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

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

Second order differential equations are a key topic in the WJEC GCSE Further Mathematics specification. They extend the idea of first order equations by including the second derivative, allowing us to model more complex systems such as oscillations, mechanical vibrations and electrical circuits. This revision guide covers all the core methods you need: solving homogeneous linear equations with constant coefficients, handling the three cases of the auxiliary equation, finding particular integrals for non‑homogeneous equations, and using initial conditions to pin down arbitrary constants.

二阶微分方程是 WJEC GCSE 进阶数学大纲中的重点内容。它们在包含一阶导数的基础上引入了二阶导数，使我们能够描述振动、机械运动和电路等更复杂的系统。本文涵盖了全部核心方法：求解常系数齐次线性方程、处理辅助方程的三种情况、为非齐次方程求特解，以及运用初始条件确定任意常数。

1. Introduction to Second Order ODEs | 二阶常微分方程简介

nary differential equation (ODE) involves an unknown function y(x), its first derivative y' and its second derivative y''. The most general form for the equations we will study is a y'' + b y' is a given function. When f(x) = 0 the equation is called homogeneous; otherwise it is non‑homogeneous.

二阶常微分方程包含未知函数 y(x) 及其一阶导数 y' 和二阶导数 y''。我们所要学习的方程的最一般形式为 a y'' + b y' + c y = f(x)，其中 a、b、c 为常数，f(x) 为已知函数。当 f(x) = 0 时，方程称为齐次方程，否则为非齐次方程。

The solution to a second order ODE always contains two arbitrary constants. To find a unique solution we need two extra conditions, usually given as initial values for y and y' at a certain point, or as boundary conditions.

二阶常微分方程的通解总是含有两个任意常数。要得到唯一解，我们需要两个附加条件，通常给出在某点的 y 和 y' 的初始值，或给出边界条件。

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2. Standard Form of Homogeneous Equations | 齐次方程的标准形式

A homogeneous linear second order ODE with constant coefficients can be written as:

常系数齐次线性二阶微分方程可写为：

a y'' + b y' + c y = 0

where a ≠ 0, and b and c are constants. For the methods that follow it is essential that the equation is in this form, with all terms on one side and zero on the other. If a term like 3y'' appears, divide through by the coefficient so that the equations match the standard solving formulas.

其中 a ≠ 0，b 和 c 为常数。对于后续的方法，方程必须保持这种形式：所有项在一边，另一边为零。如果出现类似 3y'' 的项，应将整个方程除以该系数，使其符合标准的求解公式。

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3. The Characteristic (Auxiliary) Equation | 特征（辅助）方程

To solve the homogeneous equation we try a solution of the form y = e^mx. Substituting y = e^mx, y' = m e^mx and y'' = m² e^mx into a y'' + b y' + c y = 0 yields:

为求解齐次方程，我们假设解的形式为 y = e^mx。将 y = e^mx，y' = m e^mx 和 y'' = m² e^mx 代入 a y'' + b y' + c y = 0 得到：

a m² + b m + c = 0

This quadratic in m is called the characteristic equation (or auxiliary equation). Its roots determine the form of the general solution.

这个关于 m 的二次方程称为特征方程（或辅助方程）。它的根决定了通解的形式。

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4. Solving the Characteristic Equation | 求解特征方程

The characteristic equation a m² + b m + c = 0 is an ordinary quadratic. Its discriminant Δ = b² – 4ac tells us which case we are dealing with. The three possibilities are:

特征方程 a m² + b m + c = 0 是一个普通的二次方程。其判别式 Δ = b² – 4ac 决定了我们面对的是哪一种情况。共有三种可能：

Δ > 0: two distinct real roots, m₁ and m₂.
Δ = 0: one repeated real root, m (twice).
Δ < 0: a pair of complex conjugate roots, α ± iβ.

Δ > 0：两个不相等的实根 m₁ 和 m₂。
Δ = 0：一个重复实根 m（二重根）。
Δ < 0：一对共轭复根 α ± iβ。

It is crucial to compute the discriminant carefully and then select the correct form of the general solution.

