Second-Order Differential Equations for CIE A-Level | CIE A-Level 二阶微分方程考点精讲

📚 Second-Order Differential Equations for CIE A-Level | CIE A-Level 二阶微分方程考点精讲

Second-order differential equations play a central role in CIE A-Level Mathematics (typically within the Further Mathematics syllabus). They involve the second derivative of an unknown function and appear frequently in modelling oscillations, circuits, and mechanical vibrations. This article covers the key concepts, solution methods, and typical exam permutations that you must master: homogeneous equations with constant coefficients, the auxiliary equation, handling cases of real distinct roots, repeated roots, and complex conjugate roots, finding particular integrals for non-homogeneous equations, and applying initial or boundary conditions to obtain specific solutions.

二阶微分方程是 CIE A-Level 数学（通常属于 Further Mathematics 大纲）的核心内容。方程中涉及未知函数的二阶导数，广泛应用于振动、电路和力学系统的建模。本文覆盖你必须掌握的重要概念与解题方法：常系数齐次线性方程、辅助方程、相异实根、重根、共轭复根三种情形的处理、非齐次方程的特解求法，以及如何利用初始条件或边界条件得到特解。

1. Basic Form and Classification | 基本形式与分类

A second-order ordinary differential equation (ODE) contains a second derivative d²y/dx², possibly a first derivative dy/dx, the function y itself, and may include a function of x on the right-hand side. The general linear second-order ODE with constant coefficients can be written as:

二阶常微分方程中包含二阶导数 d²y/dx²，可能会有一阶导数 dy/dx、函数 y 本身，以及右边关于 x 的函数。常系数线性二阶 ODE 的一般形式为：

a d²y/dx² + b dy/dx + c y = f(x)

where a, b, c are constants (a ≠ 0) and f(x) is a given function of x. If f(x) = 0, the equation is said to be homogeneous; otherwise it is non‑homogeneous. This distinction is crucial because the solution strategy splits into two parts: the complementary function (CF), which solves the homogeneous equation, and the particular integral (PI), which is any solution of the full non‑homogeneous equation.

其中 a, b, c 是常数（a ≠ 0），f(x) 是 x 的给定函数。如果 f(x) = 0，方程称为齐次的；否则为非齐次的。这一区别至关重要，因为求解策略分为两部分：补函数（CF），即齐次方程的解；以及特解（PI），是非齐次方程的任意一个解。

2. Homogeneous Equations and the Auxiliary Equation | 齐次方程与辅助方程

For the homogeneous case a d²y/dx² + b dy/dx + c y = 0, we look for solutions of the form y = e^(mx). Substituting this trial solution gives the auxiliary equation (also called the characteristic equation):

对于齐次情形 a d²y/dx² + b dy/dx + c y = 0，我们寻找形如 y = e^(mx) 的解。代入这一试探解可以得到辅助方程（也称特征方程）：

a m² + b m + c = 0

The roots m = m₁ and m = m₂ determine the form of the complementary function. This quadratic equation arises because differentiating y = e^(mx) twice pulls down a factor of m², once pulls down m, and the original function provides a constant multiple. The auxiliary equation is the single most important algebraic step in solving any constant‑coefficient linear ODE.

辅助方程的根 m = m₁ 与 m = m₂ 决定了补函数的形式。产生这一二次方程的原因是：对 y = e^(mx) 求导两次会得到因子 m²，一次得到 m，原函数提供常数倍。辅助方程是求解任何常系数线性 ODE 时最为关键的代数步骤。

3. Case 1: Real and Distinct Roots | 情形一：相异实根

If the auxiliary equation yields two distinct real roots m₁ and m₂, the complementary function is:

若辅助方程有两个不同的实根 m₁ 与 m₂，则补函数为：

y_c = A e^(m₁x) + B e^(m₂x)

where A and B are arbitrary constants. This is the superposition of two linearly independent solutions. The functions e^(m₁x) and e^(m₂x) are independent because their ratio is not constant. Almost all homogeneous second‑order problems that yield real distinct roots can be solved directly using this formula; exam questions will then ask you to find A and B from given conditions.

