📚 Differential Equations of the Form dy/dx = f(x) | 形如 dy/dx = f(x) 的微分方程
Differential equations are equations that involve derivatives and describe how a quantity changes with respect to another variable. In AQA A-Level Mathematics, the simplest class of differential equations has the form dy/dx = f(x), where the derivative is given explicitly as a function of x alone. Solving such an equation is a matter of reversing differentiation through integration.
微分方程是包含导数的方程，描述一个量相对于另一个变量的变化规律。在 AQA A-Level 数学中，最简单的一类微分方程具有 dy/dx = f(x) 的形式，即导数被显式地表示为仅含 x 的函数。求解这类方程的核心就是通过积分来逆转微分过程。
1. What Is a Differential Equation? | 什么是微分方程？
A differential equation is any equation that contains a derivative such as dy/dx, d²y/dx², or dV/dt. The order of a differential equation is the highest derivative present. For example, dy/dx = 3x² is a first-order differential equation, while d²y/dx² = 6x is second-order. The solution of a differential equation is a function (or family of functions) that satisfies the equation for all values of the independent variable.
微分方程是包含导数（如 dy/dx、d²y/dx² 或 dV/dt）的方程。微分方程的阶是其中出现的最高阶导数。例如，dy/dx = 3x² 是一阶微分方程，而 d²y/dx² = 6x 是二阶微分方程。微分方程的解是一个函数（或一族函数），它在自变量的所有取值上都满足该方程。
In this article, we focus on equations of the form dy/dx = f(x). Here the right-hand side depends only on x, not on y. This makes the equation directly integrable without any substitution or separation technique. You simply integrate both sides with respect to x to obtain y.
本文重点讨论形如 dy/dx = f(x) 的方程。在这里，等号右边只依赖于 x，而不依赖于 y。这使得方程可以直接积分而不需要任何换元或分离变量的技巧。你只需对两边关于 x 积分即可得到 y。
2. Solving dy/dx = f(x) by Direct Integration | 通过直接积分求解 dy/dx = f(x)
To solve dy/dx = f(x), integrate both sides with respect to x:
要求解 dy/dx = f(x)，对两边关于 x 积分：
y = ∫ f(x) dx
This gives y = F(x) + c, where F(x) is any antiderivative of f(x) and c is an arbitrary constant of integration. The presence of c reflects the fact that infinitely many curves share the same gradient function; shifting a curve vertically preserves its slope at every point.
这样得到 y = F(x) + c，其中 F(x) 是 f(x) 的任意一个原函数，c 是任意积分常数。c 的存在反映了这样一个事实：无穷多条曲线具有相同的斜率函数；将一条曲线垂直平移不会改变其在每一点的斜率。
Example 1: Solve dy/dx = 4x³ + 2x − 1.
例 1：求解 dy/dx = 4x³ + 2x − 1。
Integrating term by term:
逐项积分：
y = ∫ (4x³ + 2x − 1) dx = x⁴ + x² − x + c
Always add the constant of integration unless you are given an initial condition. Forgetting the + c is one of the most common errors in this topic.
除非题目给出了初始条件，否则务必加上积分常数。漏掉 + c 是本主题中最常见的错误之一。
3. General vs Particular Solutions | 通解与特解
The solution y = F(x) + c is called the general solution because it contains an arbitrary constant and represents a whole family of curves. When additional information — such as a point on the curve — is supplied, you can determine c and obtain a particular solution. The extra condition is often called an initial condition or boundary condition.
解 y = F(x) + c 被称为通解，因为它包含任意常数并代表一族曲线。当题目额外提供信息（例如曲线上的一个点）时，你可以确定 c 的值并得到特解。这个额外条件通常称为初始条件或边界条件。
Example 2: Given dy/dx = 6x² − 4 and y = 7 when x = 1, find y in terms of x.
例 2：已知 dy/dx = 6x² − 4，且当 x = 1 时 y = 7，求 y 关于 x 的表达式。
First integrate: y = ∫ (6x² − 4) dx = 2x³ − 4x + c.
