📚 Solving Differential Equations of the Form dy/dx = f(x)g(y) | 求解 dy/dx = f(x)g(y) 型微分方程
Many AQA A-level Mathematics problems ask you to solve first-order differential equations in which the variables can be separated. The most common general form is dy/dx = f(x)g(y), where f(x) depends only on x and g(y) depends only on y. Once you can recognise this structure, you can integrate both sides to find a general solution and then use initial conditions to find a particular solution.
许多 AQA A-level 数学题目要求求解可分离变量的一阶微分方程。最常见的一般形式是 dy/dx = f(x)g(y)，其中 f(x) 仅依赖于 x，g(y) 仅依赖于 y。一旦你能识别这种结构，就可以对两边积分求出通解，再利用初始条件求出特解。
1. Recognising Separable Differential Equations | 识别可分离变量微分方程
A differential equation is separable if it can be written in the form dy/dx = f(x)g(y). In this expression, f(x) is a function of x only, and g(y) is a function of y only. The key test is to ask whether the right-hand side can be factorised into a product of two factors, one containing only x and the other containing only y.
如果一个微分方程可以写成 dy/dx = f(x)g(y) 的形式，那么它就是可分离变量的。在这个表达式中，f(x) 是只含 x 的函数，g(y) 是只含 y 的函数。关键的判断方法是：等号右边能否分解为两个因子的乘积，其中一个只含 x，另一个只含 y。
For example, dy/dx = 3x²y is separable because 3x² depends only on x and y depends only on y. Similarly, dy/dx = (x + 1)/y² is separable because it can be written as (x + 1) × (1/y²). In contrast, dy/dx = x + y is not separable because x + y cannot be written as a product of a function of x and a function of y.
例如，dy/dx = 3x²y 是可分离的，因为 3x² 只依赖于 x，y 只依赖于 y。同样，dy/dx = (x + 1)/y² 也是可分离的，因为它可以写成 (x + 1) × (1/y²)。相反，dy/dx = x + y 不是可分离的，因为 x + y 无法写成只含 x 的函数与只含 y 的函数的乘积。
dy/dx = f(x)g(y) ⇔ (1/g(y)) dy/dx = f(x)
This equivalent form is the starting point of the separation of variables method. You always need to check that g(y) is not zero before dividing by it, because division by zero is not allowed and may hide constant solutions.
这个等价形式是分离变量法的起点。在除以 g(y) 之前，你始终需要检查 g(y) 是否为零，因为除以零是不允许的，而且可能会隐藏常数解。
2. The Separation of Variables Method | 分离变量法
The method of separation of variables uses the chain rule in reverse. After rewriting the equation in the form (1/g(y)) dy/dx = f(x), you integrate both sides with respect to x. On the left-hand side, the dx effectively cancels, leaving an integral in y only.
分离变量法反过来使用了链式法则。将方程改写为 (1/g(y)) dy/dx = f(x) 后，你对两边关于 x 积分。在左边，dx 实际上被消去，只剩下只含 y 的积分。
- Write the equation as (1/g(y)) dy/dx = f(x).
- Multiply both sides by dx to obtain (1/g(y)) dy = f(x) dx.
- Integrate both sides: ∫ (1/g(y)) dy = ∫ f(x) dx.
- Add one constant of integration, usually on the x side.
- Rearrange the result to give y explicitly if the question requires it.
- 将方程写成 (1/g(y)) dy/dx = f(x)。
- 两边乘以 dx，得到 (1/g(y)) dy = f(x) dx。
- 对两边积分：∫ (1/g(y)) dy = ∫ f(x) dx。
- 只加一个积分常数，通常加在含 x 的一边。
- 如果题目要求，整理结果，给出 y 的显式表达式。
∫ (1/g(y)) dy = ∫ f(x) dx
It is important to write only one constant of integration. If you add a constant to both sides, the two constants can be combined into a single constant, so adding one is enough. In AQA exams, you should always show the integration step clearly before moving terms around.
只写一个积分常数非常重要。如果你在两边都加常数，这两个常数可以合并为一个常数，所以只加一个就够了。在 AQA 考试中，你应该在移项之前清楚地展示积分步骤。
3. Step-by-Step Worked Example | 逐步解题示例
Worked example: Solve the differential equation dy/dx = 6x²y for y > 0.
解题示例：求解微分方程 dy/dx = 6x²y，其中 y > 0。
Step 1: Rewrite the equation in separated form. Since y > 0, we can divide by y safely to obtain (1/y) dy/dx = 6x².
