3. Particular Solutions: Exceptional Cases | 3. 特解：例外情形

📚 3. Particular Solutions: Exceptional Cases | 3. 特解：例外情形

When solving second-order linear non-homogeneous differential equations with constant coefficients, the method of undetermined coefficients is a powerful technique to find a particular solution. However, the standard trial function may fail if its form overlaps with the complementary solution. This overlap leads to exceptional cases where we must modify the guess by multiplying by powers of x. Understanding these resonant situations is essential for IB Higher Level Mathematics students tackling differential equations in the calculus option.

在求解常系数二阶线性非齐次微分方程时，待定系数法是寻找特解的重要工具。然而，如果所设的试探函数形式与齐次通解发生重叠，标准方法就会失效。这种重叠导致例外情形，我们必须通过乘上 x 的幂来修正猜测形式。对于学习微积分选项中微分方程的 IB 高级数学学生来说，理解这些共振情形至关重要。

1. The Method of Undetermined Coefficients | 待定系数法概述

s a polynomial, exponential, sine, cosine, or a product of these. First, we solve the homogeneous equation to obtain the complementary function y_c. Then we propose a particular solution y_p that resembles f(x) but with unknown coefficients. Substituting y_p into the ODE determines these coefficients.

待定系数法适用于形如 ay′′ + by′ + cy = f(x) 的方程，其中 f(x) 是多项式、指数函数、正弦、余弦或它们的乘积。首先求解齐次方程得到齐次通解 y_c。然后我们假设一个与 f(x) 形式相似、但系数待定的特解 y_p，代入原方程即可确定这些系数。

For example, if f(x) = 5e^3x, we would try y_p = Ae^3x al plus a trigonometric combination. The standard trial forms are well established, but they can break down when the suggested y_p is already a solution of the associated homogeneous equation.

例如，若 f(x) = 5e^3x，我们可设 y_p = Ae^3x。若 f(x) = x² + sin 2x，则设想多项式加三角组合。这些标准试探形式已经固定，但当拟设的 y_p 恰好是对应齐次方程的解时，就会出现问题。

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2. The Complementary Function and Characteristic Equation | 齐次通解与特征方程

For the homogeneous equation ay′′ + by′ + cy = 0, we assume solutions of the form y = e^rx istic equation ar² + br + c = 0. The roots r₁ and r₂ determine the complementary function:

对于齐次方程 ay′′ + by′ + cy = 0，我们设解为 y = e^rx，代入后得特征方程 ar² + br + c = 0。由根 r₁ 和 r₂ 可确定齐次通解：

• If r₁ ≠ r₂, both real: y_c = C₁ e^r₁x + C₂ e^r₂x
• If r₁ = r₂ = r (real): y_c = (C₁ + C₂ x) e^rx
• If r = α ± βi: y_c = e^αx (C₁ cos βx + C₂ sin βx)

中文对应：若 r₁ ≠ r₂ 实根：y_c = C₁ e^r₁x + C₂ e^r₂x；若重实根 r：y_c = (C₁ + C₂ x) e^rx；若共轭复根 α ± βi：y_c = e^αx (C₁ cos βx + C₂ sin βx)。

An exceptional case in the method of undetermined coefficients occurs precisely when the trial function for y_p contains any term that also appears in y_c. Since y_c solves the homogeneous equation, any constant multiple of it vanishes upon substitution, making it impossible to determine coefficients unless the guess is modified.

待定系数法的例外情形恰好发生在 y_p 的试探形式包含某个也出现在 y_c 中的项。因为 y_c 满足齐次方程，任何常数倍在代入后都会消去，如果不修改猜测形式，就无法求出系数。

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3. Why Exceptional Cases Arise – Resonance | 例外情形的成因——共振

The overlap between f(x) and y_c is analogous to resonance in a physical system. If the forcing term has the same form as a natural frequency, the particular solution must grow in amplitude. Mathematically, this translates to multiplying the standard trial function by x (or x²) to break the linear dependence with the complementary solution.

f(x) 与 y_c 之间的重叠类似于物理系统中的共振。若强迫项的形式与固有频率相同，特解的振幅必然增大。在数学上，这体现为将标准试探函数乘以 x（或 x²），以打破与齐次通解的线性相关性。

In IB examinations, three main exceptional patterns are routinely tested: when f(x) is an exponential that matches a simple root, a double root, or when a trigonometric forcing term matches complex roots. Recognising these cases quickly saves time and prevents algebraic dead ends.

