Edexcel Maths: Second Order Differential Equations | Edexcel 数学：二阶微分方程考点精讲

📚 Edexcel Maths: Second Order Differential Equations | Edexcel 数学：二阶微分方程考点精讲

Second order differential equations form a crucial part of the Edexcel A Level Mathematics syllabus. Mastering them requires a systematic approach: recognise the type of equation, solve the homogeneous part using the auxiliary equation, find a suitable particular integral for non-homogeneous cases, and finally apply boundary or initial conditions to pin down the arbitrary constants. This revision guide unpacks each of these steps, highlights common examiner pitfalls, and provides you with the confidence to tackle any second order ODE that appears in your exam.

二阶微分方程是 Edexcel A Level 数学课程中的核心内容。掌握它们需要一套系统的方法：识别方程类型、利用特征方程求解齐次部分、为非齐次情形设定合适的特解形式，最后代入边界条件或初始条件确定任意常数。本考点精讲将逐一拆解这些步骤，突出阅卷人经常扣分的易错点，帮助你在考试中自信地应对任何出现的二阶常微分方程。

1. Introduction to Second Order Differential Equations | 二阶微分方程概述

A second order ordinary differential equation (ODE) involves an unknown function y(x) and its derivatives up to the second order. The standard linear form with constant coefficients is a d²y/dx² + b dy/dx + c y = f(x), where a, b, c are constants and a ≠ 0. If f(x) = 0, the equation is homogeneous; otherwise it is non-homogeneous. The general solution to a non-homogeneous equation is the sum of the complementary function (CF) – the general solution of the associated homogeneous equation – and a particular integral (PI): y = y_CF + y_PI.

二阶常微分方程涉及未知函数 y(x) 及其最高至二阶的导数。具有常系数的标准线性形式为 a d²y/dx² + b dy/dx + c y = f(x)，其中 a, b, c 为常数且 a ≠ 0。如果 f(x) = 0，方程为齐次方程；否则为非齐次方程。非齐次方程的通解是余函数（CF，即对应的齐次方程的通解）与特积分（PI）之和：y = y_CF + y_PI。

In Edexcel exams, you will only meet linear equations with constant coefficients. The key is to follow a structured thought process: is the equation homogeneous? If yes, solve the auxiliary equation. If not, solve the homogeneous part first, then choose a trial PI whose form mirrors f(x). Finally, use given conditions to evaluate the unknown constants A and B.

在 Edexcel 考试中，你只会遇到常系数线性方程。关键在于遵循结构化的思路：方程是齐次的吗？如果是，求解特征方程。如果不是，先求解齐次部分，再根据 f(x) 的形式猜测试探特解。最后利用给定条件求出未知常数 A 和 B。

2. Homogeneous Equations & the Auxiliary Equation | 齐次方程与特征方程

For a homogeneous second order linear ODE with constant coefficients a d²y/dx² + b dy/dx + c y = 0, we look for solutions of the form y = e^mx. Substituting this trial solution yields the auxiliary (or characteristic) equation: a m² + b m + c = 0. The nature of the roots m determines the form of the complementary function.

对于常系数齐次线性二阶常微分方程 a d²y/dx² + b dy/dx + c y = 0，我们寻找形如 y = e^mx 的解。将这一试探解代入后得到特征方程（辅助方程）：a m² + b m + c = 0。根 m 的性质决定了余函数的形式。

The auxiliary equation may produce three types of roots: real and distinct, real and repeated, or complex conjugates. Each case gives rise to a different structure for y_CF. Memorising these three templates is essential, but you must also understand why they are used – the structure ensures that the two parts of the solution are linearly independent.

特征方程可能产生三种类型的根：相异实根、重实根、共轭复根。每种情形对应不同的 y_CF 结构。记住这三种模板至关重要，但你也需要理解为什么——这些结构保证了解的两个部分线性无关。

3. Real Distinct Roots | 相异实根

When the auxiliary equation yields two distinct real roots m₁ and m₂ (m₁ ≠ m₂), the complementary function is y_CF = A e^m₁x + B e^m₂x, where A and B are arbitrary constants. This is the simplest case, but always check that your roots are correct by factorising or using the quadratic formula.

当特征方程产生两个相异实根 m₁ 和 m₂ (m₁ ≠ m₂) 时，余函数为 y_CF = A e^m₁x + B e^m₂x，其中 A 和 B 为任意常数。这是最简单的情形，但务必通过因式分解或求根公式验证你的根是否正确。

Example: Solve y” − 5y’ + 6y = 0. Auxiliary eqn: m² − 5m + 6 = 0 → (m − 2)(m − 3) = 0 → m = 2, 3. Hence y_CF = A e^2x + B e^3x. Always present your answer in terms of constants, unless initial conditions are given.

