📚 GCSE AQA Maths: Differential Equations Essentials | GCSE AQA 数学：微分方程 考点精讲
A differential equation links a function with one or more of its derivatives. At GCSE level, particularly for AQA Higher Tier and Further Maths, you are expected to form simple differential equations, solve first‑order separable equations, and interpret solutions in real‑world contexts such as population growth, cooling, and radioactive decay. This article breaks down every key point you need to master.
微分方程将一个函数与其一个或多个导数联系在一起。在 GCSE 层次，尤其是 AQA 高阶和进阶数学中，你需要建立简单的微分方程、求解一阶可分离方程，并解释其在现实情境中的解，例如人口增长、冷却和放射性衰变。本文分解你需要掌握的所有关键要点。
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 the rate of change of a quantity. For example, dy/dx = 3x + 2 is a first‑order differential equation because it only involves the first derivative.
微分方程是包含导数的方程，如 dy/dx、d²y/dx² 或某个量的变化率。例如，dy/dx = 3x + 2 是一个一阶微分方程，因为它只涉及一阶导数。
In GCSE, we mostly deal with first‑order ordinary differential equations that represent direct proportion between a rate of change and a variable, e.g. dP/dt = kP for population P.
在 GCSE 中，我们主要处理一阶常微分方程，表示变化率与某个变量之间的正比关系，例如人口 P 满足 dP/dt = kP。
The solution of a differential equation is not a number but a function (or a family of functions) that satisfies the equation. You will often find a particular solution using an initial condition.
微分方程的解不是一个数字，而是一个满足方程的函数（或函数族）。你通常会使用初始条件求出特解。
2. Direct Integration: dy/dx = f(x) | 直接积分法：dy/dx = f(x)
When the derivative is given purely in terms of x, such as dy/dx = 3x², you can find y by integrating both sides with respect to x. This gives y = ∫ f(x) dx + C, where C is an arbitrary constant. You must always include +C.
当导数纯粹用 x 表示时，例如 dy/dx = 3x²，你可以通过对两边关于 x 积分求出 y。这样得到 y = ∫ f(x) dx + C，其中 C 是任意常数。你必须始终加上 +C。
For example, solve dy/dx = 4x + 1. Integrating, y = 2x² + x + C. If you are told that y = 5 when x = 1, substitute: 5 = 2(1)² + 1 + C → C = 2, so the particular solution is y = 2x² + x + 2.
例如，求解 dy/dx = 4x + 1。积分得 y = 2x² + x + C。如果已知 x = 1 时 y = 5，代入：5 = 2(1)² + 1 + C → C = 2，因此特解为 y = 2x² + x + 2。
3. Separating Variables | 分离变量法
If the differential equation can be written so that all y terms (including dy) are on one side and all x terms (including dx) are on the other, you can solve by separating variables. The form is dy/dx = g(x)h(y). Rearrange to 1/h(y) dy = g(x) dx, then integrate both sides.
如果微分方程可以写成所有含 y 项（包括 dy）在一边、所有含 x 项（包括 dx）在另一边，就可以用分离变量法求解。形式为 dy/dx = g(x)h(y)。重新排列为 1/h(y) dy = g(x) dx，然后两边积分。
For example, solve dy/dx = 2xy. Separate: 1/y dy = 2x dx. Integrate: ln|y| = x² + C. Exponentiate: |y| = e^(x²+C) = e^C e^(x²). Letting A = ± e^C, we get y = A e^(x²).
例如，求解 dy/dx = 2xy。分离变量：1/y dy = 2x dx。积分：ln|y| = x² + C。取指数：|y| = e^(x²+C) = e^C e^(x²)。令 A = ± e^C，得到 y = A e^(x²)。
GCSE questions usually avoid absolute value complications by giving contexts where y > 0. Just apply the technique carefully.
GCSE 题目通常通过设定 y > 0 的背景来避免绝对值复杂性。只要仔细运用该方法即可。
4. The Natural Law of Growth and Decay | 自然增长与衰减定律
The most important GCSE differential equation is dN/dt = kN, where N is the quantity (population, mass, temperature difference) and k is a constant. If k > 0 it is growth; if k < 0 it is decay.