其中 A 和 B 是任意常数。这是两个线性无关解的叠加。因为 e^(m₁x) 与 e^(m₂x) 的比值不是常数，所以它们线性无关。几乎所有出现相异实根的齐次二阶问题都能直接使用该公式求解；考试题通常会要求你根据给定条件求出 A 和 B。

4. Case 2: Repeated Real Root | 情形二：重实根

When the discriminant b² − 4ac = 0, the auxiliary equation has one repeated real root m (twice). A single exponential e^(mx) only gives one independent solution; we need a second solution linearly independent of the first. The standard result is that the second solution is x e^(mx). Hence the complementary function takes the form:

当判别式 b² − 4ac = 0 时，辅助方程有一个二重实根 m。单纯的指数函数 e^(mx) 只给出一个独立解；我们需要与第一个解线性无关的第二个解。标准结果是第二个解为 x e^(mx)。因此补函数的形式为：

y_c = (A + B x) e^(mx)

This can be derived using the method of reduction of order, but at A‑Level it is taken as a standard result. It is important to remember the extra factor x – without it, the general solution would be incomplete. In exam conditions, spotting a repeated root quickly not only saves time but also guarantees the correct structure of the answer.

可以通过降阶法推导，不过在 A‑Level 阶段将其视为标准结论。记住要多出一个因子 x 至关重要——缺少它通解就不完整。在考试中，迅速识别重根不仅能节省时间，还能确保答案结构正确。

5. Case 3: Complex Conjugate Roots | 情形三：共轭复根

If the auxiliary equation has complex conjugate roots m = α ± iβ, where α and β are real and β > 0, the complementary function is expressed in terms of real functions as:

若辅助方程有共轭复根 m = α ± iβ，其中 α 与 β 为实数且 β > 0，则补函数用实函数表示为：

y_c = e^(αx) ( C cos(βx) + D sin(βx) )

where C and D are arbitrary constants. This form avoids complex exponentials and is directly usable for applying initial conditions. The real part α determines the exponential growth or decay, while the imaginary part β controls the frequency of oscillation. Systems that exhibit damped oscillations, for instance, lead precisely to this type of solution.

其中 C 和 D 为任意常数。这一形式避免了复指数函数，可直接用于代入初始条件。实部 α 决定了指数增长或衰减，虚部 β 控制振荡频率。例如，表现出阻尼振动的系统恰好就会得到这类解。

6. The Non‑Homogeneous Equation and Particular Integral | 非齐次方程与特解

The general solution of a non‑homogeneous equation a d²y/dx² + b dy/dx + c y = f(x) is the sum of the complementary function y_c (solution of the corresponding homogeneous equation) and any particular integral y_p that satisfies the full equation:

非齐次方程 a d²y/dx² + b dy/dx + c y = f(x) 的通解是对应齐次方程的补函数 y_c 与满足原方程的一个特解 y_p 的和：

y = y_c + y_p

The complementary function contains two arbitrary constants, which will later be fixed by initial or boundary conditions. The particular integral does not contain any arbitrary constants. The method of undetermined coefficients is the standard A‑Level technique for finding y_p; it involves making an intelligent guess for the form of y_p based on the type of f(x).

补函数中含有两个任意常数，后续会通过初始条件或边界条件确定。特解中不含任意常数。待定系数法是 A‑Level 中求 y_p 的标准方法，需要根据 f(x) 的类型明智地猜测 y_p 的形式。

7. Method of Undetermined Coefficients | 待定系数法

The idea is to propose a trial function for y_p that has the same general form as f(x), but with unknown coefficients. Substituting this trial function into the differential equation yields algebraic equations for the coefficients, which can then be solved. The general rule is:

其思路是提出一个与 f(x) 一般形式相同的试探函数 y_p，但带有待定系数。将试探函数代入微分方程会得到关于这些系数的代数方程，然后可求解。一般规则如下：

If f(x) is a polynomial of degree n, try a general polynomial of the same degree, for example y_p = Pₙ(x).

若 f(x) 是 n 次多项式，试用相同次数的一般多项式，例如 y_p = Pₙ(x)。
If f(x) = k e^(px), try y_p = C e^(px) (unless e^(px) already appears in the complementary function, in which case multiply by x or x² as appropriate).