在 IB 考试中，通常考查三种主要的例外模式：f(x) 的指数与单根重合、与重根重合，或三角强迫项与复根匹配。迅速识别这些情形能节省时间，避免代数死胡同。

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4. Exceptional Case 1: Overlap with a Simple Root | 例外情形一：与单特征根重叠

The simplest exceptional case occurs when the non-homogeneous term f(x) = ke^px and p equals one of the two distinct roots of the characteristic equation. The standard trial y_p = Ae^px fails because it is a solution of the homogeneous equation. Instead, we must multiply by x and use y_p = Axe^px.

最简单的例外情形发生在非齐次项 f(x) = ke^px 且 p 等于特征方程两个不同根之一。标准试探 y_p = Ae^px 失效，因为它已是齐次解。此时应乘以 x，采用 y_p = Axe^px。

Consider y′′ − 4y′ + 3y = 2e^x. The characteristic equation r² − 4r + 3 = 0 gives r = 1 and r = 3. Thus y_c = C₁ e^x + C₂ e^3x. The forcing term e^x shares a root with r = 1, so we try y_p = Axe^x. Substituting and solving yields A = −1, so y_p = −xe^x.

考虑 y′′ − 4y′ + 3y = 2e^x。特征方程 r² − 4r + 3 = 0 的根为 r = 1 和 r = 3，故 y_c = C₁ e^x + C₂ e^3x。强迫项 e^x 与 r = 1 根重合，因此我们设 y_p = Axe^x。代入求解得 A = −1，y_p = −xe^x。

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5. Worked Example for Case 1 | 情形一示例详解

Equation: y′′ − y′ − 2y = 4e^−x.
Step 1: Characteristic equation r² − r − 2 = 0 ⇒ r = 2, r = −1. So y_c = C₁ e^2x + C₂ e^−x.
Step 2: f(x) = 4e^−x matches the root −1, so standard y_p = Ae^−x is replaced by y_p = Axe^−x.
Step 3: Differentiate: y_p′ = Ae^−x − Axe^−x, y_p′′ = −Ae^−x − Ae^−x + Axe^−x = −2Ae^−x + Axe^−x.
Step 4: Substitute into LHS: (−2Ae^−x + Axe^−x) − (Ae^−x − Axe^−x) − 2(Axe^−x) = −2Ae^−x + Axe^−x − Ae^−x + Axe^−x − 2Axe^−x = −3Ae^−x.
Equating to 4e^−x gives −3A = 4, thus A = −4/3. Hence y_p = −(4/3) x e^−x.

方程： y′′ − y′ − 2y = 4e^−x。
第一步： 特征方程 r² − r − 2 = 0 ⇒ r = 2, r = −1，得 y_c = C₁ e^2x + C₂ e^−x。
第二步： f(x) = 4e^−x 与根 −1 匹配，故将标准 y_p = Ae^−x 替换为 y_p = Axe^−x。
第三步： 求导：y_p′ = Ae^−x − Axe^−x，y_p′′ = −2Ae^−x + Axe^−x。
第四步： 代入左边并化简得 −3Ae^−x，与右边 4e^−x 比较得 A = −4/3，故特解 y_p = −(4/3) x e^−x。

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6. Exceptional Case 2: Overlap with a Double Root | 例外情形二：与重特征根重叠

When the characteristic equation has a repeated root r (multiplicity two) and f(x) contains e^rx, both y_c = (C₁ + C₂ x) e^rx and the standard trial Ae^rx plus Axe^rx are already captured by the homogeneous solution. In this case, we must multiply by x² and use y_p = Ax² e^rx.

当特征方程具有重根 r，且 f(x) 包含 e^rx 时，齐次通解 y_c = (C₁ + C₂ x) e^rx 已经涵盖了标准的 Ae^rx 及 Axe^rx 形式。此时必须乘以 x²，尝试 y_p = Ax² e^rx。

For instance, solve y′′ − 6y′ + 9y = 4e^3x. The characteristic equation r² − 6r + 9 = 0 has double root r = 3. y_c = (C₁ + C₂ x) e^3x. Since e^3x and xe^3x are already in y_c, we need y_p = Ax² e^3x. Differentiation and substitution will yield the correct coefficient.

例如，求解 y′′ − 6y′ + 9y = 4e^3x。特征方程 r² − 6r + 9 = 0 有二重根 r = 3，y_c = (C₁ + C₂ x) e^3x。由于 e^3x 与 xe^3x 均已在齐次解中，我们需要设 y_p = Ax² e^3x。通过求导代入得到正确系数。

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7. Detailed Double Root Example | 重根情形详细计算

Take y′′ − 6y′ + 9y = 4e^3x and let y_p = Ax² e^3x.
y_p′ = 2Ax e^3x + 3Ax² e^3x = A e^3x (2x + 3x²).
y_p′′ = A e^3x (2 + 6x) + 3A e^3x (2x + 3x²) = A e^3x (2 + 12x + 9x²).
Now compute LHS: y_p′′ − 6y_p′ + 9y_p = A e^3x [(2 + 12x + 9x²) − 6(2x + 3x²) + 9x²] = A e^3x [2 + 12x + 9x² − 12x − 18x² + 9x²] = 2A e^3x.
Equating to 4e^3x gives 2A = 4 ⇒ A = 2. Therefore y_p = 2x² e^3x.