最重要的 GCSE 微分方程是 dN/dt = kN，其中 N 是数量（人口、质量、温度差），k 是常数。若 k > 0，表示增长；若 k < 0，表示衰减。
The general solution is N = N₀ e^(kt), where N₀ is the initial value at t = 0. This exponential solution describes many real‑world processes, from bacteria growth to radioactive half‑life.
通解为 N = N₀ e^(kt)，其中 N₀ 是 t = 0 时的初始值。这个指数解描述了许多现实过程，从细菌生长到放射性半衰期。
You must be confident moving between the differential form dN/dt = kN and the exponential solution. They are equivalent.
你必须能熟练地在微分形式 dN/dt = kN 与指数解之间转换。它们是等价的。
5. Solving Growth/Decay with Initial Conditions | 利用初始条件求解增长/衰减问题
Given dN/dt = kN and N = N₀ at t = 0, you can find the particular solution. Often the value of k is not given directly, but you can find it using a second data point.
已知 dN/dt = kN 且 t = 0 时 N = N₀，你可以求得特解。常数 k 通常不会直接给出，但你可用第二个数据点求出它。
For instance, a bacterial colony starts with 500 bacteria and grows at a rate proportional to its size. After 2 hours there are 800. Find the population after 5 hours. Set N = 500 e^(kt). Use t=2, N=800: 800 = 500 e^(2k) → e^(2k) = 1.6 → 2k = ln 1.6 → k ≈ 0.235. Then N(5) = 500 e^(0.235×5) ≈ 500 × 3.24 ≈ 1620.
例如，一个细菌菌落初始有500个细菌，增长速率与其大小成正比。2小时后有800个。求5小时后的数量。设 N = 500 e^(kt)。利用 t=2, N=800：800 = 500 e^(2k) → e^(2k) = 1.6 → 2k = ln 1.6 → k ≈ 0.235。然后 N(5) = 500 e^(0.235×5) ≈ 500 × 3.24 ≈ 1620。
6. Half‑Life and Doubling Time | 半衰期与倍增时间
For decay, the half‑life T is the time taken for half the substance to remain. Using N = N₀ e^(kt), set N = N₀/2 and solve: ½ = e^(kT) → ln(½) = kT → T = −ln2 / k (since k < 0, T is positive).
对于衰减，半衰期 T 是剩余一半物质所需的时间。利用 N = N₀ e^(kt)，设 N = N₀/2 求解：½ = e^(kT) → ln(½) = kT → T = −ln2 / k（因 k < 0，T 为正）。
For growth, the doubling time D satisfies 2 = e^(kD) → D = ln2 / k. You might be asked to find k given half‑life, or find how long it takes to decay to a certain percentage.
对于增长，倍增时间 D 满足 2 = e^(kD) → D = ln2 / k。你可能需要根据半衰期求 k，或者求衰减到某一百分比所需的时间。
7. Newton’s Law of Cooling | 牛顿冷却定律
Newton’s Law of Cooling 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), where k > 0.
牛顿冷却定律指出，物体温度的变化率与其温度 T 和环境温度 Tₐ 之差成正比：dT/dt = −k (T − Tₐ)，其中 k > 0。
Let θ = T − Tₐ. Then dθ/dt = −k θ, whose solution is θ = θ₀ e^(−kt). So T − Tₐ = (T₀ − Tₐ) e^(−kt). This is a direct application of the exponential decay model.
令 θ = T − Tₐ，则 dθ/dt = −k θ，其解为 θ = θ₀ e^(−kt)。所以 T − Tₐ = (T₀ − Tₐ) e^(−kt)。这是指数衰减模型的直接应用。
A typical GCSE question gives initial temperature, ambient temperature, and another temperature after a time, then asks for k or the temperature at another time. Set up the equation and solve for the unknown.
典型的 GCSE 题目会给出初始温度、环境温度以及一段时间后的另一个温度，然后要求求 k 或另一时刻的温度。建立方程并解出未知数。
8. Setting Up Differential Equations from Context | 根据上下文建立微分方程
Expect to interpret a written statement and form a differential equation. Phrases like “the rate of increase is directly proportional to…” translate to dQ/dt ∝ Q → dQ/dt = kQ.
你需要解读文字叙述并建立微分方程。“增加速率与……成正比”这类语句翻译为 dQ/dt ∝ Q → dQ/dt = kQ。
If the rate is “proportional to the square root of the population”, you write dP/dt = k √P. If it says “the rate of decrease is proportional to the cube of the mass”, write dM/dt = −k M³. Pay close attention to the proportionality constant and sign.