若 f(x) = k e^(px)，试用 y_p = C e^(px)（除非 e^(px) 已在补函数中出现，此时需要酌情乘以 x 或 x²）。
If f(x) = M cos(qx) + N sin(qx), try y_p = P cos(qx) + Q sin(qx).

若 f(x) = M cos(qx) + N sin(qx)，试用 y_p = P cos(qx) + Q sin(qx)。

When f(x) is a sum of such functions, the trial particular integral is simply the sum of the corresponding individual trial forms. Always check whether any term in the trial function coincides with a term in the complementary function; if it does, a modification (multiplying by x) is necessary to maintain independence.

当 f(x) 是这些函数的和时，试探特解就是对应各试探形式的和。务必检查试探函数中的项是否与补函数的项重合；若有重合，就需要通过乘以 x 进行修正以保持独立性。

8. Standard Forms of Particular Integral | 特解的标准形式

CIE exam questions regularly feature f(x) drawn from a short list, and the expected trial y_p is always of the following types. Using the correct starting form avoids unnecessary rearrangements.

CIE 考试题目中 f(x) 通常出自有限的几种类型，预期的试探特解 y_p 总是以下形式。使用正确的起步形式可以避免不必要的中间调整。

f(x)	Trial y_p	Notes
k (constant)	C	Provided c ≠ 0; if c = 0, use Cx.
k xⁿ (n integer)	A₀ + A₁x + … + Aₙxⁿ	Full polynomial of same degree.
e^(px)	C e^(px)	Multiply by x if e^(px) is in CF.
cos(qx), sin(qx)	P cos(qx) + Q sin(qx)	Always include both cosine and sine, even if only one appears.
e^(px)cos(qx) or e^(px)sin(qx)	e^(px)(P cos(qx) + Q sin(qx))	Combine exponential and trigonometric trial.

Care must be taken when the trial form overlaps with the complementary function. In such instances, multiply the trial function by x (or x² if a double overlap occurs). This “resonance” case appears frequently in past papers.

当试探形式与补函数重叠时需要小心。此时应将试探函数乘以 x（若出现双重重叠则乘以 x²）。这种“共振”情况在历年试题中经常出现。

9. Fixing Constants with Initial Conditions | 利用初始条件确定常数

Once the general solution y = y_c + y_p has been written, two conditions are needed to determine the arbitrary constants A and B (or C and D). Usually these are initial values, such as y(0) and y'(0). The procedure is straightforward: substitute x = 0 into the general solution to get the first equation, then differentiate the general solution and substitute x = 0 to get the second equation. Solving the simultaneous equations gives the specific solution satisfying the given data.

一旦写出通解 y = y_c + y_p，就需要两个条件来确定任意常数 A 和 B（或 C 和 D）。通常这些条件是初始值，如 y(0) 与 y'(0)。步骤十分直接：将 x = 0 代入通解得到第一个方程，再对通解求导并代入 x = 0 得到第二个方程。联立求解即可得到满足给定条件的具体解。

10. Worked Example 1 – Real Distinct Roots | 例题一：相异实根

Solve the differential equation y” − 3y’ − 10y = 0 with y(0) = 2, y'(0) = −1.

求解微分方程 y” − 3y’ − 10y = 0，且 y(0) = 2，y'(0) = −1。

The auxiliary equation is m² − 3m − 10 = 0 → (m − 5)(m + 2) = 0 → m = 5, −2. Hence the complementary function is y = A e^(5x) + B e^(−2x). Using y(0) = 2 gives A + B = 2. Differentiating: y’ = 5A e^(5x) − 2B e^(−2x), and y'(0) = 5A − 2B = −1. Solving the system A + B = 2 and 5A − 2B = −1 yields A = 3/7, B = 11/7. Thus the specific solution is y = (3/7) e^(5x) + (11/7) e^(−2x).

