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 \(a\frac{d^2y}{dx^2} + b\frac{dy}{dx} + cy = f(x)\), where \(a\), \(b\), and \(c\) are real constants. When \(f(x) = 0\), the equation is homogeneous; when \(f(x) \neq 0\), it is non-homogeneous. The complete solution is written as \(y = \text{CF} + \text{PI}\).

我们研究的方程形式为 \(a\frac{d^2y}{dx^2} + b\frac{dy}{dx} + cy = f(x)\)，其中 \(a\)、\(b\)、\(c\) 为实常数。当 \(f(x) = 0\) 时，方程为齐次的；当 \(f(x) \neq 0\) 时，方程为非齐次的。完整的解写作 \(y = \text{CF} + \text{PI}\)。

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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, \(a\frac{d^2y}{dx^2} + b\frac{dy}{dx} + cy = 0\) leads to the auxiliary equation \(am^2 + bm + c = 0\). 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 \(a\frac{d^2y}{dx^2} + b\frac{dy}{dx} + cy = 0\) 对应辅助方程 \(am^2 + bm + c = 0\)。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 \(m_1\) and \(m_2\), the complementary function is \(y_c = Ae^{m_1x} + Be^{m_2x}\). 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 \(m_1\) 和 \(m_2\)，余函数为 \(y_c = Ae^{m_1x} + Be^{m_2x}\)。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: \(m^2 – 3m + 2 = 0 \Rightarrow m = 1,\ 2\)
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: \(y” + 6y’ + 9y = 0\)
\(m^2 + 6m + 9 = 0 \Rightarrow m = -3\) (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 \(m = p \pm qi\), the CF is expressed using sine and cosine: \(y_c = e^{px}(A\cos(qx) + B\sin(qx))\). This form is essential for modelling oscillatory systems such as springs and circuits.

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

Example: \(y” – 2y’ + 5y = 0\)
\(m = 1 \pm 2i\)
CF: \(y_c = e^{x}(A\cos 2x + B\sin 2x)\)

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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 \(y_p = a_nx^n + \cdots + a_1x + a_0\). If \(f(x) = ke^{mx}\), try \(y_p = Ce^{mx}\). If \(f(x) = k\cos(\omega x)\) or \(k\sin(\omega x)\), 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 \(y_p = a_nx^n + \cdots + a_1x + a_0\)。若 \(f(x) = ke^{mx}\)，则设 \(y_p = Ce^{mx}\)。若 \(f(x) = k\cos(\omega x)\) 或 \(k\sin(\omega x)\)，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: \(m^2 + m – 2 = 0 \Rightarrow m = 1, -2\), 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: \(y_c = A\cos 2x + B\sin 2x\).
Since \(\cos 2x\) is in the CF, try \(y_p = x(C\cos 2x + D\sin 2x)\).
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: \(y” – y = x^2 + e^{3x}\)
PI for \(x^2\): \(y_{p1} = ax^2 + bx + c\)
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 \(m_1, m_2\) | 相异实根	\(Ae^{m_1x} + Be^{m_2x}\)
Repeated real \(m\) | 重根	\((A + Bx)e^{mx}\)
Complex \(p \pm qi\) | 复根	\(e^{px}(A\cos qx + B\sin qx)\)

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