Complementary Functions and Particular Integrals | 互补函数与特解积分

📚 Complementary Functions and Particular Integrals | 互补函数与特解积分

c y = f(x), where a, b and c are constants. The complete solution has two parts: the complementary function, which comes from the homogeneous equation, and a particular integral, which is a specific solution to the full non-homogeneous equation. This article explains how to find both parts, how to choose the correct trial form, and how to combine them into a general solution.

在 AQA 进阶数学中，学习二阶微分方程的主要目标是求解形如 a d²y/dx² + b dy/dx + c y = f(x) 的方程，其中 a、b、c 为常数。方程的解由两部分组成：互补函数来自齐次方程，特解积分则是非齐次方程的一个特定解。本文将解释如何求出这两部分，如何选择正确的试解形式，以及如何将它们组合成通解。

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1. The Structure of the Problem | 问题的结构

n is linear because y, dy/dx and d²y/dx² each appear only to the first power and are never multiplied together. The term f(x) on the right-hand side is called the forcing term or the non-homogeneous term. If f(x) = 0, the equation is homogeneous; otherwise it is non-homogeneous and requires both a complementary function and a particular integral.

考虑一个常系数二阶线性微分方程，写作 a d²y/dx² + b dy/dx + c y = f(x)，其中 a ≠ 0。左边是线性的，因为 y、dy/dx 和 d²y/dx² 都只以一次方出现，并且没有彼此相乘。右边的 f(x) 称为强迫项或非齐次项。如果 f(x) = 0，方程是齐次的；否则该方程是非齐次的，需要同时求出互补函数和特解积分。

a d²y/dx² + b dy/dx + c y = f(x)

Two features make these equations examinable: the coefficients are constants, so the auxiliary equation is a quadratic, and the right-hand side is usually a polynomial, an exponential, a sine or cosine, or a sum of such terms. Each type of forcing term suggests a predictable trial form for the particular integral.

这类方程适合考试的两个特征是：系数为常数，因此辅助方程是一个二次方程；右端通常是多项式、指数函数、正弦或余弦函数，或者这些项的和。每一种强迫项都对应一种可预测的特解试解形式。

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2. Why the General Solution Has Two Parts | 为什么通解由两部分组成

The general solution of a non-homogeneous linear equation is written as y = y_c + y_p, where y_c is the complementary function and y_p is a particular integral. The reason is the superposition principle. If y_c satisfies the homogeneous equation and y_p satisfies the full non-homogeneous equation, then substituting y_c + y_p into the left-hand side gives 0 + f(x) = f(x). Therefore y_c + y_p is also a solution.

非齐次线性方程的通解写作 y = y_c + y_p，其中 y_c 是互补函数，y_p 是特解积分。原因是叠加原理。如果 y_c 满足齐次方程，而 y_p 满足完整的非齐次方程，那么把 y_c + y_p 代入左边就得到 0 + f(x) = f(x)。因此 y_c + y_p 也是解。

y = y_c + y_p

The complementary function contains the arbitrary constants A and B. These constants allow the general solution to satisfy two initial or boundary conditions. The particular integral contains no arbitrary constants; it is one specific function that reproduces the forcing term after substitution.

互补函数包含任意常数 A 和 B。这些常数使通解能够满足两个初始条件或边界条件。特解积分不包含任意常数；它是一个特定的函数，代入后能够产生右端的强迫项。

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3. The Complementary Function and the Auxiliary Equation | 互补函数与辅助方程

y = 0. Since the coefficients are constant, we look for trial solutions of the form y = e^mx. Differentiating gives dy/dx = m e^mx and d²y/dx² = m² e^mx. Substituting these into the homogeneous equation produces (a m² + b m + c) e^mx = 0. Because e^mx b m + c = 0. The roots of this quadratic determine the form of the complementary function.

互补函数求解相应的齐次方程 a d²y/dx² + b dy/dx + c y = 0。由于系数是常数，我们可以尝试形如 y = e^mx 的解。对其求导得到 dy/dx = m e^mx 和 d²y/dx² = m² e^mx。将这些代入齐次方程，得到 (a m² + b m + c) e^mx = 0。因为 e^mx 永不为零，所以辅助方程必然是 a m² + b m + c = 0。这个二次方程的根决定了互补函数的形式。

a m² + b m + c = 0

hich of the three standard cases applies. A positive discriminant gives two distinct real roots, a zero discriminant gives a repeated real root, and a negative discriminant gives a pair of complex conjugate roots. Each case produces a different but standard complementary function.