示例：求解 y” − 5y’ + 6y = 0。特征方程：m² − 5m + 6 = 0 → (m − 2)(m − 3) = 0 → m = 2, 3。因此 y_CF = A e^2x + B e^3x。除非给出初始条件，否则答案中要保留常数。

4. Repeated Roots | 重根

If the auxiliary equation has a repeated root m (i.e. discriminant Δ = b² − 4ac = 0), the complementary function becomes y_CF = (A + B x) e^mx. The x factor is necessary to ensure linear independence of the two solution components. A common mistake is to write only A e^mx + B e^mx, which collapses to a single constant multiple.

如果特征方程具有重根 m（即判别式 Δ = b² − 4ac = 0），余函数变为 y_CF = (A + B x) e^mx。引入 x 因子是为了保证两个解分量线性无关。常见错误是只写成 A e^mx + B e^mx，这实际上只相当于一个常数倍。

Example: y” − 4y’ + 4y = 0 → m² − 4m + 4 = 0 → (m − 2)² = 0 → m = 2 (repeated). CF: y = (A + B x) e^2x. In exam questions, always verify the discriminant to avoid misclassifying the root type.

示例：y” − 4y’ + 4y = 0 → m² − 4m + 4 = 0 → (m − 2)² = 0 → m = 2（重根）。CF：y = (A + B x) e^2x。在考试中，务必验证判别式以免错误归类根的类型。

5. Complex Roots | 复根

When the auxiliary equation yields complex conjugate roots m = α ± iβ, the complementary function takes the oscillatory form y_CF = e^αx (A cos βx + B sin βx). Notice that the real part α appears in the exponential multiplier, dictating growth or decay, while the imaginary part β gives the angular frequency of the trigonometric terms.

当特征方程产生共轭复根 m = α ± iβ 时，余函数取振荡形式 y_CF = e^αx (A cos βx + B sin βx)。注意实部 α 出现在指数因子中，支配增长或衰减；虚部 β 则给出三角项的角频率。

A frequent algebraic slip occurs when extracting α and β from the quadratic formula. For m² + 2m + 5 = 0, we get m = −1 ± 2i, so α = −1, β = 2. The CF is y = e^−x (A cos 2x + B sin 2x). Avoid writing the real part inside the trigonometric functions – keep them separate.

从求根公式中提取 α 和 β 时，经常出现代数失误。对于 m² + 2m + 5 = 0，解得 m = −1 ± 2i，因此 α = −1，β = 2。CF 为 y = e^−x (A cos 2x + B sin 2x)。不要将实部混入三角函数中——务必分开表示。

6. Non-Homogeneous Equations: Form of f(x) | 非齐次方程：f(x) 的形式

For a non-homogeneous equation a y” + b y’ + c y = f(x), the particular integral y_PI is chosen according to the function f(x). In the Edexcel syllabus, f(x) is typically a polynomial, an exponential, a trigonometric function (sine/cosine), or a simple combination of these. The principle is to try a similar form with undetermined coefficients.

对于非齐次方程 a y” + b y’ + c y = f(x)，特积分 y_PI 根据函数 f(x) 的形式进行选取。在 Edexcel 考纲中，f(x) 通常为多项式、指数函数、三角函数（正弦/余弦），或它们的简单组合。其原则是试设一个包含待定系数的相似形式的解。

The table below summarises the standard trial PI forms. Remember: if your trial PI duplicates any term in the CF, you must multiply the trial PI by x (or by x² if the duplication is repeated) to obtain an independent particular integral.

下表总结了标准的试探特解形式。记住：如果试探特解与余函数中的某项重复，你必须将试探特解乘以 x（若重复为二重，则乘以 x²），以获得独立的特积分。

f(x)	Trial y_PI (if no overlap with CF)	中文说明
Polynomial of degree n	p_nxⁿ + … + p₀	同次多项式
k e^px	C e^px	同指数形式的指数函数
k cos ωx or k sin ωx	P cos ωx + Q sin ωx	包含正、余弦的线性组合
Product e^px × trig	e^px (P cos ωx + Q sin ωx)	指数与三角函数的乘积

When f(x) is a sum, use superposition: treat each term separately, find its PI, and add them.

当 f(x) 为和式时，使用叠加原理：分别处理每一项，求出各自的 PI 再相加。

7. Finding Particular Integral: Polynomial f(x) | 求特解：多项式形式

If f(x) is a polynomial, set y_PI as a general polynomial of the same degree. For example, if f(x) = 3x², try y_PI = Px² + Qx + R. Substitute into the ODE and equate coefficients of like powers of x to determine P, Q, R. If the constant term c = 0 in the ODE, a lower-degree PI may work, but always start with the full polynomial to be safe.