先积分：y = ∫ (6x² − 4) dx = 2x³ − 4x + c。
Substitute x = 1, y = 7: 7 = 2(1)³ − 4(1) + c = 2 − 4 + c = c − 2, so c = 9.
代入 x = 1、y = 7：7 = 2(1)³ − 4(1) + c = 2 − 4 + c = c − 2，因此 c = 9。
y = 2x³ − 4x + 9
Notice that the particular solution is unique: it is the one member of the family of curves that passes through the given point (1, 7).
注意特解是唯一的：它是曲线族中恰好通过给定点 (1, 7) 的那一条曲线。
4. Finding the Constant of Integration | 求积分常数
In examinations, the initial condition may be written in different ways. You might see “y = 5 when x = 0”, or “the curve passes through the point (2, −1)”, or “y(0) = 5”. All of these mean the same thing: substitute the given values into the general solution to solve for c.
在考试中，初始条件可能有不同的表述方式。你可能会看到 “y = 5 when x = 0″（当 x = 0 时 y = 5），或 “the curve passes through the point (2, −1)”（曲线经过点 (2, −1)），或 “y(0) = 5″。这些表达的含义相同：将给定值代入通解以解出 c。
Example 3: The gradient of a curve is given by dy/dx = 3√x − 5/x², and the curve passes through (1, 2). Find its equation.
例 3：某曲线的斜率为 dy/dx = 3√x − 5/x²，且曲线过点 (1, 2)。求其方程。
Rewrite the terms as powers before integrating:
在积分前先将各项改写为幂的形式：
dy/dx = 3x1/2 − 5x⁻²
Integrating gives y = 3 × (2/3)x3/2 − 5 × (−1) x⁻¹ + c = 2x3/2 + 5/x + c.
积分得到 y = 3 × (2/3)x3/2 − 5 × (−1) x⁻¹ + c = 2x3/2 + 5/x + c。
Using x = 1, y = 2: 2 = 2(1) + 5(1) + c = 7 + c, hence c = −5.
代入 x = 1、y = 2：2 = 2(1) + 5(1) + c = 7 + c，因此 c = −5。
y = 2x3/2 + 5/x − 5
Always check the domain: because x⁻² appears, x ≠ 0, so the solution is valid for x > 0.
始终检查定义域：因为出现了 x⁻²，所以 x ≠ 0，因此该解在 x > 0 时有效。
5. Equations of the Form d²y/dx² = f(x) | 形如 d²y/dx² = f(x) 的方程
Second-order differential equations of the form d²y/dx² = f(x) are solved by integrating twice. Each integration introduces its own constant, so the general solution contains two arbitrary constants, usually denoted A and B or c₁ and c₂.
形如 d²y/dx² = f(x) 的二阶微分方程通过积分两次来求解。每次积分都会引入一个常数，因此通解包含两个任意常数，通常记作 A 和 B 或 c₁ 和 c₂。
Example 4: Solve d²y/dx² = 12x + 2, given that y = 4 and dy/dx = 6 when x = 0.
例 4：求解 d²y/dx² = 12x + 2，已知当 x = 0 时 y = 4 且 dy/dx = 6。
First integration:
第一次积分：
dy/dx = ∫ (12x + 2) dx = 6x² + 2x + c₁
Using dy/dx = 6 when x = 0 gives c₁ = 6, so dy/dx = 6x² + 2x + 6.
利用 x = 0 时 dy/dx = 6，得到 c₁ = 6，所以 dy/dx = 6x² + 2x + 6。
Second integration:
第二次积分：
y = ∫ (6x² + 2x + 6) dx = 2x³ + x² + 6x + c₂
Using y = 4 when x = 0 gives c₂ = 4. Therefore the particular solution is y = 2x³ + x² + 6x + 4.