以 y′′ − 6y′ + 9y = 4e^3x 为例，设 y_p = Ax² e^3x。
y_p′ = 2Ax e^3x + 3Ax² e^3x = A e^3x (2x + 3x²)。
y_p′′ = A e^3x (2 + 6x) + 3A e^3x (2x + 3x²) = A e^3x (2 + 12x + 9x²)。
计算左边：y_p′′ − 6y_p′ + 9y_p 化简后得到 2A e^3x。与右边 4e^3x 比较得 A = 2，所以 y_p = 2x² e^3x。

This example illustrates the rule: the power of x in the multiplier equals the multiplicity of the root. For a double root, we multiply by x².

此例说明了规则：乘子中 x 的幂等于根的重数。对于二重根，乘 x²。

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8. Exceptional Cases with Trigonometric Forcing | 含三角强迫项的例外情形

When f(x) includes sin βx or cos βx, the standard trial is y_p ccurs if the homogeneous characteristic equation has purely imaginary roots ±βi, meaning y_c already contains cos βx and sin βx. The trial must then be multiplied by x: y_p = x (A cos βx + B sin βx).

当 f(x) 包含 sin βx 或 cos βx 时，标准试探函数为 y_p = A cos βx + B sin βx。若齐次特征方程具有纯虚根 ±βi，即 y_c 已含 cos βx 与 sin βx，则出现例外情形，必须乘以 x：y_p = x (A cos βx + B sin βx)。

More generally, if f(x) = e^αx cos βx and α + βi is a root of the characteristic equation, the complementary solution involves e^αx cos βx and e^αx sin βx, creating an overlap. We then use y_p = x e^αx (A cos βx + B sin βx). This situation combines both exponential and trigonometric resonance.

更一般地，若 f(x) = e^αx cos βx 且 α + βi 是特征根，齐次解包含 e^αx cos βx 与 e^αx sin βx，从而产生重叠。此时我们用 y_p = x e^αx (A cos βx + B sin βx)，这一情况结合了指数共振与三角共振。

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9. Example of Trigonometric Resonance | 三角共振示例

Consider y′′ + 9y = 5 sin 3x. The characteristic equation r² + 9 = 0 gives r = ±3i, so y_c = C₁ cos 3x + C₂ sin 3x. The standard trial A sin 3x + B cos 3x fails, so we modify to y_p = x (A cos 3x + B sin 3x). Substitute and simplify: y_p′′ = −6A sin 3x + 6B cos 3x − 9x(A cos 3x + B sin 3x). Adding 9y_p e last term, leaving −6A sin 3x + 6B cos 3x = 5 sin 3x. Hence B = 0 and A = −5/6, giving y_p = −(5/6) x cos 3x.

考虑 y′′ + 9y = 5 sin 3x。特征方程 r² + 9 = 0 的根为 ±3i，y_c = C₁ cos 3x + C₂ sin 3x。标准试探 A sin 3x + B cos 3x 失效，改为 y_p = x (A cos 3x + B sin 3x)。代入并化简后得 −6A sin 3x + 6B cos 3x = 5 sin 3x，因此 B = 0，A = −5/6，特解为 y_p = −(5/6) x cos 3x。

Notice how the x factor introduces a linearly growing amplitude, reflecting the resonant energy buildup. In a mass-spring system, this corresponds to undamped forcing at the natural frequency.

注意到 x 因子引入了线性增长的振幅，反映了共振能量的累积。在弹簧–质量系统中，这对应于以固有频率施加的无阻尼强迫。

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10. Combining Polynomial, Exponential, and Trigonometric Terms | 多项式、指数与三角项的组合

When f(x) is a product or sum of different function types, the trial y_p is constructed by superposition. For example, if f(x) = x² e^3x cos 2x, the standard trial would be (Ax² + Bx + C) e^3x cos 2x + (Dx² + Ex + F) e^3x sin 2x. Exceptional modifications apply only to the part whose form overlaps with y_c.

当 f(x) 是不同函数类型的乘积或和时，通过叠加原理构建试探 y_p。例如，若 f(x) = x² e^3x cos 2x，标准试探函数为 (Ax² + Bx + C) e^3x cos 2x + (Dx² + Ex + F) e^3x sin 2x。例外修正仅适用于与 y_c 重叠的部分。

If the complex number 3 + 2i happens to be a root of the

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