步骤 1：将方程改写为分离形式。因为 y > 0，我们可以安全地除以 y，得到 (1/y) dy/dx = 6x²。
Step 2: Multiply both sides by dx: (1/y) dy = 6x² dx.
步骤 2：两边乘以 dx：(1/y) dy = 6x² dx。
Step 3: Integrate both sides: ∫ (1/y) dy = ∫ 6x² dx. This gives ln y = 2x³ + C, where C is the constant of integration.
步骤 3：两边积分：∫ (1/y) dy = ∫ 6x² dx。得到 ln y = 2x³ + C，其中 C 是积分常数。
Step 4: Solve for y. Exponentiate both sides: y = e^(2x³ + C) = e^C · e^(2x³). Since e^C is a positive constant, we can replace it by A, where A > 0.
步骤 4：解出 y。两边取指数：y = e^(2x³ + C) = e^C · e^(2x³)。因为 e^C 是正常数，我们可以将其记为 A，其中 A > 0。
y = A e^(2x³), A > 0
This is the general solution. It contains one arbitrary constant, exactly as expected for a first-order differential equation.
这就是通解。它包含一个任意常数，这正是一阶微分方程所期望的结果。
4. Handling the Constant of Integration | 处理积分常数
The constant of integration is often the source of small errors. For example, if you write ln|y| = 2x³ + C, then exponentiating gives |y| = e^(2x³ + C) = e^C e^(2x³). Since e^C is a positive constant, the absolute value sign allows y to be either positive or negative after rewriting.
积分常数常常是产生小错误的根源。例如，如果你写 ln|y| = 2x³ + C，那么两边取指数得到 |y| = e^(2x³ + C) = e^C e^(2x³)。由于 e^C 是正常数，绝对值符号允许 y 在重写后既可为正也可为负。
ln|y| = 2x³ + C ⇒ y = A e^(2x³), A ≠ 0
In this form, A can be any non-zero constant. If the original equation has g(y) = 0 for some constant value of y, such as y = 0, then that value may also be a solution. You should check for these constant solutions before dividing by g(y).
在这种形式下，A 可以是任意非零常数。如果原方程在某个常数 y 值处有 g(y) = 0，例如 y = 0，那么该值也可能是解。在除以 g(y) 之前，你应该检查这些常数解。
In AQA exams, the question often specifies a condition such as y > 0 or y < 0, which removes the need for the absolute value. If no condition is given, you should keep |y| and then decide the sign using the initial condition.
在 AQA 考试中，题目通常会给出 y > 0 或 y < 0 这样的条件，这就无需使用绝对值。如果没有给出条件，你应该保留 |y|，然后根据初始条件确定符号。
5. Finding Particular Solutions from Initial Conditions | 由初始条件求特解
An initial condition is a pair of values, such as y(0) = 5, that tells you the value of y when x takes a specific value. Substituting this into the general solution allows you to find the value of the arbitrary constant A.
初始条件是一对值，例如 y(0) = 5，它告诉你当 x 取某个特定值时 y 的值。将其代入通解，你就能求出任意常数 A 的值。
Using the previous example, the general solution is y = A e^(2x³). If y(0) = 5, substitute x = 0 and y = 5 into the general solution: 5 = A e^(0) = A. Therefore A = 5.
使用前面的例子，通解为 y = A e^(2x³)。如果 y(0) = 5，将 x = 0 和 y = 5 代入通解：5 = A e^(0) = A。因此 A = 5。
y = 5 e^(2x³)
This is the particular solution that satisfies both the differential equation and the initial condition. Always substitute the initial condition into the final general solution, not into the differential equation before integration.
这就是同时满足微分方程和初始条件的特解。一定要将初始条件代入最终的通解，而不是代入积分前的微分方程。
6. Verifying a Solution | 验证解
Verification is a quick way to check your answer. Differentiate your particular solution and confirm that substituting y and dy/dx back into the original differential equation gives a true statement.
验证是检查答案的一种快速方法。对你的特解求导，然后将 y 和 dy/dx 代回原微分方程，确认得到一个恒等式。
For y = 5 e^(2x³), differentiate with respect to x: dy/dx = 5 e^(2x³) · 6x² = 6x² · 5 e^(2x³). Since 5 e^(2x³) is exactly y, this becomes dy/dx = 6x²y, which matches the original equation. Also, y(0) = 5 e^(0) = 5, so the initial condition is satisfied.
