IB Edexcel Mathematics: Differential Equations Key Points | IB Edexcel 数学：微分方程考点精讲

📚 IB Edexcel Mathematics: Differential Equations Key Points | IB Edexcel 数学：微分方程考点精讲

Differential equations are a cornerstone of calculus, appearing frequently in IB and Edexcel A-level Mathematics. They allow us to model relationships between changing quantities and their rates of change. This guide covers the essential techniques and exam strategies you need to master differential equations at this level.

微分方程是微积分的重要组成部分，在IB和Edexcel A-level数学中频繁出现。它们使我们能够对变化量及其变化率之间的关系进行建模。本指南涵盖了掌握这一级别微分方程所必需的关键技巧和考试策略。

1. What are Differential Equations? | 什么是微分方程？

A differential equation is an equation that contains an unknown function and one or more of its derivatives. The order of a differential equation is the highest derivative that appears in it. A solution to a differential equation is a function that satisfies the equation; the general solution contains arbitrary constants (e.g. + C), while a particular solution is obtained by applying given initial or boundary conditions.

微分方程是包含未知函数及其导数的方程。微分方程的阶数是其中出现的最高阶导数。解是指满足该方程的函数；通解包含任意常数（例如 + C），而特解则是通过给定的初始条件或边界条件求得的。

For example, dy/dx = 3x² + C. If we also know that y(0) = 5, then C = 5 and the particular solution is y = x³ + 5.

例如，dy/dx = 3x² 是一个一阶微分方程。其通解为 y = x³ + C。若再已知 y(0) = 5，则 C = 5，特解为 y = x³ + 5。

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2. Forming a Differential Equation | 建立微分方程

Sometimes you are given a family of curves containing one or more arbitrary constants, and you must form the differential equation that represents that family. The general method is to differentiate the given relation as many times as there are constants, and then eliminate the constants algebraically.

有时题目会给出含有任意常数的曲线族，要求你建立代表该族曲线的微分方程。一般方法是根据常数的个数对关系式进行相应次数的微分，然后通过代数方法消去常数。

For instance, given the family y = A e^2x + B e^−x, differentiate twice: y' = 2A e^2x − B e^−x and y'' = 4A e^2x + B e^−x. Combine the expressions for y, y' and y'' to eliminate A and B, eventually obtaining y'' − y' − 2y = 0.

例如，给定曲线族 y = A e^2x + B e^−x，微分两次：y' = 2A e^2x − B e^−x，y'' = 4A e^2x + B e^−x。将 y、y' 和 y'' 的表达式组合消去 A 和 B，最终得到 y'' − y' − 2y = 0。

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3. Separable Differential Equations | 可分离变量的微分方程

A first-order differential equation of the form dy/dx = g(x)h(y) is called separable dx and then integrate both sides. Always remember to include the constant of integration.

形如 dy/dx = g(x)h(y) 的一阶微分方程称为可分离变量方程。解法是将其改写为 (1/h(y)) dy = g(x) dx，然后两边积分。务必记住加上积分常数。

y²/2 = x² + C. Alternatively, express the general solution as y² = 2x² + K, where K = 2C.

考虑 dy/dx = (2x)/(y)。分离变量：y dy = 2x dx。积分：∫ y dy = ∫ 2x dx → y²/2 = x² + C。也可将通解表示为 y² = 2x² + K，其中 K = 2C。

rticular constant. For example, if y(1) = 3, then 3² = 2(1)² + K → K = 7, so y² = 2x² + 7.

当给出初始条件时，代入即可求出特定常数。例如，若 y(1) = 3，则 3² = 2(1)² + K → K = 7，因此 y² = 2x² + 7。

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4. First-Order Linear Equations and Integrating Factor | 一阶线性微分方程与积分因子

A first-order linear ODE can be written in the standard form dy/dx + P(x)y = Q(x). To solve it, we multiply through by the integrating factor μ(x) = e^{∫ P(x) dx}. The left-hand side then becomes the derivative of μ(x)y.

