📚 The General Solutions of Forced Equations | 受迫方程的通解
Second-order linear differential equations with a non-zero forcing term appear throughout the IB Analysis & Approaches HL syllabus. Learning to construct their general solution by combining the complementary function and a particular integral is an essential skill, and it forms the mathematical backbone for modelling oscillatory systems, electrical circuits and many other dynamic processes.
带有非零强制项的二阶线性微分方程贯穿 IB 分析和方法（AA）高等级课程。学会将互补函数与特积分组合来构造通解是一项核心技能，它构成了振动系统、电路和许多其他动态过程建模的数学基础。
1. Understanding Forced Differential Equations | 理解受迫微分方程
In IB Mathematics, a second-order linear ODE is usually written as a y'' + b y' + c y = f(x), where a, b, c are constants and a ≠ 0. When the right-hand side f(x) is identically zero, the equation is homogeneous; otherwise it is called a forced or non-homogeneous equation. The function f(x) represents an external driving influence – a force, a voltage source or an input signal.
在 IB 数学中，二阶线性常微分方程通常写作 a y'' + b y' + c y = f(x)，其中 a、b、c 为常数且 a ≠ 0。当右端 f(x) 恒为零时，方程为齐次方程；否则称为受迫方程或非齐次方程。函数 f(x) 表示外部驱动力——可以是力、电压源或输入信号。
The term “forced” originates from mechanics, where an external force causes a system to oscillate. In the IB syllabus you will meet forced equations in problems ranging from spring–mass systems to simple electronic circuits. The same solution strategy applies to all such equations, regardless of the physical context.
“受迫”一词来源于力学，即外力使系统发生振动。在 IB 课程中，你会遇到从弹簧–质量系统到简单电子电路等各类受迫方程。无论物理背景如何，相同的求解策略适用于所有这类方程。
2. Structure of the General Solution | 通解的结构
The general solution of a forced equation is always the sum of two parts: y(x) = yc(x) + yp(x). The complementary function yc solves the associated homogeneous equation a y'' + b y' + c y = 0 and contains two arbitrary constants. The particular integral yp is any one function that satisfies the full non-homogeneous equation.
受迫方程的通解总是由两部分之和构成：y(x) = yc(x) + yp(x)。互补函数 yc 满足对应的齐次方程 a y'' + b y' + c y = 0，包含两个任意常数。特积分 yp 是满足完整非齐次方程的任意一个函数。
This additive structure works because the ODE is linear – the operator L[y] = a y'' + b y' + c y satisfies L[yc + yp] = L[yc] + L[yp ten asked to write the general solution and then use initial conditions to find the particular solution that fits a given scenario.
这种可加性源于方程是线性的——算子 L[y] = a y'' + b y' + c y 满足 L[yc + yp] = L[yc] + L[yp] = 0 + f(x)。在考试中，经常要求先写出通解，再通过初始条件求出符合具体情景的特解。
3. The Complementary Function yc | 互补函数 yc
To find yc we solve the homogeneous equation. Assume a trial solution y = eλx (using the exponential form). Substituting yields the auxiliary equation a λ² + b λ + c = 0. The nature of the roots λ determines the form of yc.
为求得 yc，需解齐次方程。假设试解为 y = eλx，代入可得辅助方程 a λ² + b λ + c = 0。根 λ 的性质决定了 yc 的形式。
There are three classic cases:
- Two distinct real roots λ₁, λ₂: yc = C₁ eλ₁x + C₂ eλ₂x.
- Repeated real root λ: yc = (C₁ + C₂ x) eλx.
- Complex conjugate roots α ± iβ: yc = eαx (C₁ cos βx + C₂ sin βx).
三种经典情况：
- 两个不等实根 λ₁、λ₂：yc = C₁ eλ₁x + C₂ eλ₂x。
- 重实根 λ：yc = (C₁ + C₂ x) eλx。
- 共轭复根 α ± iβ：yc = eαx (C₁ cos βx + C₂ sin βx)。
The constants C₁ and C₂ are only determined once we have the full general solution and can apply initial or boundary conditions. In forced problems, the homogeneous part describes the system’s natural (transient) behaviour, while the particular integral captures the steady-state response to the forcing term.
常数 C₁ 和 C₂ 只能在得到完整通解并利用初始或边界条件后才能确定。在受迫问题中，齐次部分描述系统固有的（暂态）行为，而特积分则反映系统对受迫项的稳态响应。
4. Particular Integral: Method of Undetermined Coefficients | 特积分：待定系数法
We construct yp by looking at the form of f(x) and choosing a trial function that has the same functional shape but with unknown coefficients. The trial function is then substituted into the ODE and the coefficients are found by equating terms.
我们通过观察 f(x) 的形式来构造 yp，选取与其函数形状相同但带有未知系数的试解函数。然后将试解代入原方程，并逐项比较系数以确定未知值。
The table below summarises the standard trial choices that are most common in IB problems:
| Form of f(x) | Trial yp (first guess) |
|---|---|
| Polynomial of degree n | General polynomial of degree n: An xⁿ + … + A₁ x + A₀ |
| C ekx | A ekx |
| C sin(ωx) or D cos(ωx) or both | A sin(ωx) + B cos(ωx) |
下表中总结了 IB 习题中最常见的标准试解选择：
| f(x) 的形式 | 试解 yp（初步猜测） |
|---|---|
| n 次多项式 | n 次一般多项式：An xⁿ + … + A₁ x + A₀ |
| C ekx | A ekx |
| C sin(ωx) 或 D cos(ωx) 或两者组合 | A sin(ωx) + B cos(ωx) |
A crucial extra rule: if any part of the trial yp already appears in the complementary function yc, multiply the whole trial by x (or by x² if the root is repeated) until it no longer clashes with yc. This modification is needed for resonance and repeated-root cases.
一条关键的额外规则：若试解 yp 中的任何部分已在互补函数 yc 中出现，则需将整个试解乘以 x（若根为重根则乘以 x²），直至与 yc 不再冲突。共振和重根情形都必须进行这一修正。
5. Worked Example: Polynomial Forcing | 案例详解：多项式强制项
Consider the equation y'' - 3y' + 2y = 4x². First, the auxiliary equation λ² - 3λ + 2 = 0 gives λ = 1, 2, so yc = C₁ eˣ + C₂ e²ˣ.
考虑方程 y'' - 3y' + 2y = 4x²。首先，辅助方程 λ² - 3λ + 2 = 0 解得 λ = 1, 2，因此 yc = C₁ eˣ + C₂ e²ˣ。
Since f(x) = 4x², a second-degree polynomial, we try yp = A x² + B x + C. Substitute into the ODE: yp'' = 2A, yp' C) = 4x². Collecting coefficients: 2A x² + ( -6A + 2B ) x + (2A - 3B + 2C) = 4x² + 0x + 0.
由于 f(x) = 4x² 为二次多项式，我们设试解 yp = A x² + B
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