仔细计算判别式并选择正确的通解形式至关重要。

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5. Case 1: Real Distinct Roots | 情况 1：实且不相等的根

When the characteristic equation gives two different real numbers m₁ and m₂, the general solution of the homogeneous equation is:

当特征方程给出两个不同的实数 m₁ 和 m₂ 时，齐次方程的通解为：

y = A e^m₁x + B e^m₂x

where A and B are arbitrary constants. Both terms are real exponential functions. If m₁ or m₂ is negative, the corresponding term decays as x increases; if positive, that term grows.

其中 A 与 B 为任意常数。这两项均为实指数函数。如果 m₁ 或 m₂ 为负，相应的项随 x 增大而衰减；若为正，该项将增长。

For example, if the characteristic equation is m² – 5m + 6 = 0 with roots m = 2 and m = 3, the general solution is y = A e^2x + B e^3x.

例如，若特征方程为 m² – 5m + 6 = 0，根为 m = 2 和 m = 3，则通解为 y = A e^2x + B e^3x。

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6. Case 2: Repeated Roots | 情况 2：重根

When Δ = 0 the characteristic equation has a single repeated root m = –b/(2a). The general solution is then:

当 Δ = 0 时，特征方程有一个重根 m = –b/(2a)。此时通解为：

y = (A + B x) e^mx

Notice the extra factor of x multiplying one of the arbitrary constants. This ensures the two parts of the solution are linearly independent.

注意这里多了一个因子 x 与一个任意常数相乘。这确保了解的两个部分线性无关。

For example, the equation y'' – 6y' + 9y = 0 gives characteristic equation m² – 6m + 9 = 0, with repeated root m = 3. The general solution is y = (A + B x) e^3x.

例如，方程 y'' – 6y' + 9y = 0 的特征方程为 m² – 6m + 9 = 0，重根 m = 3，通解为 y = (A + B x) e^3x。

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7. Case 3: Complex Conjugate Roots | 情况 3：共轭复根

ant is negative, the roots are complex and occur as a conjugate pair m = α ± iβ, where α = –b/(2a) and β = √4ac – b²/(2a). The general solution is written in terms of real functions as:

若判别式为负，根为复数，且以共轭对 m = α ± iβ 的形式出现，其中 α = –b/(2a)，β = √4ac – b²/(2a)。通解用实函数表示为：

y = e^αx (A cos(βx) + B sin(βx))

The expression e^αx controls the amplitude growth or decay, while the sine and cosine terms give oscillatory behaviour.

其中 e^αx 控制振幅的增减，而正弦与余弦项提供振荡行为。

For example, y'' + 4y' + 13y = 0 gives m² + 4m + 13 = 0 with roots m = –2 ± 3i. The general solution is y = e^–2x (A cos(3x) + B sin(3x)).

例如，y'' + 4y' + 13y = 0 的特征方程为 m² + 4m + 13 = 0，根为 m = –2 ± 3i，通解为 y = e^–2x (A cos(3x) + B sin(3x))。

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8. General Solution Summary | 通解总结

The three cases can be summarised in a single table for quick reference:

三种情况可总结为下表以便快速查阅：

Discriminant Δ	Roots of a m2 + b m + c = 0	General Solution
Δ > 0	m₁, m₂ (real, distinct)	y = A em₁x + B em₂x
Δ = 0	m (repeated)	y = (A + B x) emx
Δ < 0	α ± iβ	y = eαx (A cos βx + B sin βx)

Always write the solution with the arbitrary constants A and B; these will later be determined by initial or boundary conditions.

解中必须始终写出任意常数 A 与 B，随后将由初始条件或边界条件对其加以确定。

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9. Non‑Homogeneous Equations: Particular Integrals | 非齐次方程：特解

When the ODE has a non‑zero right‑hand side, we must find the complete solution as the sum of two parts:

当常微分方程右端非零时，完全解必须表示为两部分之和：

y = y_c + y_p

Here y_c ary function, i.e. the general solution of the associated homogeneous equation (f(x)=0). y_p is a particular integral, which is any single function that satisfies the full non‑homogeneous equation.