若 f(x) 为多项式，将 y_PI 设为相同次数的一般多项式。例如，若 f(x) = 3x²，尝试 y_PI = Px² + Qx + R。代入原微分方程，比较 x 同次幂的系数以确定 P、Q、R。如果方程中常数项 c = 0，可能较低次多项式即可，但为保险起见，通常从完整多项式开始。

Worked illustration: solve y” − 3y’ + 2y = 4x. CF is A e^x + B e^2x. Trial PI: y = Px + Q. Then y’ = P, y” = 0. Substitute: 0 − 3(P) + 2(Px + Q) = 2Px + (−3P + 2Q) = 4x. Equating coefficients: 2P = 4 ⇒ P = 2; −3(2) + 2Q = 0 ⇒ Q = 3. So y_PI = 2x + 3. General solution: y = A e^x + B e^2x + 2x + 3.

示例：求解 y” − 3y’ + 2y = 4x。CF 为 A e^x + B e^2x。试探 PI：y = Px + Q。则 y’ = P，y” = 0。代入：0 − 3P + 2(Px + Q) = 2Px + (−3P + 2Q) = 4x。比较系数：2P = 4 ⇒ P = 2；−3(2) + 2Q = 0 ⇒ Q = 3。因此 y_PI = 2x + 3。通解：y = A e^x + B e^2x + 2x + 3。

8. Finding Particular Integral: Exponential f(x) | 求特解：指数形式

When f(x) = C e^kx, try y_PI = D e^kx provided k is not a root of the auxiliary equation. If k equals one of the CF roots, multiply the trial PI by x; if k is the repeated root, multiply by x². This modification prevents the PI from being absorbed into the CF.

当 f(x) = C e^kx 时，只要 k 不是特征方程的根，就尝试 y_PI = D e^kx。若 k 等于 CF 的一个单根，将试探 PI 乘以 x；若 k 为重根，则乘以 x²。这一修正可防止特解被余函数吸收。

Example: y” − 3y’ + 2y = 5 e^2x. CF = A e^x + B e^2x. Here k = 2 is a root, so try y_PI = D x e^2x. Differentiate using the product rule: y’ = D e^2x(1 + 2x), y” = D e^2x(4 + 4x). Substituting and simplifying leads to D = 5. Therefore y_PI = 5x e^2x. The general solution is y = A e^x + B e^2x + 5x e^2x.

示例：y” − 3y’ + 2y = 5 e^2x。CF = A e^x + B e^2x。此处 k = 2 是单根，故尝试 y_PI = D x e^2x。用乘积法则求导：y’ = D e^2x(1 + 2x)，y” = D e^2x(4 + 4x)。代入化简后可得 D = 5。因此 y_PI = 5x e^2x。通解为 y = A e^x + B e^2x + 5x e^2x。

9. Finding Particular Integral: Trigonometric f(x) | 求特解：三角函数形式

For f(x) involving sin ωx and/or cos ωx, use the trial PI y_PI = P cos ωx + Q sin ωx. Even if f(x) contains only a sine or only a cosine, you must include both trigonometric terms because the derivatives will mix sines and cosines. When ω causes the trial PI to coincide with the CF (i.e. when α = 0 and β = ω in complex roots), multiply the trial PI by x.

当 f(x) 包含 sin ωx 和/或 cos ωx 时，使用试探特解 y_PI = P cos ωx + Q sin ωx。即使 f(x) 只含有正弦或余弦，也必须同时包含两项，因为求导会使正弦和余弦交叉出现。当 ω 导致试探 PI 与 CF 重合时（即复根中 α = 0 且 β = ω），需将试探 PI 乘以 x。

Illustration: y” + 4y = 3 sin 2x. CF: auxiliary m² + 4 = 0 → m = ±2i, so y_CF = A cos 2x + B sin 2x. Since f(x) = 3 sin 2x overlaps with CF, use y_PI = x (P cos 2x + Q sin 2x). After differentiation and substitution, you will find P = −¾ and Q = 0, giving y_PI = −¾ x cos 2x.

示例：y” + 4y = 3 sin 2x。CF：特征方程 m² + 4 = 0 → m = ±2i，故 y_CF = A cos 2x + B sin 2x。因 f(x) = 3 sin 2x 与 CF 重合，使用 y_PI = x (P cos 2x + Q sin 2x)。求导并代入后，可得 P = −¾，Q = 0，因此 y_PI = −¾ x cos 2x。

10. Superposition & General Solution Strategy | 叠加原理与一般解法

When f(x) is a sum of different function types,

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Edexcel Maths: Second Order Differential Equations | Edexcel 数学：二阶微分方程 考点精讲