对于 y = 5 e^(2x³)，对 x 求导：dy/dx = 5 e^(2x³) · 6x² = 6x² · 5 e^(2x³)。由于 5 e^(2x³) 正好是 y，所以得到 dy/dx = 6x²y，与原方程一致。同时，y(0) = 5 e^(0) = 5，满足初始条件。
In an exam, if you have time, this verification step can catch errors in signs, powers, or constants. It is especially useful when you have to choose between two possible signs after removing an absolute value.
在考试中，如果有时间，这个验证步骤可以发现符号、幂或常数上的错误。当你去掉绝对值后需要在两个可能的符号之间做选择时，它尤其有用。
7. Common Pitfalls and How to Avoid Them | 常见错误与避免方法
Several mistakes appear regularly in AQA differential equations questions. Recognising them in advance will help you avoid losing marks.
在 AQA 微分方程题目中，有些错误会反复出现。提前识别它们有助于避免失分。
- Forgetting to add the constant of integration immediately after integrating.
- Adding two constants, one on each side, which is unnecessary.
- Dividing by g(y) without considering the case g(y) = 0, which may give extra constant solutions.
- Losing the absolute value when integrating 1/y to obtain ln|y|.
- Substituting the initial condition into the wrong stage of the solution.
- Confusing the roles of x and y when separating variables.
- 积分后忘记立即加上积分常数。
- 在两边各加一个常数，这是不必要的。
- 除以 g(y) 时没有考虑 g(y) = 0 的情况，这可能会产生额外的常数解。
- 对 1/y 积分得到 ln|y| 时丢失绝对值。
- 将初始条件代入了解题过程中的错误阶段。
- 分离变量时混淆了 x 和 y 的角色。
To avoid these errors, always write the separated form first, integrate carefully, add a single constant, and then use the initial condition on the general solution. Check any constant solutions separately before dividing by g(y).
为避免这些错误，请先写出分离形式，仔细积分，只加一个常数，然后在通解上使用初始条件。在除以 g(y) 之前，单独检查任何常数解。
8. Modelling with Separable Differential Equations | 可分离变量微分方程建模
In AQA A-level Mathematics, separable differential equations often appear in modelling questions. Common contexts include population growth, radioactive decay, and Newton’s law of cooling. In these problems, you are usually given a rate of change that is proportional to a quantity or to a difference of quantities.
在 AQA A-level 数学中，可分离变量微分方程经常出现在建模题中。常见的背景包括人口增长、放射性衰变和牛顿冷却定律。在这些问题中，通常会给出某个量的变化率与该量或某些量的差成比例。
For example, if a population P grows at a rate proportional to P, the model is dP/dt = kP, where t is time and k is a positive constant. This is separable because kP can be written as k × P, with k depending only on t? Actually k is a constant, P depends only on P. Separating variables gives ∫ (1/P) dP = ∫ k dt, so ln|P| = kt + C, and hence P = A e^(kt).
例如，如果人口 P 以与 P 成正比的速度增长，则模型为 dP/dt = kP，其中 t 是时间，k 是正常数。这个方程是可分离的，因为 kP 可以写成常数 k 与 P 的乘积。分离变量得到 ∫ (1/P) dP = ∫ k dt，于是 ln|P| = kt + C，因此 P = A e^(kt)。
dP/dt = kP ⇒ P = A e^(kt)
If P(0) = P₀, then A = P₀, so P = P₀ e^(kt). This is the standard exponential growth model. If k is negative, the same form describes exponential decay.
如果 P(0) = P₀，则 A = P₀，所以 P = P₀ e^(kt)。这是标准的指数增长模型。如果 k 为负值，同样的形式描述指数衰减。
9. Exponential Growth and Decay Models | 指数增长与衰减模型
A differential equation of the form dy/dt = ky, where k is a constant, always has the general solution y = A e^(kt). The sign of k determines the behaviour: if k > 0, y grows exponentially; if k < 0, y decays exponentially.
形如 dy/dt = ky 的微分方程，其中 k 为常数，其通解总是 y = A e^(kt)。k 的符号决定其行为：如果 k > 0，y 呈指数增长；如果 k < 0，y 呈指数衰减。
For decay models, you are sometimes asked to find the half-life, which is the time taken for the quantity to halve. Starting from y = y₀ e^(-kt), set y = y₀/2 and solve for t. This gives y₀/2 = y₀ e^(-kt), so 1/2 = e^(-kt), and taking natural logs gives t = ln 2 / k.
对于衰减模型，有时会要求你求半衰期，即数量减半所需的时间。从 y =
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