📚 IB Mathematics: Particular Solutions of Differential Equations under Standard Forcing Terms | IB数学：标准强迫项下微分方程的特解
When studying differential equations in IB Mathematics Analysis and Approaches HL, one of the most practical skills is finding a particular solution for a non-homogeneous linear differential equation. This is especially important when the forcing term—the non-zero side of the equation—belongs to a small family of “standard” functions.
在IB数学分析与方法（AA）HL中，求解非齐次线性微分方程时，最实用的技能之一就是寻找特解。尤其是当强迫项（方程中非零的那一侧）属于一小类“标准”函数时，特解的形式可以直接确定。
1. The Goal: Particular Solutions | 目标：求解特解
A non-homogeneous linear differential equation with constant coefficients has the general form
a y” + b y’ + c y = f(x)
where f(x) is called the forcing term. The solution is the sum of the complementary function (the general solution of the homogeneous equation) and one particular solution yₚ.
常系数非齐次线性微分方程的一般形式为
a y” + b y’ + c y = f(x)
其中 f(x) 称为强迫项。其通解等于齐次方程的通解（互补函数）加上一个特解 yₚ。
This article focuses on how to construct yₚ quickly and correctly when f(x) is a polynomial, an exponential, sine/cosine, or a product of these—the standard forcing terms in IB.
本文重点讨论当 f(x) 为多项式、指数函数、正弦/余弦函数，或它们的乘积时，如何快速且正确地构造 yₚ，这些正是IB考试中的标准强迫项。
2. The Standard Forcing Terms | 标准强迫项
The IB syllabus usually expects you to recognise the following families of forcing terms:
- Polynomials: e.g. 5x² + 2x, x³, constants.
- Exponentials: e.g. 3eˣ, 5e⁻²ˣ.
- Sine and cosine: e.g. sin 2x, cos(3x), 4 sin x − 2 cos x.
- Products of these: e.g. x eˣ, eˣ sin 2x, x² cos x.
IB课程大纲通常要求你识别以下类型的强迫项：
- 多项式：如 5x² + 2x、x³、常数。
- 指数函数：如 3eˣ、5e⁻²ˣ。
- 正弦和余弦：如 sin 2x、cos(3x)、4 sin x − 2 cos x。
- 这些函数的乘积：如 x eˣ、eˣ sin 2x、x² cos x。
For each family, we have a predetermined “form” for yₚ, with unknown coefficients called undetermined coefficients.
对于每一类强迫项，我们都有一个预设的 yₚ 形式，其中的未知系数称为“待定系数”。
3. General Solution = Homogeneous Solution + Particular Solution | 通解 = 齐次解 + 特解
Before guessing yₚ, you must first solve the homogeneous equation a y” + b y’ + c y = 0. Its general solution is called the complementary function y_c.
在猜测 yₚ 之前，必须先解齐次方程 a y” + b y’ + c y = 0，其通解称为互补函数 y_c。
For example, if the characteristic equation ar² + br + c = 0 has two distinct real roots r₁ and r₂, then y_c = C₁ e^{r₁x} + C₂ e^{r₂x}. If it has a repeated root r, then y_c = (C₁ + C₂x)e^{rx}. If it has complex roots p ± q i, then y_c = e^{px}(C₁ cos qx + C₂ sin qx).
例如，若特征方程 ar² + br + c = 0 有两个不同实根 r₁ 和 r₂，则 y_c = C₁ e^{r₁x} + C₂ e^{r₂x}。若有重根 r，则 y_c = (C₁ + C₂x)e^{rx}。若有共轭复根 p ± q i，则 y_c = e^{px}(C₁ cos qx + C₂ sin qx)。
The particular solution yₚ is any one function that satisfies the original non-homogeneous equation. Then the general solution is y = y_c + yₚ.
特解 yₚ 是满足原非齐次方程的任意一个函数。通解为 y = y_c + yₚ。
4. Undetermined Coefficients: Core Idea | 待定系数法：核心思想
The method of undetermined coefficients is based on a simple observation: if f(x) is made of standard functions, then derivatives of suitable trial functions are also made of the same types of functions.