利用 x = 0 时 y = 4，得到 c₂ = 4。因此特解为 y = 2x³ + x² + 6x + 4。
In kinematics, this structure appears naturally: acceleration a = d²s/dt², velocity v = ds/dt, and displacement s. Given an acceleration function, you integrate once to find velocity and twice to find displacement, using initial values of v and s to fix the constants.
在运动学中，这种结构自然出现：加速度 a = d²s/dt²，速度 v = ds/dt，位移 s。给定加速度函数后，积分一次求速度，积分两次求位移，并用 v 和 s 的初始值来确定常数。
6. Modelling with Differential Equations | 微分方程建模
Many real-world problems describe a rate of change directly. For instance, the rate at which a population grows, the rate at which a liquid cools, or the rate at which a car’s displacement changes (its velocity) can all be represented as differential equations. The phrase “rate of change” translates mathematically to a derivative.
许多现实问题直接描述变化率。例如，人口增长的速度、液体冷却的速度或汽车位移变化的速度（即速度）都可以用微分方程表示。”变化率”这一说法在数学上就是导数。
Example 5: The rate of change of the radius r cm of a circular ripple on a pond, with respect to time t seconds, is given by dr/dt = 4/(t + 1)². Given r = 2 when t = 0, find r in terms of t.
例 5：池塘中圆形涟漪的半径 r 厘米关于时间 t 秒的变化率为 dr/dt = 4/(t + 1)²。已知当 t = 0 时 r = 2，求 r 关于 t 的表达式。
Integrating both sides with respect to t:
对两边关于 t 积分：
r = ∫ 4(t + 1)⁻² dt = 4 × (t + 1)⁻¹ / (−1) + c = −4/(t + 1) + c
Using t = 0, r = 2: 2 = −4/1 + c = c − 4, so c = 6.
代入 t = 0、r = 2：2 = −4/1 + c = c − 4，因此 c = 6。
r = 6 − 4/(t + 1)
This model shows that the radius approaches 6 cm as t becomes large, but never exceeds it — a meaningful physical interpretation.
该模型表明，随着 t 增大，半径趋近于 6 厘米但永远不会超过它——这是一个有意义的物理解释。
7. Rates of Change and Connected Rates | 变化率与关联变化率
In some problems, the rate given is not dy/dx directly, but a related rate such as dV/dt or dA/dt. The chain rule then connects these rates. If you know dV/dt and you want dV/dr, you use dV/dt = dV/dr × dr/dt. This is a common source of confusion, so always identify which variables are involved and which rate is being sought.
在有些问题中，给出的变化率并不是 dy/dx 本身，而是相关的速率，如 dV/dt 或 dA/dt。此时链式法则将这些速率联系起来。如果你知道 dV/dt 而要求 dV/dr，则使用 dV/dt = dV/dr × dr/dt。这是常见的困惑来源，因此务必明确涉及哪些变量以及题目要求哪个变化率。
Example 6: A spherical balloon is being inflated so that its volume increases at a constant rate of 8 cm³/s. Find the rate at which the radius is increasing when the radius is 2 cm. The volume of a sphere is V = (4/3)πr³.
例 6：一个球形气球被充气，其体积以 8 cm³/s 的恒定速率增加。求当半径为 2 cm 时半径的增长速率。球的体积为 V = (4/3)πr³。
Differentiate V with respect to r: dV/dr = 4πr². By the chain rule, dV/dt = dV/dr × dr/dt.
对 V 关于 r 求导：dV/dr = 4πr²。根据链式法则，dV/dt = dV/dr × dr/dt。
Substituting dV/dt = 8 and r = 2: 8 = 4π(2)² × dr/dt = 16π × dr/dt.
代入 dV/dt = 8 和 r = 2：8 = 4π(2)² × dr/dt = 16π × dr/dt。
dr/dt = 8 / (16π) = 1 / (2π) cm/s
Solving such problems requires careful formal rearrangement, and the chain rule must be applied rather than simply dividing by dV/dr.
