📚 Second Order Differential Equations | 二阶微分方程考点精讲
Second order differential equations are a core topic in the Edexcel IGCSE Further Pure Mathematics syllabus. These equations involve the second derivative of a function and appear frequently in mechanics, oscillations, and electrical circuits. Mastering the solution techniques for linear equations with constant coefficients is essential for exam success, as questions often integrate algebraic manipulation, trigonometric identities, and initial conditions. This article will walk you through the key concepts, from the auxiliary equation to particular integrals, using clear step-by-step methods.
二阶微分方程是 Edexcel IGCSE 进阶纯数学大纲中的核心内容。这类方程包含函数的二阶导数，在力学、振动以及电路分析中经常出现。掌握常系数线性方程的求解技巧对于考试至关重要，因为考题往往综合了代数运算、三角恒等式和初始条件。本文将带你逐一攻克从辅助方程到特积分的关键概念，并用清晰的分步方法加以讲解。
1. Introduction to Second Order ODEs | 二阶常微分方程简介
A second order ordinary differential equation (ODE) involves an unknown function y(x) and its derivatives up to the second order, typically written as a relation among x, y, dy/dx, and d²y/dx². In the IGCSE Further Pure syllabus, we focus on linear equations with constant coefficients, where the terms only contain y and its derivatives multiplied by constants, plus a possible function of x on the right-hand side.
二阶常微分方程涉及未知函数 y(x) 以及它的一阶和二阶导数，通常表示为 x、y、dy/dx 和 d²y/dx² 之间的关系。在 IGCSE 进阶纯数学大纲中，我们重点研究常系数线性方程，即方程中 y 及其导数项只乘以常数，等号右边可能还有一个关于 x 的函数。
These equations are called ‘linear’ because y and its derivatives appear to the first power and are not multiplied together. The general form is a(d²y/dx²) + b(dy/dx) + cy = f(x), where a, b, and c are constants, and f(x) is a given function. If f(x) = 0, the equation is homogeneous; otherwise, it is non-homogeneous.
这类方程之所以称为“线性”，是因为 y 及其导数均以一次幂出现，且彼此不相互乘在一起。其一般形式为 a(d²y/dx²) + b(dy/dx) + cy = f(x)，其中 a、b、c 为常数，f(x) 为给定函数。若 f(x) = 0，方程是齐次的；否则为非齐次方程。
2. General Form and Classification | 一般形式与分类
The standard form we work with is a d²y/dx² + b dy/dx + c y = f(x). Here a, b, c are real constants with a ≠ 0. The order is determined by the highest derivative present—in this case, the second derivative. The equation is homogeneous when f(x) = 0; otherwise, it is non-homogeneous. Homogeneous equations are solved using the auxiliary equation, while non-homogeneous equations require an additional particular integral.
我们学习的一般形式为 a d²y/dx² + b dy/dx + c y = f(x)。其中 a、b、c 为实常数，且 a ≠ 0。方程的阶数由出现的最高阶导数决定——此处为二阶导数。当 f(x) = 0 时，方程为齐次方程；否则为非齐次方程。齐次方程可通过辅助方程求解，而非齐次方程还需要额外求出一个特积分。
Classification of solutions depends on the roots of the auxiliary equation ar² + br + c = 0. The roots may be real and distinct, real and equal (repeated), or complex conjugates. Each case yields a different form for the complementary function, which forms the general solution of the homogeneous equation.
解的分类取决于辅助方程 ar² + br + c = 0 的根。根可以是两不等实根、两相等实根（重根）或一对共轭复根。每种情况给出的互补函数形式不同，互补函数构成了齐次方程的通解。
3. Homogeneous Equations with Constant Coefficients | 常系数齐次方程
For a homogeneous equation a d²y/dx² + b dy/dx + c y = 0, we assume a solution of the form y = e^(rx), where r is a constant to be determined. Substituting this trial solution gives the auxiliary equation a r² + b r + c = 0. Solving this quadratic gives the values of r that determine the complementary function yc.
对于齐次方程 a d²y/dx² + b dy/dx + c y = 0，我们假设解的形式为 y = e^(rx)，其中 r 为待定常数。将此试解代入方程可得到辅助方程 a r² + b r + c = 0。求解该二次方程便能得到 r 的值，从而确定互补函数 yc。
This method works because the exponential function is proportional to its own derivatives. The complementary function is the general solution to the homogeneous equation and contains two arbitrary constants, reflecting the second order of the equation. The constants are later determined if initial or boundary conditions are given.