待定系数法基于一个简单的事实：如果 f(x) 由标准函数组成，那么合适的试探函数的导数也由同类型的函数组成。
We choose a trial form for yₚ that contains unknown constants (A, B, C, …), substitute it into the differential equation, and then equate coefficients to determine those constants.
我们为 yₚ 选择一个包含未知常数（A、B、C……）的试探形式，将其代入微分方程，然后比较系数以确定这些常数。
The table below shows the basic trial forms for the most common standard forcing terms.
下表展示了最常见标准强迫项对应的基本试探形式。
| Forcing term f(x) | Trial form for yₚ |
| Pₙ(x) (degree n polynomial) | xᵏ · Qₙ(x), where Qₙ is a general polynomial of degree n |
| A e^{mx} | xᵏ · C e^{mx} |
| A sin(ωx) + B cos(ωx) | xᵏ · (C sin(ωx) + D cos(ωx)) |
In the table, the integer k is chosen to be the smallest non-negative integer that makes yₚ not a solution of the homogeneous equation. This is the “resonance correction” discussed later.
表中整数 k 选取为使得 yₚ 不是齐次方程解的最小非负整数，即后面讨论的“共振修正”。
5. Polynomial Forcing Terms | 多项式强迫项
If f(x) is a polynomial of degree n, we normally try yₚ = xᵏ Qₙ(x), where Qₙ is a polynomial of degree n with unknown coefficients.
如果 f(x) 是 n 次多项式，我们通常尝试 yₚ = xᵏ Qₙ(x)，其中 Qₙ 是 n 次多项式，系数未知。
The value of k is determined by how many times 0 is a root of the characteristic equation. If c ≠ 0, then 0 is not a root, so k = 0. If c = 0 but b ≠ 0, then 0 is a simple root, so k = 1. If b = c = 0, then 0 is a double root, so k = 2.
k 的值由特征方程中 0 作为根的重数决定。若 c ≠ 0，则 0 不是根，k = 0。若 c = 0 但 b ≠ 0，则 0 是单根，k = 1。若 b = c = 0，则 0 是二重根，k = 2。
Example: For y” − 3y’ + 2y = 6x, the characteristic roots are 1 and 2, so 0 is not a root. Thus yₚ = Ax + B.
示例：对于 y” − 3y’ + 2y = 6x，特征根为 1 和 2，因此 0 不是根，所以 yₚ = Ax + B。
Substituting: yₚ’ = A, yₚ” = 0. So 0 − 3A + 2(Ax + B) = 2Ax + (−3A + 2B) = 6x. Matching coefficients gives 2A = 6 ⇒ A = 3, and −3A + 2B = 0 ⇒ B = 9/2. Hence yₚ = 3x + 9/2.
代入：yₚ’ = A，yₚ” = 0。所以 0 − 3A + 2(Ax + B) = 2Ax + (−3A + 2B) = 6x。比较系数得 2A = 6 ⇒ A = 3，且 −3A + 2B = 0 ⇒ B = 9/2。因此 yₚ = 3x + 9/2。
6. Exponential Forcing Terms | 指数函数强迫项
For f(x) = A e^{mx}, we try yₚ = C xᵏ e^{mx}. The value of k is the multiplicity of m as a root of the characteristic equation.
对于 f(x) = A e^{mx}，我们尝试 yₚ = C xᵏ e^{mx}。k 的值是 m 作为特征方程根的重数。
If m is not a root, k = 0. If m is a simple root, k = 1. If m is a repeated root, k = 2. This correction prevents the trial function from being annihilated by the homogeneous operator.
如果 m 不是根，k = 0。如果 m 是单根，k = 1。如果 m 是重根，k = 2。这种修正避免试探函数被齐次算子化为零。
Example: Solve the homogeneous problem first: y” − 3y’ + 2y = 0 has characteristic roots 1 and 2. For f(x) = 4eˣ, since m = 1 is a simple root, choose yₚ = A x eˣ.