解决此类问题需要仔细的形式化变换，必须运用链式法则，而不是简单除以 dV/dr。
8. Exponential Differential Equations dy/dx = ky | 指数微分方程 dy/dx = ky
Although this article focuses on dy/dx = f(x), AQA also examines the special case dy/dx = ky, where the rate of change is proportional to y itself. This equation models exponential growth and decay, such as radioactive decay, cooling, and continuously compounded interest. Its general solution is y = Aekx.
虽然本文重点讨论 dy/dx = f(x)，但 AQA 还考查特殊情况 dy/dx = ky，即变化率与 y 本身成正比。此方程模拟指数增长与衰减，如放射性衰变、冷却和连续复利。其通解为 y = Aekx。
Note that dy/dx = ky is not of the form f(x) because the right-hand side contains y. To solve it, you separate the variables:
注意 dy/dx = ky 不属于 f(x) 的形式，因为等号右边含有 y。要求解它，你需要分离变量：
∫ (1/y) dy = ∫ k dx
This gives ln|y| = kx + c, and exponentiating yields y = Aekx, where A = ±ec.
由此得到 ln|y| = kx + c，取指数后得到 y = Aekx，其中 A = ±ec。
Example 7: The rate of decay of a radioactive substance is such that dN/dt = −0.02N, where N is the number of atoms and t is time in years. Given N = 1000 initially, find N after 10 years.
例 7：某放射性物质的衰变速率为 dN/dt = −0.02N，其中 N 为原子数，t 为以年为单位的时间。已知初始 N = 1000，求 10 年后的 N。
Separating variables and integrating gives N = Ae−0.02t. Since N = 1000 when t = 0, A = 1000, so N = 1000e−0.02t.
分离变量并积分得到 N = Ae−0.02t。由于 t = 0 时 N = 1000，所以 A = 1000，即 N = 1000e−0.02t。
N(10) = 1000e−0.2 ≈ 818.7 atoms
This exponential form is central to many A-Level modelling questions and may appear alongside the direct-integration type.
这种指数形式在许多 A-Level 建模题中处于核心地位，并可能与直接积分类型同时出现。
9. Common Exam Pitfalls | 常见考试陷阱
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Forgetting the constant of integration after the first (or second) integration. Always write + c after every indefinite integral.
在第一次（或第二次）积分后忘记写积分常数。每次不定积分后务必写上 + c。
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Not simplifying powers before integrating. Terms like 1/x² must be rewritten as x⁻² before applying the power rule.
积分前没有先将幂化简。像 1/x² 这样的项必须先改写为 x⁻²，然后才能套用幂法则。
-
Using incorrect initial conditions for second-order equations: one condition fixes the first constant and the other fixes the second; do not swap them.
对二阶方程使用了错误的初始条件：一个条件确定第一个常数，另一个条件确定第二个常数；不能将二者颠倒。
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Confusing ds/dt, dv/dt and d²s/dt² in kinematics. Write down which quantity each derivative represents before starting.
在运动学中混淆 ds/dt、dv/dt 和 d²s/dt²。动笔前先写清楚每一个导数代表哪个物理量。
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Dropping absolute values when integrating 1/y; although final answers may not show them, the intermediate step ln|y| is mathematically correct.
对 1/y 积分时丢掉绝对值符号；虽然最终答案可能不出现它们，但中间步骤 ln|y| 在数学上是严谨的。
10. Worked Exam-Style Example | 考试风格例题精讲
Question: A curve is such that dy/dx = 3x² − 2x + 1. Given that the curve passes through the point (1, 5), find:
题目：某曲线满足 dy/dx = 3x² − 2x + 1。已知曲线经过点 (1, 5)，求：
(a) the equation of the curve; (b) the gradient of the curve at x = 2; (c) the x-coordinates of any stationary points.
(a) 曲线的方程；(b) 曲线在 x = 2 处的斜率；(c) 任何驻点的 x 坐标。
Solution (a): Integrate to find the general solution.