该方法之所以有效，是因为指数函数与其各阶导数成比例。互补函数是齐次方程的通解，包含两个任意常数，这与方程的二阶性质相对应。如果给出了初始条件或边界条件，这两个常数随后便可确定。
4. The Auxiliary Equation | 辅助方程
The auxiliary equation (also called the characteristic equation) is a r² + b r + c = 0. Solving it is the crucial first step. The discriminant Δ = b² – 4ac determines the nature of the roots. Three cases arise: Δ > 0 (two distinct real roots), Δ = 0 (one repeated real root), and Δ < 0 (two complex conjugate roots).
辅助方程（也称特征方程）为 a r² + b r + c = 0。求解该方程是关键的第一步。判别式 Δ = b² – 4ac 决定了根的性质。共有三种情形：Δ > 0（两个不等实根），Δ = 0（一个重实根）以及 Δ < 0（一对共轭复根）。
For each case, the complementary function has a specific structure. It is vital to memorise these forms, as errors in setting up yc will propagate through the entire solution. Edexcel IGCSE questions often ask you to write down the auxiliary equation explicitly and then find the complementary function, so always show this step clearly.
针对每种情形，互补函数都有着特定的结构。务必牢记这些形式，因为如果在设定 yc 时出错，整个求解过程都将出错。Edexcel IGCSE 考题经常会要求考生明确写出辅助方程，然后求出互补函数，因此一定要清晰展示这一步骤。
5. Real and Distinct Roots | 实且不等根
When the auxiliary equation yields two distinct real roots r₁ and r₂, the complementary function is yc = A e^(r₁ x) + B e^(r₂ x), where A and B are arbitrary constants. This is the simplest case. For example, to solve d²y/dx² – 5 dy/dx + 6y = 0, the auxiliary equation r² – 5r + 6 = 0 gives roots r = 2 and r = 3. Hence the general solution is y = A e^(2x) + B e^(3x).
当辅助方程有两个不等实根 r₁ 和 r₂ 时，互补函数为 yc = A e^(r₁ x) + B e^(r₂ x)，其中 A 与 B 为任意常数。这是最简单的情形。例如，要求解 d²y/dx² – 5 dy/dx + 6y = 0，辅助方程 r² – 5r + 6 = 0 的根为 r = 2 和 r = 3。因此通解为 y = A e^(2x) + B e^(3x)。
In exam problems, you might need to find the constants A and B by applying initial conditions such as y(0) = 2 and y'(0) = 5. Always differentiate the general solution first before substituting conditions for the derivative. Clearly label your working to avoid arithmetic slips.
在考题中，你可能需要通过应用初始条件（如 y(0) = 2 和 y'(0) = 5）来求出常数 A 和 B。在代入关于导数的条件之前，务必先对通解求导。清晰地标注计算过程，避免运算失误。
6. Repeated Roots | 重根
If the discriminant is zero, the auxiliary equation has a repeated real root r = –b/(2a). In this case, simply writing y = A e^(r x) + B e^(r x) would not give two independent solutions. Instead, the complementary function is yc = (A + B x) e^(r x). The extra factor of x is essential for linear independence.
若判别式为零，辅助方程有一个重实根 r = –b/(2a)。此时，若只写成 y = A e^(r x) + B e^(r x) 将无法得到两个独立的解。正确的互补函数形式为 yc = (A + B x) e^(r x)。这里多出来的因子 x 对于保证解的线性无关性至关重要。
For example, for d²y/dx² – 4 dy/dx + 4y = 0, the auxiliary equation r² – 4r + 4 = 0 gives (r – 2)² = 0, so r = 2 (repeated). The general solution is y = (A + B x) e^(2x). This form appears frequently, and many students forget the x term, so double-check whenever Δ = 0.
例如，对于方程 d²y/dx² – 4 dy/dx + 4y = 0，辅助方程 r² – 4r + 4 = 0 化为 (r – 2)² = 0，所以 r = 2（重根）。通解为 y = (A + B x) e^(2x)。这种形式经常出现，而许多学生会忘记 x 这一项，因此每当 Δ = 0 时都要仔细核对。
7. Complex Roots | 复数根
When Δ < 0, the auxiliary equation has complex conjugate roots of the form r = α ± i β, where α and β are real numbers and i = √(–1). The complementary function is then yc = e^(α x) (A cos(β x) + B sin(β x)). This combination arises from Euler’s formula and is the standard real-valued form required in the syllabus.