示例：先解齐次问题：y” − 3y’ + 2y = 0 的特征根为 1 和 2。对于 f(x) = 4eˣ，由于 m = 1 是单根，选择 yₚ = A x eˣ。
Compute yₚ’ = A eˣ + A x eˣ = A eˣ (1 + x), yₚ” = A eˣ (2 + x). Substitute into y” − 3y’ + 2y = 4eˣ:
计算 yₚ’ = A eˣ + A x eˣ = A eˣ(1 + x)，yₚ” = A eˣ(2 + x)。代入 y” − 3y’ + 2y = 4eˣ：
A eˣ[(2 + x) − 3(1 + x) + 2x] = A eˣ(2 + x − 3 − 3x + 2x) = −A eˣ = 4eˣ
So A = −4, and yₚ = −4x eˣ. The general solution is y = C₁eˣ + C₂e²ˣ − 4x eˣ.
因此 A = −4，yₚ = −4x eˣ。通解为 y = C₁eˣ + C₂e²ˣ − 4x eˣ。
7. Trigonometric Forcing Terms | 三角函数强迫项
For f(x) = A sin(ωx) + B cos(ωx), we try yₚ = xᵏ (C sin(ωx) + D cos(ωx)). The value of k is the multiplicity of iω as a root of the characteristic equation.
对于 f(x) = A sin(ωx) + B cos(ωx)，我们尝试 yₚ = xᵏ (C sin(ωx) + D cos(ωx))。k 的值是 iω 作为特征方程根的重数。
In IB, k is usually 0 or 1. If iω is not a root, k = 0. If iω is a root, k = 1. This occurs when the characteristic equation has pure imaginary roots equal to ±iω.
在IB中，k 通常为 0 或 1。如果 iω 不是根，k = 0。如果 iω 是根，k = 1。当特征方程有纯虚根 ±iω 时就会发生共振。
Example: For y” + y = sin x, the characteristic equation is r² + 1 = 0, so r = ±i. Here iω = i, which is a simple root, so k = 1. Thus yₚ = x(C sin x + D cos x).
示例：对于 y” + y = sin x，特征方程为 r² + 1 = 0，所以 r = ±i。这里 iω = i，是单根，因此 k = 1，所以 yₚ = x(C sin x + D cos x)。
Differentiating: yₚ’ = C sin x + C x cos x + D cos x − D x sin x. Then yₚ” is computed and substituted. Simplification leads to 2C cos x − 2D sin x = sin x, so C = 0, D = −1/2. Therefore yₚ = −1/2 x cos x.
求导：yₚ’ = C sin x + C x cos x + D cos x − D x sin x。再求 yₚ” 并代入，化简得 2C cos x − 2D sin x = sin x，所以 C = 0，D = −1/2。因此 yₚ = −1/2 x cos x。
8. Combined Standard Forms | 组合标准形式
If f(x) is a product of the standard families, for example e^{mx} Pₙ(x) or e^{mx} sin(ωx), the trial form is the product of the corresponding trial forms, with the same k determined by the multiplicity of the relevant “root” m or m + iω.
如果 f(x) 是标准类型的乘积，例如 e^{mx} Pₙ(x) 或 e^{mx} sin(ωx)，则试探形式是对应试探形式的乘积，k 仍由相关“根” m 或 m + iω 的重数决定。
The table below gives the complete trial form for a product forcing term.
下表给出乘积型强迫项的完整试探形式。
| Forcing term f(x) | Trial form for yₚ |
| Pₙ(x) e^{mx} | xᵏ Qₙ(x) e^{mx} |
| e^{mx} sin(ωx) or e^{mx} cos(ωx) | xᵏ e^{mx} (C sin(ωx) + D cos(ωx)) |
| Pₙ(x) sin(ωx) or Pₙ(x) cos(ωx) | xᵏ [Qₙ(x) sin(ωx) + Rₙ(x) cos(ωx)] |
Example: For y” − 3y’ + 2y = x eˣ, we first note that the characteristic roots are 1 and 2. Since m = 1 is a simple root, k = 1. The forcing term is a polynomial of degree 1 times eˣ, so yₚ = x(Ax + B)eˣ = (Ax² + Bx)eˣ.
示例：对于 y” − 3y’ + 2y = x eˣ，首先特征根为 1 和 2。由于 m = 1 是单根，k = 1。强迫项是一次多项式乘以 eˣ，所以 yₚ = x(Ax + B)eˣ = (Ax² + Bx)eˣ。
After differentiation and substitution, you would find A = −1 and B = −1, giving yₚ = (−x² − x)eˣ. The constants depend on the exact equation; here we omit the algebra for brevity.