一阶线性常微分方程可以写成标准形式 dy/dx + P(x)y = Q(x)。求解时，两边同乘积分因子 μ(x) = e^{∫ P(x) dx}，此时左边变为 μ(x)y 的导数。

Q(x). Integrating both sides gives the general solution. For example, solve dy/dx + 2y = e^x. Here P(x) = 2, so μ = e^{∫ 2 dx} = e^2x. Multiplying: e^2x dy/dx + 2e^2x y = e^3x, which is d/dx (e^2x y) = e^3x. Integrate: e^2x y = (1/3)e^3x + C, so y = (1/3)e^x + C e^−2x.

因此，d/dx [μ(x)y] = μ(x)Q(x)。两边积分即得通解。例如，求解 dy/dx + 2y = e^x。这里 P(x) = 2，所以 μ = e^{∫ 2 dx} = e^2x。乘之：e^2x dy/dx + 2e^2x y = e^3x，即 d/dx (e^2x y) = e^3x。积分：e^2x y = (1/3)e^3x + C，故 y = (1/3)e^x + C e^−2x。

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5. Homogeneous First-Order Equations | 齐次一阶微分方程

A first-order equation that can be written as dy/dx = F(y/x) is called homogeneous. The standard technique is to use the substitution y = vx x. Then dy/dx = v + x dv/dx, which converts the equation into a separable one in v and x.

能够写成 dy/dx = F(y/x) 形式的一阶方程称为齐次方程。标准的求解技巧是作代换 y = vx，其中 v 是 x 的函数。于是 dy/dx = v + x dv/dx，从而将原方程转化为关于 v 和 x 的可分离变量方程。

+ v²) / v. Simplify: x dv/dx = (1 + v²)/v − v = 1/v. So v dv = dx/x. Integrate: v²/2 = ln|x| + C. Substitute back v = y/x to get the general solution in x and y.

例如，求解 dy/dx = (x² + y²) / (xy)。将右边改写为 (1 + (y/x)²) / (y/x)。令 y = vx，则 v + x dv/dx = (1 + v²) / v。化简：x dv/dx = (1 + v²)/v − v = 1/v。于是 v dv = dx/x。积分：v²/2 = ln|x| + C。将 v = y/x 代回即可得到关于 x 和 y 的通解。

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6. Second-Order Linear Homogeneous Equations with Constant Coefficients | 二阶常系数线性齐次方程

A second-order linear homogeneous ODE has the form a d²y/dx² + b dy/dx + c y = 0, where a, b, c are constants. The solution method begins by writing the auxiliary equation: a m² + b m + c = 0. The nature of the roots determines the form of the general solution.

二阶线性齐次常微分方程具有形式 a d²y/dx² + b dy/dx + c y = 0，其中 a, b, c 为常数。解法首先写出辅助方程：a m² + b m + c = 0。根的性质决定了通解的形式。

the auxiliary equation has two distinct real roots m₁, m₂, the general solution is y = A e^m₁x + B e^m₂x. If it has a repeated real root m, then y = (A + Bx) e^mx. If the roots are complex conjugates α ± iβ, then y = e^αx (C cos βx + D sin βx).

若辅助方程有两个不相等的实根 m₁、m₂，则通解为 y = A e^m₁x + B e^m₂x。若有重实根 m，则 y = (A + Bx) e^mx。若根为一对共轭复根 α ± iβ，则 y = e^αx (C cos βx + D sin βx)。

For instance, solve y'' − 5y' + 6y = 0. Auxiliary equation: m² − 5m + 6 = 0 → m = 2, 3. Hence general solution: y = A e^2x + B e^3x.

例如，求解 y'' − 5y' + 6y = 0。辅助方程：m² − 5m + 6 = 0 → m = 2, 3。通解为：y = A e^2x + B e^3x。

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7. Finding Particular Solutions with Initial Conditions | 利用初始条件求特解

nough; for second-order equations you will need two conditions, typically y(x₀) = y₀ and y'(x₀) = y'₀.

要求出特解，需利用给定的初始条件（或边界条件）来确定任意常数的值。对于一阶方程，一个条件即可；对于二阶方程，需要两个条件，通常为 y(x₀) = y₀ 和 y'(x₀) = y'₀。

Given y'' + 4y' ry equation m² + 4m + 4 = 0 gives repeated root m = −2, so general solution y = (A + Bx) e^−2x. If y(0) = 1 and y'(0) = 0, differentiate to find y' e to y' then substitute to solve for A and B. This yields A = 1, B = 2, hence particular solution y = (1 + 2x) e^−2x.

