acceleration a is the derivative of velocity: a = dv/dt. Differential equations can link these quantities.

在运动学中，速度 v 是位移 s 对时间 t 的导数：v = ds/dt。加速度 a 是速度的导数：a = dv/dt。微分方程可将这些量联系起来。

You may be given ds/dt = 3t² – 4 and asked to find the displacement after 2 seconds given that s = 0 at t = 0. Integrate: s = t³ – 4t + C. Using s(0)=0 gives C=0, so s(2) = 8 – 8 = 0.

你可能遇到 ds/dt = 3t² – 4，要求 t=0 时 s=0，求 t=2 秒后的位移。积分：s = t³ – 4t + C。利用 s(0)=0 得 C=0，所以 s(2)=8-8=0。

Another common scenario: a particle’s velocity is proportional to the distance travelled. This leads to a differential equation dv/dt = kv, which you solve using the exponential idea above.

另一种常见场景：粒子速度与已行驶距离成正比。这会导出微分方程 dv/dt = kv，用上述指数方法求解。

9. Geometrical Interpretation: Slope Fields | 几何解释：斜率场

A differential equation defines a slope at every point in the plane. By drawing short line segments with the gradient given by dy/dx, you can visualise the family of solution curves without solving the equation. At GCSE, you may be asked to sketch a particular solution through a given point on such a diagram.

微分方程定义了平面上每一点处的斜率。通过在每点绘制与 dy/dx 给出的梯度一致的短线，你可将解曲线族可视化而无需真正求解方程。在 GCSE 中，可能要求你在这种图上画出经过已知点的特解曲线。

For dy/dx = x + y, at the point (0,1) the slope is 0+1=1. You can plot a short line there with gradient 1. This can help confirm your algebraic solution makes sense.

对于 dy/dx = x + y，在点 (0,1) 处的斜率为 0+1=1。你可以在此画一条梯度为 1 的短线。这有助于验证你的代数解是否合理。

10. Common Mistakes and Exam Tips | 常见错误与考试技巧

Forgetting the +C is the most frequent error. Even if the question does not explicitly ask for a particular solution, you must include the constant in the general solution. OCR will deduct marks for its omission.

忘记 +C 是最常见的错误。即使题目未明确要求特解，通解中也必须包含常数。OCR 会因漏写而扣分。

Another pitfall is integrating incorrectly when the derivative involves powers. Double-check your reverse power rule: increase the exponent by 1 and divide by the new exponent.

另一个陷阱是当导数含幂函数时积分出错的。反复检查你的逆幂法则：指数加 1，再除以新指数。

When solving dy/dx = ky, many candidates mistakenly write y = eᵏᐟ instead of y = Aeᵏᐟ. Always insert the constant and then use initial conditions to find A.

在解 dy/dx = ky 时，很多考生错误地写成 y = eᵏᐟ 而不是 y = Aeᵏᐟ。务必先写出常数，再利用初始条件求出 A。

In application questions, match your variables carefully. If you use s for displacement and t for time, keep the notation consistent and state your final answer with correct units.

在应用题中，仔细匹配变量。如果用 s 表示位移、t 表示时间，保持符号一致，并写明最终答案的正确单位。

Finally, always verify your solution by differentiating it and substituting back. It takes seconds and guarantees the mark.

最后，一定要通过求导并代入回原来检验你的解。只需几秒，却能确保得分。

Published by TutorHao | GCSE OCR Maths Revision Series | aleveler.com

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