当 Δ < 0 时，辅助方程具有一对共轭复根，形式为 r = α ± i β，其中 α 和 β 为实数，i = √(–1)。互补函数则为 yc = e^(α x) (A cos(β x) + B sin(β x))。这一组合源自欧拉公式，也是大纲要求的实数形式标准答案。
For instance, the equation d²y/dx² + 2 dy/dx + 5y = 0 gives auxiliary equation r² + 2r + 5 = 0 with roots r = –1 ± 2i. Here α = –1, β = 2, so the general solution is y = e^(–x) (A cos(2x) + B sin(2x)). Do not leave the answer in terms of complex exponentials unless the question specifically allows it.
例如，方程 d²y/dx² + 2 dy/dx + 5y = 0 的辅助方程为 r² + 2r + 5 = 0，其根为 r = –1 ± 2i。此时 α = –1，β = 2，因此通解为 y = e^(–x) (A cos(2x) + B sin(2x))。除非题目明确允许，否则不要将答案保留为复指数形式。
8. Particular Integrals for Non-Homogeneous Equations | 非齐次方程的特解
For a non-homogeneous equation a d²y/dx² + b dy/dx + c y = f(x), the general solution is y = yc + yp, where yc is the complementary function (found by setting f(x) = 0) and yp is any particular integral that satisfies the full equation. The particular integral depends on the form of f(x).
对于非齐次方程 a d²y/dx² + b dy/dx + c y = f(x)，其通解为 y = yc + yp，其中 yc 是互补函数（令 f(x) = 0 求得），而 yp 是满足完整方程的任意一个特积分。特积分的形式取决于 f(x) 的形式。
The method of undetermined coefficients is used to find yp. You guess a form for yp based on f(x), with undetermined constants, substitute into the differential equation, and equate coefficients to find the constants. Common f(x) types include polynomials, exponentials, sine/cosine, and combinations of these.
我们使用待定系数法来求特积分 yp。依据 f(x) 的形式假设一个带有待定常数的 yp 试解，将其代入微分方程，然后通过比较系数解出常数。常见的 f(x) 类型包括多项式、指数函数、正弦/余弦以及它们的组合。
9. Method of Undetermined Coefficients | 待定系数法
This method requires a suitable trial function. For a polynomial f(x) of degree n, try a polynomial of the same degree with unknown coefficients. For f(x) = e^(kx), try yp = C e^(kx), unless e^(kx) already appears in yc, in which case multiply by x (or x²). For f(x) = p cos(ω x) + q sin(ω x), try yp = C cos(ω x) + D sin(ω x), again multiplying by x if resonance occurs.
该方法需要选用一个合适的试函数。若 f(x) 是 n 次多项式，可尝试同次数的带有未知系数的多项式。若 f(x) = e^(kx)，则尝试 yp = C e^(kx)，除非 e^(kx) 已经出现在 yc 中，此时需乘以 x（或 x²）。若 f(x) = p cos(ω x) + q sin(ω x)，则尝试 yp = C cos(ω x) + D sin(ω x)，同样，若发生共振则需乘以 x。
After substituting yp and its derivatives, gather like terms and equate coefficients of corresponding functions on both sides of the equation. Solve the resulting simultaneous equations for the unknown constants. For example, to find the particular integral for d²y/dx² – 3 dy/dx + 2y = 4 e^(3x), try yp = C e^(3x), leading to (9C – 9C + 2C) e^(3x) = 4 e^(3x), giving C = 2, so yp = 2 e^(3x).
代入 yp 及其导数后，合并同类项，并让等式两边对应函数的系数相等。解出由此产生的关于未知常数的方程组。例如，要求 d²y/dx² – 3 dy/dx + 2y = 4 e^(3x) 的特积分，可设 yp = C e^(3x)，代入后得 (9C – 9C + 2C) e^(3x) = 4 e^(3x)，解出 C = 2，因此 yp = 2 e^(3x)。
10. Initial and Boundary Conditions | 初值条件与边界条件
To determine the arbitrary constants in the general solution, additional information is required. In IGCSE Further Pure, you will encounter initial conditions, such as y(x₀) = y₀ and y'(x₀) = y₁, typically given at x = 0. Boundary conditions specify values at two different points, but initial conditions are more common.