给定 y'' + 4y' + 4y = 0，辅助方程 m² + 4m + 4 = 0 得到重根 m = −2，故通解为 y = (A + Bx) e^−2x。若 y(0) = 1 且 y'(0) = 0，求导得到 y' 后代入求解 A 和 B。得 A = 1, B = 2，因此特解为 y = (1 + 2x) e^−2x。

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8. Applications: Exponential Growth and Decay | 实际应用：指数增长与衰减

Many natural phenomena are modelled by the differential equation dP/dt = kP if k < 0, exponential decay (e.g. radioactive decay). The general solution is P(t) = P₀ e^kt, where P₀ is the initial quantity.

许多自然现象可用微分方程 dP/dt = kP 来建模，其中 k 为常数。若 k > 0，表示指数增长（如人口增长）；若 k < 0，表示指数衰减（如放射性衰变）。其通解为 P(t) = P₀ e^kt，其中 P₀ 为初始量。

ass, and half of it remains after 5 years, set up dM/dt = −kM. The solution is M(t) = M₀ e^−kt. Using M(5) = M₀/2, find k = (ln 2)/5. Then you can predict the mass at any time t.

例如，若某放射性物质衰变速率与其质量成正比，且 5 年后只剩下一半，建立 dM/dt = −kM。解为 M(t) = M₀ e^−kt。利用 M(5) = M₀/2，求得 k = (ln 2)/5，然后可预测任何时刻 t 的质量。

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9. Modelling with Differential Equations: Cooling and Mixing | 微分方程建模：冷却与混合问题

Newton' ifference between its temperature and the ambient temperature. This gives dT/dt = −k(T − T_amb). The solution is typically T(t) = T_amb + (T₀ − T_amb) e^−kt.

牛顿冷却定律指出，物体温度的变化率与物体温度和环境温度的差成正比。由此得到 dT/dt = −k(T − T_amb)。其解通常为 T(t) = T_amb + (T₀ − T_amb) e^−kt。

ount of substance x(t) in the tank is often dx/dt = (rate in) − (rate out). Rate in = conc_in × flow rate; rate out = (x(t)/volume) × flow rate. This yields a first-order linear ODE.

在混合问题中，容器内装有某种物质的溶液，另一种浓度的溶液流入，同时混合液被排出。描述容器内物质量 x(t) 的微分方程常为 dx/dt = （注入速率）− （流出速率）。注入速率 = 流入浓度 × 流量；流出速率 = (x(t)/体积) × 流量。这构成一阶线性常微分方程。

2)(5) − (x/100)(5) = 1 − 0.05x. This is a linear equation with initial condition x(0) = 0.

例如，一容器最初装有 100 升纯水。浓度为 0.2 kg/L 的盐溶液以 5 L/min 流入，充分混合的溶液以 5 L/min 流出。体积保持 100 L 不变。设 x(t) 为盐的质量（kg），则 dx/dt = (0.2)(5) − (x/100)(5) = 1 − 0.05x。此为一线性方程，初始条件为 x(0) = 0。

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

Many errors come from misidentifying the type of differential equation. Always check if it is separable, linear, homogeneous, or second-order with constant coefficients before choosing a method. Do not forget the constant of integration; losing a mark for a missing +C is very common.

许多错误源于对方程类型的误判。在选择解法之前，务必检查它是可分离的、线性的、齐次的，还是二阶常系数的。不要忘记积分常数；因漏掉 +C 而丢分十分常见。

When using the integrating factor, ensure the equation is divided through to make the coefficient of dy/dx equal to 1. For second-order ODE with complex roots, remember to write the solution using sine and cosine, not complex exponentials (unless specified). In modelling questions, carefully define your variables and check whether the volume is constant or changing.

使用积分因子时，确保方程已除至 dy/dx 系数为 1。对于具有复根的二阶常微分方程，记得用正弦和余弦表出解，而非复指数形式（除非题目要求）。在建模问题中，仔细定义变量，并检查体积是否恒定。

g the quadratic fully. And always, after finding a solution, test it against the given initial conditions to confirm it works.

练习快速写出辅助方程，并根据判别式（Δ = b² − 4ac）直接写出通解，而不必完整求解二次方程。此外，求出解后，务必用给定的初始条件验证，以确保解正确。

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