📚 Particular Solutions for Standard Forcing Terms | 标准强迫项的特解
In IB Mathematics Analysis & Approaches HL, solving non-homogeneous linear differential equations with constant coefficients is a core skill. The general solution consists of the complementary function (solving the homogeneous equation) plus a particular integral (any solution of the full non-homogeneous equation). This article focuses on finding particular solutions using the method of undetermined coefficients when the forcing term is a standard function — polynomials, exponentials, sines, cosines, and their combinations.
在IB数学分析与方法HL课程中，求解常系数非齐次线性微分方程是一项核心技能。通解由补函数（对应齐次方程的解）加上特解（满足完整非齐次方程的任意一个解）组成。本文重点介绍当强迫项为标准函数——多项式、指数、正弦、余弦以及它们的组合时，如何用待定系数法求出特解。
1. The Big Picture of Non-Homogeneous ODEs | 非齐次常微分方程总览
A second-order linear ODE with constant coefficients is typically written as y'' + a y' + b y = r(x), where r(x) is the forcing term. The general solution is y = yc + yp, where yc is the complementary function obtained from the characteristic equation, and yp is any particular solution that satisfies the original equation. Our task is to find yp efficiently when r(x) belongs to a familiar class.
典型的二阶常系数线性常微分方程可写作 y'' + a y' + b y = r(x)，其中 r(x) 是强迫项。通解为 y = yc + yp，yc 是由特征方程得出的补函数，yp 是满足原方程的任意一个特解。当 r(x) 属于某个常见类型时，我们的任务就是高效地求出 yp。
2. The Method of Undetermined Coefficients | 待定系数法
The idea is straightforward: guess a form for yp that resembles the forcing term, containing unknown constants. Substitute this guess into the ODE and equate coefficients of like terms to determine the constants. The guess must be linearly independent from the complementary function; otherwise we multiply by an appropriate power of x (the modification rule). This method works for r(x) which is a polynomial, exponential, sine/cosine, or sum/product of these.
核心思想很简单：根据强迫项的形式猜测一个含有待定常数的特解 yp，将其代入原方程，通过比较同类项的系数求出待定常数。猜测试探解时必须保证它与补函数线性无关，否则需要乘以适当的 x 的幂次（即修正规则）。该方法适用于多项式、指数、正弦/余弦以及它们的和与积等强迫项。
3. Forcing Term Is a Polynomial | 强迫项为多项式
If r(x) = Pn(x), a polynomial of degree n, try a general polynomial of the same degree: yp = Anxn + An−1xn−1 + ... + A1x + A0. Substitute into the ODE, collect powers of x, and solve the resulting system. Remember to include all lower-degree terms even if the forcing term misses some powers.
如果 r(x) = Pn(x) 是一个 n 次多项式，则尝试设 yp 为同次的一般多项式：yp = Anxn + An−1xn−1 + ... + A1x + A0。代入方程后合并 x 的各次幂，解出待定系数。即使强迫项中缺少某些低次幂，试探解仍必须包含所有低于 n 次的项。
4. Forcing Term Is an Exponential | 强迫项为指数函数
When r(x) = C ekx, try yp = A ekx, provided k is not a root of the characteristic equation. Substituting gives a simple algebraic equation for A. If k is a single root, multiply by x: try A x ekx. If k is a double root, try A x2 ekx. The same logic extends to any order.
当 r(x) = C ekx 时，如果 k 不是特征方程的根，则可设 yp = A ekx。代入后会得到关于 A 的简单代数方程。如果 k 是单根，则试探解乘以 x，即设为 A x ekx；如果 k 是二重根，则设为 A x2 ekx。这一思路适用于任意阶方程。
5. Forcing Term Is Sine or Cosine | 强迫项为正弦或余弦
r a linear combination, the trial particular solution must contain both sine and cosine: yp cing term has only sine or only cosine, we must include both, because derivatives of sine produce cosine and vice versa. Substitute and equate coefficients of sin(ωx) and cos(ωx) to find A and B.