如果速率“与人口的平方根成正比”，写为 dP/dt = k √P。如果说是“减少的速率与质量的立方成正比”，写为 dM/dt = −k M³。要密切关注比例常数的符号。
9. Linking to Gradients and Direction Fields | 与梯度和方向场的联系
Although GCSE does not require you to sketch full slope fields, you should understand that dy/dx = f(x,y) gives the gradient of the solution curve at any point (x,y). A given point (x₀,y₀) can be used as an initial condition to trace the specific curve.
虽然 GCSE 不要求你画出完整的斜率场，但你应理解 dy/dx = f(x,y) 给出了解曲线上任意一点 (x,y) 处的梯度。给定的点 (x₀,y₀) 可作为初始条件，用来追踪特定的曲线。
This idea helps when you are asked to verify that a function is a solution or to match differential equations with their solution curves.
当要求验证某个函数是否为解，或者将微分方程与其解曲线匹配时，这一概念会有帮助。
10. Verifying a Solution | 验证一个解
To check if a given function y = f(x) is a solution of a differential equation, differentiate it to find dy/dx (and higher derivatives if needed) and substitute into the equation. If the left‑hand side equals the right‑hand side identically, it is a solution.
要检验给定的函数 y = f(x) 是否为微分方程的解，对其求导得到 dy/dx（如有需要还包括高阶导数），然后代入方程。如果左边恒等于右边，那么它就是一个解。
Example: Show that y = 3e^(2x) is a solution of dy/dx = 2y. LHS = dy/dx = 6e^(2x); RHS = 2y = 6e^(2x). They match, so it is a solution.
例如：证明 y = 3e^(2x) 是 dy/dx = 2y 的解。左边 = dy/dx = 6e^(2x)；右边 = 2y = 6e^(2x)。两者相等，故它是一个解。
11. Common Mistakes and Tips | 常见错误与技巧
Never forget the constant of integration. When you integrate both sides, you only need to add one constant (usually on the side with the independent variable). Combining constants correctly is vital.
切勿忘记积分常数。对两边积分时，只需添加一个常数（通常加在自变量一侧）。正确合并常数至关重要。
When separating variables dy/dx = g(x)h(y), ensure h(y) ≠ 0 to divide. Also, be careful with signs: if the problem states decay, the equation should include a minus sign, e.g. dM/dt = −λM.
分离变量 dy/dx = g(x)h(y) 时，要确保 h(y) ≠ 0 才能相除。此外，注意符号：如果问题叙述为衰减，方程中应包含负号，如 dM/dt = −λM。
Always interpret final answers in context: round appropriately, give units, and check if a value is sensible (e.g. time cannot be negative).
始终结合上下文解释最终答案：合理取舍、给出单位，并检查数值是否合理（如时间不能为负数）。
12. Exam‑Style Question Walkthrough | 考试题型解析
Question: A population P grows at a rate proportional to P. Initially P = 200. After 3 years P = 500. (a) Write the differential equation. (b) Find the population after 6 years. (c) Find the doubling time.
题目： 某种群数量 P 以与 P 成正比的速率增长。初始时 P = 200。3年后 P = 500。(a) 写出微分方程。(b) 求6年后的种群数量。(c) 求倍增时间。
Solution: (a) dP/dt = kP. (b) General solution P = 200 e^(kt). Using t=3, P=500: 500 = 200 e^(3k) → e^(3k) = 2.5 → 3k = ln 2.5 → k ≈ 0.3054. So P = 200 e^(0.3054t). For t=6, P = 200 e^(0.3054×6) = 200 e^(1.8324) ≈ 200 × 6.25 = 1250. (c) Doubling time D = ln2 / k = 0.6931 / 0.3054 ≈ 2.27 years.
解答： (a) dP/dt = kP。(b) 通解为 P = 200 e^(kt)。代入 t=3, P=500：500 = 200 e^(3k) → e^(3k) = 2.5 → 3k = ln 2.5 → k ≈ 0.3054。所以 P = 200 e^(0.3054t)。对于 t=6，P = 200 e^(0.3054×6) = 200 e^(1.8324) ≈ 200 × 6.25 = 1250。(c) 倍增时间 D = ln2 / k = 0.6931 / 0.3054 ≈ 2.27 年。
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