辅助方程为 m² − 3m − 10 = 0 → (m − 5)(m + 2) = 0 → m = 5, −2。因此补函数为 y = A e^(5x) + B e^(−2x)。利用 y(0) = 2 得 A + B = 2。求导：y’ = 5A e^(5x) − 2B e^(−2x)，且 y'(0) = 5A − 2B = −1。联立方程组 A + B = 2 与 5A − 2B = −1，解得 A = 3/7，B = 11/7。因此特解为 y = (3/7) e^(5x) + (11/7) e^(−2x)。

11. Worked Example 2 – Non‑Homogeneous with Exponential f(x) | 例题二：指数型非齐次方程

Find the general solution of y” + y’ − 6y = 10 e^(2x).

求 y” + y’ − 6y = 10 e^(2x) 的通解。

First, the homogeneous part: auxiliary equation m² + m − 6 = 0 → (m + 3)(m − 2) = 0 → m = −3, 2. Therefore y_c = A e^(−3x) + B e^(2x). Notice that e^(2x) is already present in y_c, so the standard trial y_p = C e^(2x) would fail. Multiply by x: try y_p = C x e^(2x). Differentiate: y_p‘ = C e^(2x) + 2C x e^(2x), y_p” = 4C e^(2x) + 4C x e^(2x). Substitute into the ODE: (4C e^(2x) + 4C x e^(2x)) + (C e^(2x) + 2C x e^(2x)) − 6(C x e^(2x)) = 10 e^(2x). Collect e^(2x) terms: (4C + C) = 5C → 5C = 10, so C = 2. The x e^(2x) terms cancel as expected. Thus y_p = 2x e^(2x). General solution: y = A e^(−3x) + B e^(2x) + 2x e^(2x).

首先处理齐次部分：辅助方程 m² + m − 6 = 0 → (m + 3)(m − 2) = 0 → m = −3, 2。因此 y_c = A e^(−3x) + B e^(2x)。注意到 e^(2x) 已出现在 y_c 中，因此标准试探 y_p = C e^(2x) 会失效。乘以 x：尝试 y_p = C x e^(2x)。求导：y_p‘ = C e^(2x) + 2C x e^(2x)，y_p” = 4C e^(2x) + 4C x e^(2x)。代入 ODE：(4C e^(2x) + 4C x e^(2x)) + (C e^(2x) + 2C x e^(2x)) − 6(C x e^(2x)) = 10 e^(2x)。合并 e^(2x) 项：(4C + C) = 5C → 5C = 10，所以 C = 2。x e^(2x) 项正如预期的那样抵消了。因此 y_p = 2x e^(2x)。通解为 y = A e^(−3x) + B e^(2x) + 2x e^(2x)。

12. Common Mistakes and Exam Tips | 常见错误与应试技巧

Always write the homogeneous part first. Many candidates lose marks by trying to guess the particular integral before analysing the complementary function, leading to invalid trial forms and wasted time.

务必先求出齐次部分的补函数。许多考生在分析补函数之前就急于猜测特解形式，导致无效的试探函数和时间的浪费。
When roots are complex, remember to include both cosine and sine. Writing only the real part or forgetting the factor e^(αx) is a frequent error.

当根为复数时，记住要同时包含余弦和正弦。只写实部或遗漏因子 e^(αx) 是常见错误。
In the particular integral for trigonometric f(x), always use both P cos(qx) and Q sin(qx), even if f(x) contains only sine or only cosine. The derivative generates both.

对于三角函数 f(x) 的特解，即便 f(x) 只含正弦或只含余弦，也要同时使用 P cos(qx) 和 Q sin(qx)，因为求导会生成两者。
Check for overlap between y_p trial and y_c. If any term in your trial appears in the complementary function, multiply the entire trial by x (or x² for repeated overlap).

检查试探特解 y_p 与 y_c 是否有重叠。若试探函数中的任何项出现在补函数中，则将整个试探式乘以 x（重复重叠则乘以 x²）。
When initial conditions are given, don’t forget to include the particular integral in the differentiation. Losing the PI contribution in y’ is a very common error when solving for constants.

当给出初始条件时，不要忘记在求导时包含特解部分。在求解常数时漏掉 y’ 中的特解贡献是一个非常常见的错误。

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Second-Order Differential Equations for CIE A-Level | CIE A-Level 二阶微分方程考点精讲