经过求导和代入后，可求得 A = −1，B = −1，因此 yₚ = (−x² − x)eˣ。具体常数取决于方程，这里略去代数过程。
9. The Resonance Correction: Multiply by x | 共振修正：乘以 x
The most common mistake is forgetting to multiply by x (or x²) when the forcing term overlaps with a solution of the homogeneous equation. This overlap is called resonance.
最常见的错误是当强迫项与齐次方程的解重叠时，忘记乘以 x（或 x²）。这种重叠称为“共振”。
If the unmodified trial form yₚ satisfies the homogeneous equation L(yₚ) = 0, then substituting it into the non-homogeneous equation gives no useful information. Multiplying by xᵏ, where k is the multiplicity of the overlapping root, breaks the resonance.
如果未经修正的试探形式 yₚ 满足齐次方程 L(yₚ) = 0，则将其代入非齐次方程无法得到有用信息。乘以 xᵏ（k 为重叠根的重数）可以打破共振。
Rule of thumb: Compare the exponent (or frequency) in f(x) with the characteristic roots. If it matches a root, add one factor of x. If it matches a double root, add x².
经验法则：比较 f(x) 中的指数（或频率）与特征根是否一致。若匹配某个根，则乘以 x；若匹配二重根，则乘以 x²。
For example, y” − 2y’ + y = eˣ has characteristic root r = 1 (double). Since m = 1 is a double root, the trial form must be yₚ = C x² eˣ, not C eˣ or C x eˣ. The x² factor is essential.
例如，y” − 2y’ + y = eˣ 的特征根为 r = 1（二重根）。由于 m = 1 是二重根，试探形式必须是 yₚ = C x² eˣ，而不是 C eˣ 或 C x eˣ。x² 因子必不可少。
10. Worked Example (IB-Style) | 完整例题（IB风格）
Let us solve the complete problem: y” − 3y’ + 2y = 4x eˣ + sin x.
让我们完整求解：y” − 3y’ + 2y = 4x eˣ + sin x。
Step 1: Homogeneous solution. r² − 3r + 2 = 0 ⇒ r = 1, 2, so y_c = C₁eˣ + C₂e²ˣ.
第一步：齐次解。 r² − 3r + 2 = 0 ⇒ r = 1, 2，所以 y_c = C₁eˣ + C₂e²ˣ。
Step 2: Particular solution for 4x eˣ. Since m = 1 is a simple root, yₚ₁ = x(Ax + B)eˣ = (Ax² + Bx)eˣ. Substitute into L(y) = y” − 3y’ + 2y and equate with 4x eˣ. After simplification, we obtain A = −2 and B = −2. Therefore yₚ₁ = −2x(x + 1)eˣ.
第二步：求 4x eˣ 的特解。 由于 m = 1 是单根，yₚ₁ = x(Ax + B)eˣ = (Ax² + Bx)eˣ。将其代入 L(y) = y” − 3y’ + 2y 并与 4x eˣ 比较，化简后得 A = −2，B = −2。因此 yₚ₁ = −2x(x + 1)eˣ。
Step 3: Particular solution for sin x. Here iω = i. The characteristic roots are 1 and 2, so i is not a root. Hence yₚ₂ = C sin x + D cos x. Substitution yields (C + 3D) sin x + (D − 3C) cos x = sin x. Solving gives C = 1/10, D = 3/10. So yₚ₂ = (1/10) sin x + (3/10) cos x.
第三步：求 sin x 的特解。 这里 iω = i。特征根为 1 和 2，因此 i 不是根，所以 yₚ₂ = C sin x + D cos x。代入得 (C + 3D) sin x + (D − 3C) cos x = sin x。解得 C = 1/10，D = 3/10。因此 yₚ₂ = (1/10) sin x + (3/10) cos x。
Step 4: General solution. y = y_c + yₚ₁ + yₚ₂ = C₁eˣ + C₂e²ˣ − 2x(x + 1)eˣ + (1/10) sin x + (3/10) cos x.
第四步：通解。 y = y_c + yₚ₁ + yₚ₂ = C₁eˣ + C₂e²ˣ − 2x(x + 1)eˣ + (1/10
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