对于 r(x) = C sin(ωx) 或 C cos(ωx)，或两者的线性组合，试探特解必须同时包含正弦和余弦：yp = A cos(ωx) + B sin(ωx)。即使强迫项只有正弦或只有余弦，试探解中也必须同时存在两项，因为正弦的导数是余弦，反之亦然。代入后比较 sin(ωx) 和 cos(ωx) 的系数即可求出 A 和 B。
6. Product of Polynomial and Exponential | 多项式与指数乘积
When r(x) = Pn(x) ekx, where Pn is a polynomial of degree n, the trial solution should be yp = Qn(x) ekx, where Qn(x) is a full polynomial of degree n with unknown coefficients. In other words, we "attach" a polynomial of the same degree. If k coincides with a root of the characteristic equation, multiply the polynomial part by xs as before.
当 r(x) = Pn(x) ekx，其中 Pn 为 n 次多项式时，试探解应设为 yp = Qn(x) ekx，这里 Qn(x) 是一个系数待定的完整 n 次多项式。换言之，我们“配上”一个同次多项式。若 k 与特征方程的根重合，则按之前规则将多项式部分乘以 xs。
7. Exponential Times Sine or Cosine | 指数乘正弦或余弦
If r(x) = eαx (C cos(βx) + D sin(βx)), the trial form is yp = eαx root of the characteristic equation (resonance), multiply the whole trial expression by x.
若 r(x) = eαx (C cos(βx) + D sin(βx))，试探特解形式为 yp = eαx (A cos(βx) + B sin(βx))。代入后通过比较系数得到关于 A、B 的线性方程组。当 α ± iβ 恰好是特征方程的根（共振情况）时，将整个试探表达式乘以 x。
8. Sums of Standard Forcing Terms | 标准强迫项的和
When r(x) is a sum of independent standard terms, such as r(x) = 3x2 + e−x + sin(2x), the principle of superposition applies. We find a particular solution for each term separately and then add them. For instance, try yp = (A x2 + Bx + C) + D e−x + (E cos(2x) + F sin(2x)). No cross‑terms are needed for distinct functional families.
若 r(x) 是几个独立的标准项之和，例如 r(x) = 3x2 + e−x + sin(2x)，可以应用叠加原理。我们分别求每一项的特解，再将其相加。例如可设 yp = (A x2 + Bx + C) + D e−x + (E cos(2x) + F sin(2x))。不同函数族之间无需引入交叉项。
9. Modification Rule: When the Guess Overlaps with yc | 修正规则：探试解与补函数重叠时
The method of undetermined coefficients requires the trial yp to have no term that already satisfies the homogeneous equation. If the normal guess contains a term from the complementary function, we multiply the entire guess by x (or a higher power) until all terms are linearly independent from yc. The table below summarises standard forcing terms and the necessary multiplication.
待定系数法要求试探解 yp 中不含任何已满足齐次方程的项。如果常规猜测试探解与补函数中的项重合，就需要将整个试探解乘以 x（或 x 的更高次幂），直到所有项都与 yc 线性无关。下表总结了标准强迫项以及必需乘上的 x 的幂次。
| Forcing term r(x) | Default trial yp | If overlap with yc |
|---|---|---|
| Pn(x) | Qn(x) | xs Qn(x), s = smallest integer making no term in yc |
| C ekx | A ekx | A xs ekx, s = multiplicity of k as root |
| C cos(ωx) or D sin(ωx) | A cos(ωx) + B sin(ωx) | x (A cos(ωx) + B sin(ωx)) if ±iω are roots |
| eαx cos(βx) etc. | eαx(A cos(βx)+B sin(βx)) | xs eαx(A cos(βx)+B sin(βx)), s = multiplicity of α±iβ |
Always check the roots of the characteristic equation first. Overlooking this step is one of the most common mistakes.
一定要先求出特征方程的根。忽略这一步是最常见的错误之一。
10. Worked Example 1: Polynomial Forcing | 例题1：多项式强迫
Problem: Find the general solution of y'' − 3y' + 2y = 4x2.