判别式 Δ = b² - 4ac 告诉我们适用三种标准情况中的哪一种。判别式为正给出两个相异实根，判别式为零给出一个重实根，判别式为负给出一对共轭复根。每种情况都产生不同但标准的互补函数。

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4. Case 1: Real and Distinct Roots | 情况一：相异实根

liary equation has two distinct real roots, say m₁ and m₂. The complementary function is y_c = A e^m₁x + B e^m₂x, where A and B are arbitrary constants. These constants are determined only after the initial or boundary conditions are applied.

如果判别式 b² - 4ac 为正，辅助方程有两个相异实根，记为 m₁ 和 m₂。互补函数为 y_c = A e^m₁x + B e^m₂x，其中 A 和 B 是任意常数。这些常数要等到代入初始条件或边界条件后才能确定。

m + 2 = 0, which factorises as (m - 1)(m - 2) = 0, giving roots m₁ = 1 and m₂ = 2. Hence y_c = A e^x + B e^2x.

例如，方程 d²y/dx² - 3 dy/dx + 2y = 0 的辅助方程是 m² - 3m + 2 = 0，因式分解为 (m - 1)(m - 2) = 0，得到根 m₁ = 1 和 m₂ = 2。因此 y_c = A e^x + B e^2x。

y_c = A e^m₁x + B e^m₂x

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5. Case 2: Repeated Real Roots | 情况二：重实根

If the discriminant is zero, the auxiliary equation has one repeated real root m. The complementary function cannot simply be A e^mx + B e^mx, because that would collapse into a single term and would not provide two independent constants. Instead, the second independent solution is x e^mx, so the complementary function is y_c = (A + Bx) e^mx.

如果判别式为零，辅助方程有一个重实根 m。互补函数不能简单写成 A e^mx + B e^mx，因为那会合并成一项，无法提供两个独立常数。第二个独立解是 x e^mx，因此互补函数为 y_c = (A + Bx) e^mx。

- 4m + 4 = 0, that is (m - 2)² = 0, so m = 2 is repeated. The complementary function is y_c = (A + Bx) e^2x.

例如，d²y/dx² - 4 dy/dx + 4y = 0 的辅助方程是 m² - 4m + 4 = 0，即 (m - 2)² = 0，所以 m = 2 是重根。互补函数为 y_c = (A + Bx) e^2x。

y_c = (A + Bx) e^mx

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6. Case 3: Complex Conjugate Roots | 情况三：共轭复根

the auxiliary equation has complex conjugate roots m = α ± iβ, where i² = -1. Using Euler's formula, the complementary function can be written in real form as y_c = e^αx(A cos βx + B sin βx). Here α is the real part of the root and β is the coefficient of i, taken as positive in the standard answer.

如果判别式为负，辅助方程有共轭复根 m = α ± iβ，其中 i² = -1。利用欧拉公式，互补函数可以写成实数形式 y_c = e^αx(A cos βx + B sin βx)。这里 α 是根的实部，β 是 i 的系数，在标准答案中取正值。

4y = 0 has auxiliary equation m² + 4 = 0, so m = ±2i, giving α = 0 and β = 2. Therefore y_c If the equation were d²y/dx² + 2 dy/dx + 5y = 0, the roots would be m = -1 ± 2i, giving y_c = e^-x(A cos 2x + B sin 2x).

例如，d²y/dx² + 4y = 0 的辅助方程是 m² + 4 = 0，所以 m = ±2i，得到 α = 0，β = 2。因此 y_c = A cos 2x + B sin 2x。如果方程是 d²y/dx² + 2 dy/dx + 5y = 0，根为 m = -1 ± 2i，得到 y_c = e^-x(A cos 2x + B sin 2x)。

y_c = e^αx(A cos βx + B sin βx)

The table below summarises the three cases. In every case the complementary function contains exactly two arbitrary constants, which is what a second-order equation requires.

下表总结了三种情况。每种情况下互补函数都恰好包含两个任意常数，这正是二阶方程所需要的。

Discriminant | 判别式	Roots | 根	Complementary function | 互补函数
b² - 4ac > 0	m₁, m₂ real and distinct | 相异实根	yc = A em₁x + B em₂x
b² - 4ac = 0	m repeated | 重根	yc = (A + Bx) emx
b² - 4ac < 0	α ± iβ	yc = eαx(A cos βx + B sin βx)

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7. The Method of Undetermined Coefficients | 待定系数法

To find a particular integral, we use the method of undetermined coefficients. The form of y_p is guessed from the form of f(x), but the coefficients in that guess are left unknown. We substitute the trial y_p and its derivatives into the original non-homogeneous equation and compare coefficients on both sides.

