Complementary Functions and Particular Integrals | 余函数与特解

📚 Complementary Functions and Particular Integrals | 余函数与特解

In A-Level Further Mathematics, solving second-order linear differential equations with constant coefficients is a core skill. The general solution of a non-homogeneous equation is formed by adding two parts: the complementary function (CF), which solves the associated homogeneous equation, and a particular integral (PI), which is any single solution to the full non-homogeneous equation.

在 A-Level 进阶数学中，求解常系数二阶线性微分方程是一项核心技能。非齐次方程的通解由两部分相加组成：余函数（CF）是对应齐次方程的解，而特解（PI）是原非齐次方程的任意一个解。

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1. Setting the Stage: The General Form | 基本框架：一般形式

We study equations of the form a^d2y⁄_dx² + bdydx + cy = f(x), where a, b, and c are real constants. When f(x) = 0, the equation is homogeneous; when f(x) ≠ 0, it is non-homogeneous. The complete solution is written as y = CF + PI.

我们研究的方程形式为 a^d2y⁄_dx² + bdydx + cy = f(x)，其中 a、b、c 为实常数。当 f(x) = 0 时，方程为齐次的；当 f(x) ≠ 0 时，方程为非齐次的。完整的解写作 y = CF + PI。

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2. The Auxiliary Equation | 辅助方程

To find the complementary function, we replace each derivative by a power of a variable m. Thus, a^d2y⁄_dx² + bdydx + cy = 0 leads to the auxiliary equation am² + bm + c = 0. Solving this quadratic gives us the roots that determine the form of the CF.

为了求余函数，我们将每阶导数替换为变量 m 的幂。于是 a^d2y⁄_dx² + bdydx + cy = 0 对应辅助方程 am² + bm + c = 0。解这个二次方程得到的根决定了 CF 的形式。

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3. Two Distinct Real Roots | 两个相异实根

If the auxiliary equation has two distinct real roots m₁ and m₂, the complementary function is y_c = Ae^m₁x + Be^m₂x. Here A and B are arbitrary constants determined by boundary or initial conditions.

若辅助方程有两个相异实根 m₁ 和 m₂，余函数为 y_c = Ae^m₁x + Be^m₂x。其中 A 和 B 是由边界条件或初值条件确定的任意常数。

Example: y'' - 3y' + 2y = 0
Auxiliary: m² - 3m + 2 = 0 ⇒ m = 1, 2
CF: y_c = Ae^x + Be^2x

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4. Repeated Real Root | 重根

When the auxiliary equation has a repeated root m = r, the CF becomes y_c = (A + Bx)e^rx. The extra factor x is necessary to produce two linearly independent solutions.

当辅助方程有重根 m = r 时，CF 变为 y_c = (A + Bx)e^rx。额外的因子 x 是为了产生两个线性无关的解。

Example: y'' + 6y' + 9y = 0
m² + 6m + 9 = 0 ⇒ m = -3 (twice)
CF: y_c = (A + Bx)e^-3x

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5. Complex Conjugate Roots | 共轭复根

If the roots are complex conjugate pairs m = p ± qi, the CF is expressed using sine and cosine: y_c = e^px(Acos(qx) + Bsin(qx)). This form is essential for modelling oscillatory systems such as springs and circuits.

若根为共轭复根 m = p ± qi，CF 用正弦和余弦表示：y_c = e^px(Acos(qx) + Bsin(qx))。这种形式对于弹簧、电路等振荡系统的建模至关重要。

Example: y'' - 2y' + 5y = 0
m = 1 ± 2i
CF: y_c = e^x(Acos 2x + Bsin 2x)

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6. Locating the Particular Integral | 确定特解的形式

For a given f(x), we propose a trial solution whose form matches f(x). For example, if f(x) is a polynomial of degree n, try y_p = a_nxⁿ + ⋯ + a₁x + a₀. If f(x) = ke^mx, try y_p = Ce^mx. If f(x) = kcos(ω x) or ksin(ω x), try a combination of sine and cosine.

对于给定的 f(x)，我们尝试一个与 f(x) 形式匹配的试探解。例如，若 f(x) 是 n 次多项式，则设 y_p = a_nxⁿ + ⋯ + a₁x + a₀。若 f(x) = ke^mx，则设 y_p = Ce^mx。若 f(x) = kcos(ω x) 或 ksin(ω x)，则尝试正弦与余弦的组合。

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7. Substitution and Comparison of Coefficients | 代入与比较系数

We substitute the trial solution into the left-hand side of the differential equation and equate coefficients of like terms with f(x). This yields a system of linear equations that determines the unknown constants in the PI.

将试探解代入微分方程左端，并与 f(x) 比较同类项的系数。这样得到一组线性方程，求解这些方程即可确定特解中的未知常数。

Worked Example:
Solve y'' + y' - 2y = 4x.
CF: m² + m - 2 = 0 ⇒ m = 1, -2, so y_c = Ae^x + Be^-2x.
Try y_p = Cx + D. Then y_p' = C, y_p'' = 0.
Substitute: 0 + C - 2(Cx + D) = 4x.
Equate: -2C = 4, C - 2D = 0.
Hence C = -2, D = -1.
General solution: y = Ae^x + Be^-2x - 2x - 1.

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8. When the Standard Trial Fails: Resonance | 标准试探解失效时：共振

If the proposed PI is already part of the CF, it will not work because substitution yields zero on the left-hand side. This is called resonance. In this case, multiply the trial solution by x (or a higher power if necessary) until it is no longer contained in the CF.

若提议的特解已包含在 CF 中，代入后左端恒为零，试探将失效。这称为共振。此时需将试探解乘以 x（必要时乘以更高次幂），直到它不再被包含在 CF 中为止。

Example: y'' + 4y = cos 2x
CF: y_c = Acos 2x + Bsin 2x.
Since cos 2x is in the CF, try y_p = x(Ccos 2x + Dsin 2x).
Using this properly gives a valid PI.

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9. Special Case: f(x) Is a Sum of Terms | 特例：f(x) 为若干项之和

When f(x) consists of multiple distinct types, such as a polynomial plus an exponential, the principle of superposition applies. We find separate particular integrals for each term and add them together.

当 f(x) 由多种不同类型组成，例如多项式加指数函数时，需应用叠加原理。分别求每一项的特解，然后相加即可。

Example: y'' - y = x² + e^3x
PI for x²: y_p1 = ax² + bx + c
PI for e^3x: y_p2 = de^3x
Overall PI: y_p = y_p1 + y_p2

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10. Validity Check and Initial Conditions | 验证与初值条件

After deriving the general solution y = y_c + y_p, it is wise to substitute it back into the original equation to verify correctness. Then, if initial conditions such as y(0) and y'(0) are given, use them to determine A and B.

得到通解 y = y_c + y_p 后，建议将解代回原方程验证正确性。然后，若给定了初值条件如 y(0) 和 y'(0)，可利用这些条件确定 A 和 B。

Summary Table | 总结表

Auxiliary roots | 辅助方程根	Complementary function | 余函数
Distinct real m1, m2 | 相异实根	Aem1x + Bem2x
Repeated real m | 重根	(A + Bx)emx
Complex p ± qi | 复根	epx(Acos qx + Bsin qx)

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