这里 y_c 为余函数，即相应齐次方程 (f(x)=0) 的通解；y_p 为一个特解，它是满足整个非齐次方程的某一个特定函数。

The complementary function always contains two arbitrary constants, while the particular integral contains no arbitrary constants.

余函数总是包含两个任意常数，而特解不包含任意常数。

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10. Method of Undetermined Coefficients | 待定系数法

To find a particular integral y_p, we guess a form that resembles f(x) and contains unknown coefficients. The most common right‑hand sides and their trial functions are:

为求特解 y_p，我们需要猜想一个与 f(x) 形式相似、但包含待定系数的函数。最常见的右端项及其试函数如下：

If f(x) is a polynomial of degree n, try y_p = a polynomial of degree n with unknown coefficients.
If f(x) = k e^px, try y_p = C e^px.
If f(x) = C cos(px) + D sin(px), try y_p = E cos(px) + F sin(px).

若 f(x) 为 n 次多项式，则试取 y_p 为相同次数的多项式，其系数待定。
若 f(x) = k e^px，则试取 y_p = C e^px。
若 f(x) = C cos(px) + D sin(px)，则试取 y_p = E cos(px) + F sin(px)。

Substitute the trial y_p into the ODE and equate coefficients to find the unknown values. If the standard trial function appears in the complementary function, multiply the trial by x (or x² if it is a repeated root case). This is called the modification rule.

将试函数 y_p 代入原常微分方程，并通过比较系数求出未知值。若标准试函数恰好出现在余函数中，则需将试函数乘以 x（如果是重根情况则乘以 x²）。这称为修正规则。

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11. Applying Initial Conditions | 应用初始条件

Once the general solution y = y_c + y_p is found, we use the given conditions (often values of y and y' at x = 0) to solve for A and B. The steps are:

得到通解 y = y_c + y_p 后，我们需要利用给定条件（通常为 x = 0 时 y 与 y' 的值）来求解 A 和 B。具体步骤为：

1. Write the complete expression for y.
2. Differentiate to find y'.
3. Substitute the x‑value (often 0) into both y and y'.
4. Set up two simultaneous equations for A and B.
5. Solve to obtain the particular solution.

1. 写出 y 的完全表达式。
2. 求导得到 y'。
3. 将给定的 x 值（常为 0）代入 y 与 y'。
4. 建立关于 A 和 B 的两个联立方程。
5. 解出 A 和 B，得到特解。

Always check your final particular solution by substituting back into the original differential equation to ensure it satisfies both the ODE and the conditions.

最后务必将所得特解代回原微分方程，检查它是否同时满足方程与给定条件。

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12. Exam Tips and Common Mistakes | 考试技巧与常见错误

• Double‑check the characteristic equation: ensure you have correctly identified a, b and c, and that the ODE is set equal to zero before writing the auxiliary equation.
• Do not forget the x multiplier in the repeated root case – missing it is one of the most frequent errors.
• In the complex root case, make sure you write the solution using real sine and cosine, not complex exponentials, unless specifically asked otherwise.
• When using undetermined coefficients, check whether your trial function already appears in the complementary function. If it does, multiply by x.
• Show clear, step‑by‑step working when finding constants from initial conditions; many marks are allocated for the method.

• 仔细核对特征方程：确保正确识别 a、b 和 c，并确认已将常微分方程化为等于零的标准形式后再写辅助方程。
• 不要忘记重根情况中乘入 x 因子——遗漏这一点是极常见的错误。
• 在复根情况下，务必使用实正弦和余弦函数写出解，除非题目特别要求使用复指数形式。
• 使用待定系数法时，检查试函数是否已出现在余函数中；若已出现，应乘以 x。
• 由初始条件求常数时，要清晰地展示每一步推导过程；解题方法本身就占有大量分值。

With regular practice of each case, you will be able to handle any second order ODE that appears in the WJEC GCSE Further Mathematics examination confidently and efficiently.

通过反复练习每一种情况，你将能够在 WJEC GCSE 进阶数学考试中自信且高效地解决任何二阶常微分方程问题。

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