求特解积分时，我们使用待定系数法。y_p 的形式根据 f(x) 的形式来猜测，但猜测式中的系数保持未知。我们把试解 y_p 及其导数代入原来的非齐次方程，然后比较两边的系数。

The guiding rule is that the trial function must be able to produce f(x) after differentiation and substitution. For example, a polynomial forcing term requires a polynomial trial of the same degree, and a sine or cosine forcing term requires a combination of both sine and cosine, because differentiating one produces the other.

指导规则是：试解函数在求导和代入之后必须能够产生 f(x)。例如，多项式强迫项需要同次数的多项式试解，而正弦或余弦强迫项需要同时包含正弦和余弦的组合，因为对其中一个求导会产生另一个。

y = y_c + y_p

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8. Particular Integrals for Polynomial Forcing Terms | 多项式强迫项的特解积分

particular integral is a general polynomial of degree n. For example, if f(x) = 4x, try y_p = ax + b. If f(x) = 3x² - 1, try y_p te into the differential equation, and equate coefficients of x², x and the constant term.

如果 f(x) 是 n 次多项式，特解积分的试解就是一般的 n 次多项式。例如，如果 f(x) = 4x，试解 y_p = ax + b。如果 f(x) = 3x² - 1，试解 y_p = ax² + bx + c。对试解求导两次，代入微分方程，并比较 x²、x 和常数项的系数。

at case the polynomial trial must be multiplied by x. If m = 0 is a repeated root, the trial must be multiplied by x².

有一个重要例外。如果辅助方程有根 m = 0，那么互补函数已经包含常数项。此时多项式试解必须乘以 x。如果 m = 0 是重根，试解必须乘以 x²。

f(x) polynomial degree n ⇒ y_p polynomial degree n

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9. Particular Integrals for Exponential and Trigonometric Forcing Terms | 指数与三角函数强迫项的特解积分

If f(x) = k e^px, the standard trial is y_p = C e^px. Substituting this into the differential equation gives (a p² + b p + c) C e^px = k e^px nominator is not zero. If p is already a root of the auxiliary equation, the denominator is zero and the trial must be multiplied by x.

如果 f(x) = k e^px，标准试解是 y_p = C e^px。将其代入微分方程得到 (a p² + b p + c) C e^px = k e^px，所以 C = k / (a p² + b p + c)，前提是分母不为零。如果 p 已经是辅助方程的根，分母为零，试解必须乘以 x。

If f(x) = k cos ωx or f(x) = k sin ωx, the trial must include both sine and cosine: y_p s one into the other and the coefficients cannot be matched unless both terms are present.

如果 f(x) = k cos ωx 或 f(x) = k sin ωx，试解必须同时包含正弦和余弦：y_p = C cos ωx + D sin ωx。单独的正弦或余弦试解一般行不通，因为求导会使一个变成另一个，除非两项都在，否则无法匹配系数。

f(x) = k e^px ⇒ y_p = C e^px

If f(x) is a sum of several terms, such as f(x) = x + e^2x + sin x, the particular integral can be found term by term. This is a direct consequence of linearity: a sum of particular integrals for each separate forcing term is a particular integral for the sum. This is often called the principle of superposition for particular integrals.

如果 f(x) 是若干项的和，例如 f(x) = x + e^2x + sin x，可以逐项求特解积分。这是线性的直接结果：每个单独强迫项的特解之和，就是整个和的特解。这通常称为特解积分的叠加原理。

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10. Failure Cases: When the Particular Integral Duplicates the Complementary Function | 失效情形：特解与互补函数重复

the forcing term matches a solution of the homogeneous equation. For example, if f(x) = e^2x and the auxiliary equation has a root m = 2, then C e^2x is already part of y_c, and substitution into the left-hand side gives zero instead of e^2x.

当标准试解形式已经出现在互补函数中时，该试解就会失效。这发生在强迫项与齐次方程的解重合时。例如，如果 f(x) = e^2x，而辅助方程有根 m = 2，那么 C e^2x 已经是 y_c 的一部分，代入左边会得到零而不是 e^2x。

root is repeated, multiply by x² instead. Thus for a duplicated exponential term we try y_p = C x e^px, and if p is a repeated root we try y_p = C x² e^px. Similar adjustments apply

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