解 (a)：积分求通解。
y = ∫ (3x² − 2x + 1) dx = x³ − x² + x + c
Substitute the point (1, 5): 5 = 1 − 1 + 1 + c = 1 + c, so c = 4. The curve is y = x³ − x² + x + 4.
代入点 (1, 5)：5 = 1 − 1 + 1 + c = 1 + c，因此 c = 4。曲线为 y = x³ − x² + x + 4。
Solution (b): Substitute x = 2 into dy/dx: dy/dx = 3(2)² − 2(2) + 1 = 12 − 4 + 1 = 9.
解 (b)：将 x = 2 代入 dy/dx：dy/dx = 3(2)² − 2(2) + 1 = 12 − 4 + 1 = 9。
Solution (c): Stationary points occur where dy/dx = 0, i.e. 3x² − 2x + 1 = 0. The discriminant is (−2)² − 4(3)(1) = 4 − 12 = −8 < 0, so the quadratic has no real roots. Hence the curve has no stationary points.
解 (c)：驻点出现在 dy/dx = 0 处，即 3x² − 2x + 1 = 0。判别式为 (−2)² − 4(3)(1) = 4 − 12 = −8 < 0，因此该二次方程无实根。所以曲线没有驻点。
This example illustrates how a single differential equation can generate multiple types of exam questions: finding the equation, evaluating a gradient, and analysing stationary points.
此例说明一个微分方程可以衍生多种类型的考题：求方程、求斜率以及分析驻点。
11. Strategy for Examination Success | 考试成功策略
When tackling differential equation questions in the AQA exam, adopt a systematic routine. First, identify the form of the equation. If dy/dx is expressed purely in terms of x, integrate directly. If it involves y, consider whether separation of variables applies. Second, rewrite all terms as powers of x so that the power rule of integration can be applied without errors. Third, apply the initial conditions in the correct order when dealing with second derivatives. Finally, check your answer by differentiating: the derivative of your solution should match the original f(x).
在 AQA 考试中处理微分方程题目时，应采取系统的步骤。首先，判断方程的形式。如果 dy/dx 仅用 x 表示，则直接积分。如果含有 y，则考虑是否适用分离变量法。其次，将所有项改写为 x 的幂，以便无误地套用积分幂法则。第三，处理二阶导数时按正确顺序代入初始条件。最后，通过求导来检验答案：你的解的导数应当与原 f(x) 一致。
Time management matters. A direct-integration question typically carries 4–6 marks: one mark for integrating correctly, one for including the constant, one for substituting the condition, and one for the final simplified expression. Ensure you show each step clearly to secure all method marks even if an arithmetic slip occurs.
时间管理也很重要。直接积分类题目通常占 4–6 分：积分正确得 1 分，写上常数得 1 分，代入条件得 1 分，最终简化表达式得 1 分。即使出现计算失误，只要清晰写出每一步，也能获得相应的方法分。
12. Summary and Key Formulas | 总结与关键公式
The table below summarises the essential results for this topic. Keep it handy when revising.
下表总结了本主题的关键结果。复习时可将其放在手边。
| Equation 方程形式 | Method 解法 | General Solution 通解 |
|---|---|---|
| dy/dx = f(x) | Integrate once 积分一次 | y = ∫ f(x) dx + c |
| d²y/dx² = f(x) | Integrate twice 积分两次 | y = ∫∫ f(x) dx dx + c₁x + c₂ |
| dy/dx = ky | Separate variables 分离变量 | y = Aekx |
| dy/dt = dV/dr × dr/dt | Chain rule 链式法则 | Relate rates 关联各变化率 |
Mastering differential equations of the form dy/dx = f(x) gives you a solid foundation for more advanced topics such as separable variables, integrating factors, and numerical methods. Practice past paper questions regularly, and always verify your solutions by differentiation.
掌握形如 dy/dx = f(x) 的微分方程，将为你学习更高级的主题（如分离变量法、积分因子和数值方法）奠定坚实基础。请定期练习真题，并始终通过求导来验证你的解。
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