为确定通解中的任意常数，需要额外提供信息。在 IGCSE 进阶纯数学中，你会遇到初值条件，例如 y(x₀) = y₀ 和 y'(x₀) = y₁，通常给定在 x = 0 处。边界条件则会指定两个不同点上的值，不过初值条件更为常见。
First write the general solution y = yc + yp (with constants A and B). If conditions are at x = 0, substituting x = 0 often simplifies exponential and trigonometric terms. Then differentiate to obtain y’, apply the second condition, and solve simultaneously for A and B. Always check that your final specific solution satisfies both the equation and the conditions.
首先写出通解 y = yc + yp（含常数 A 和 B）。若条件给定在 x = 0 处，代入 x = 0 通常能简化指数项和三角项。接着求导得到 y’，代入第二个条件，并联立求解 A 和 B。务必检查最终的特解是否同时满足方程和所给条件。
11. Applications in Mechanics | 力学中的应用
Second order differential equations model simple harmonic motion (SHM) and damped oscillations. For a mass-spring system, the equation is m d²x/dt² + c dx/dt + kx = 0, where m is mass, c is damping constant, and k is spring stiffness. The roots of the auxiliary equation indicate over-damped, critically damped, or under-damped motion.
二阶微分方程可以用来描述简谐运动（SHM）和阻尼振动。对于一个质量-弹簧系统，其方程为 m d²x/dt² + c dx/dt + kx = 0，其中 m 为质量，c 为阻尼常数，k 为弹簧刚度。辅助方程的根可指示系统处于过阻尼、临界阻尼或欠阻尼状态。
Over-damping (two distinct real roots) returns to equilibrium without oscillation. Critical damping (repeated root) gives the fastest return without oscillation. Under-damping (complex roots) produces decaying oscillations, described by x = e^(–α t) (A cos(β t) + B sin(β t)). IGCSE questions may ask you to interpret these physical situations or find the displacement as a function of time.
过阻尼（两不等实根）会使系统无振荡地回到平衡位置。临界阻尼（重根）则在不发生振荡的情况下实现最快的回归。欠阻尼（复根）会产生逐渐衰减的振动，其位移可表示为 x = e^(–α t) (A cos(β t) + B sin(β t))。IGCSE 考题可能会要求你解读这些物理情境，或者求出位移关于时间的函数。
12. Exam Tips and Common Mistakes | 考试技巧与常见错误
A frequent error is miswriting the auxiliary equation, especially when a = 1 is not the case. Always write down a r² + b r + c = 0 precisely. Another pitfall is forgetting to multiply by x when the trial particular integral duplicates a term in the complementary function. Always check for overlap before finalising yp.
一个常见错误是写错辅助方程，尤其是在 a ≠ 1 的情况下。务必准确写出 a r² + b r + c = 0。另一个易错点是，当试用的特积分与互补函数中的项重复时，忘记乘以 x。在最终确定 yp 之前，一定要检查是否存在重叠。
When dealing with initial conditions, differentiate the entire general solution (including yp) before substituting. Many students mistakenly differentiate only the complementary function. Also, ensure you clearly state the general solution as y = yc + yp, not just one part. Finally, if the question asks for a specific solution, the final answer must be free of arbitrary constants A and B.
在处理初值条件时，要先对完整的通解（包括 yp）求导，然后再代入条件。许多学生错误地只对互补函数求导。此外，一定要清楚地写出通解为 y = yc + yp，而不是只写出其中一部分。最后，若题目要求给出特解，最终的答案中不能含有任意常数 A 和 B。
A systematic approach is key: (1) Write the auxiliary equation, (2) find yc based on roots, (3) determine the form of yp and find it, (4) write the general solution y = yc + yp, (5) apply conditions to find constants. Show all these steps clearly to maximise method marks.
一个系统性的解题思路至关重要：（1）列出辅助方程；（2）根据根的情况求出 yc；（3）确定 yp 的形式并求出它；（4）写出通解 y = yc + yp；（5）应用条件求出常数。清晰地写出所有这些步骤，以最大程度地获取步骤分。
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