问题：求 y'' − 3y' + 2y = 4x² 的通解。
The characteristic equation is r2 − 3r + 2 = 0, giving r = 1, 2. So yc = C1 ex + C2 e2x. The forcing term is a second-degree polynomial, and neither 1 nor x nor x2 appears in yc. Try yp = A x2 + B x + C. Compute derivatives: yp' = 2A x + B, yp'' = 2A. Substitute: 2A − 3(2A x + B) + 2(A x2 + B x + C) = 4x2. Simplify: 2A − 6A x − 3B + 2A x2 + 2B x + 2C = 2A x2 + (−6A + 2B) x + (2A − 3B + 2C). Equate coefficients: 2A = 4 ⇒ A = 2; −6A + 2B = 0 ⇒ −12 + 2B = 0 ⇒ B = 6; constant: 2A − 3B + 2C = 0 ⇒ 4 − 18 + 2C = 0 ⇒ C = 7. Hence yp = 2x2 + 6x + 7. General solution: y = C1 ex + C2 e2x + 2x2 + 6x + 7.
特征方程为 r² − 3r + 2 = 0，解得 r = 1, 2。因此 yc = C₁ eˣ + C₂ e²ˣ。强迫项是二次多项式，且 1, x, x² 均未出现在 yc 中。设 yp = A x² + B x + C。求导：yp' = 2A x + B，yp'' = 2A。代入得：2A − 3(2A x + B) + 2(A x² + B x + C) = 4x²。化简：2A x² + (−6A + 2B)x + (2A − 3B + 2C) ≡ 4x²。比较系数：2A = 4 ⇒ A = 2；−6A + 2B = 0 ⇒ B = 6；常数项：2A − 3B + 2C = 0 ⇒ C = 7。故 yp = 2x² + 6x + 7。通解为 y = C₁ eˣ + C₂ e²ˣ + 2x² + 6x + 7。
11. Worked Example 2: Resonance with Sinusoidal Forcing | 例题2：正弦强迫的共振情况
Problem: Solve y'' + 4y = 3 sin(2x).
问题：求解 y'' + 4y = 3 sin(2x)。
The characteristic equation r2 + 4 = 0 gives r = ±2i. Thus yc = C1 cos(2x) + C2 sin(2x). Notice the forcing term sin(2x) already appears in yc. Our normal guess would be A cos(2x) + B sin(2x), but that exactly matches yc. Hence we multiply by x: try yp ompute derivatives carefully. Substitute and simplify to obtain B = 0 and A = −3/4. Thus yp = −(3/4) x cos(2x). The general solution is y = C1 cos(2x) + C2 sin(2x) − (3/4) x cos(2x). The term x cos(2x) shows resonance — the amplitude grows linearly with x.
特征方程 r² + 4 = 0 的解为 r = ±2i，因此 yc = C₁ cos(2x) + C₂ sin(2x)。强迫项 sin(2x) 已经包含在补函数中。常规试探解 A cos(2x) + B sin(2x) 会与 yc 完全重合，故需乘以 x：设 yp = x (A cos(2x) + B sin(2x)) = A x cos(2x) + B x sin(2x)。仔细求导后代入方程，可解得 B = 0，A = −3/4。因此 yp = −(3/4) x cos(2x)。通解为 y = C₁ cos(2x) + C₂ sin(2x) − (3/4) x cos(2x)。x cos(2x) 项显示出共振现象——振幅随 x 线性增长。
12. Key Tips and Common Pitfalls | 关键提示与常见陷阱
Always find the complementary function first. Identify the characteristic roots before selecting the trial yp. Include all necessary terms even if the forcing term lacks some (e.g., include both sine and cosine, or all polynomial degrees). When resonance occurs, do not simply multiply the conflicting term by x; multiply the entire trial expression. Check your final yp by substitution back into the ODE. For sums, handle each part separately. Keep your working well-organized to avoid algebraic errors.
务必先求出补函数。在选定试探解 yp 之前，先确认特征根。即使强迫项缺少某些部分，试探解也必须包含所有必要的项（例如同时含正弦和余弦，或包含所有多项式次数）。遇到共振时，不能只将冲突的项乘以 x，而应将整个试探表达式乘以 x。将求得的 yp 代回原方程进行验证。遇到和式时，对每一部分分别处理。保持步骤清晰有序，以避免代数错误。
Published by TutorHao | Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图