IB Mathematics: General Solution of Forced Linear Differential Equations | IB数学：受迫线性微分方程的通解

📚 IB Mathematics: General Solution of Forced Linear Differential Equations | IB数学：受迫线性微分方程的通解

In IB Mathematics: Analysis and Approaches HL, students are often asked to find the general solution of a linear differential equation that contains a non-zero forcing function. The word “forced” simply means that the equation is not equal to zero: some external input, such as a driving force or a current source, is present on the right-hand side.

在IB数学：分析与方法高阶课程中，我们常需要求解带有非零强迫项的线性微分方程的通解。所谓“受迫”，就是指方程右端不为零，而是存在某种外部输入，例如驱动力或电流源。

1. What is a Forced Linear Differential Equation? | 什么是受迫线性微分方程？

A forced linear differential equation is a linear ordinary differential equation in which the dependent variable and its derivatives appear to the first power only, and the equation contains a non-zero term that does not involve the unknown function. This non-zero term is called the forcing term.

受迫线性微分方程是未知函数及其各阶导数都仅以一次幂出现的线性常微分方程，并且方程中含有一个与未知函数无关的非零项。这个非零项称为强迫项或驱动项。

For a second-order equation with constant coefficients, the standard form is:

对于常系数二阶方程，其标准形式是：

y″ + a y′ + b y = f(x)

where a and b are constants, and f(x) is the forcing term. If f(x) = 0, the equation is said to be unforced, or homogeneous.

其中 a 和 b 是常数，f(x) 是强迫项。若 f(x) = 0，则称该方程为无强迫或齐次方程。

2. Classification of Linear Differential Equations | 线性微分方程的分类

For a first-order linear equation, the general forced form is dy/dx + P(x)y = Q(x), where Q(x) is the forcing term. For second-order constant-coefficient equations, the forced form is y″ + a y′ + b y = f(x).

对于一阶线性方程，一般受迫形式为 dy/dx + P(x)y = Q(x)，其中 Q(x) 是强迫项。对于常系数二阶方程，受迫形式为 y″ + a y′ + b y = f(x)。

If f(x) = 0, the equation is homogeneous; its solution is called the complementary function.

若 f(x) = 0，方程为齐次方程；它的解称为余函数或补函数。
If f(x) ≠ 0, the equation is non-homogeneous, or forced; we need both a complementary function and a particular integral.

若 f(x) ≠ 0，方程为非齐次方程，也就是受迫方程；我们需要同时求出余函数和特解。

3. The Structure of the General Solution | 通解的结构

The most important idea for forced linear differential equations is that the general solution is the sum of two parts:

求解受迫线性微分方程最重要的思想是：通解由两部分叠加而成：

y(x) = y_c(x) + y_p(x)

Here y_c is the complementary function, which solves the corresponding homogeneous equation; y_p is a particular integral, which is any single solution of the full forced equation.

其中 y_c 是余函数，它满足对应的齐次方程；y_p 是特解，它是原受迫方程的任意一个解。

The reason is linearity: if L(y) = y″ + a y′ + by, then L(y_c + y_p) = L(y_c) + L(y_p) = 0 + f(x) = f(x).

原因是线性性：若 L(y) = y″ + a y′ + by，则 L(y_c + y_p) = L(y_c) + L(y_p) = 0 + f(x) = f(x)。

4. The Homogeneous Equation and the Auxiliary Equation | 齐次方程与辅助方程

To find y_c, we consider the homogeneous equation y″ + a y′ + b y = 0. We assume a solution of the form y = e^λx. Substitution gives the auxiliary equation:

为了求 y_c，我们先考虑齐次方程 y″ + a y′ + b y = 0。设解的形式为 y = e^λx，代入后得到辅助方程：

λ² + aλ + b = 0

The nature of the complementary function depends entirely on the roots of this quadratic equation. There are three cases: distinct real roots, a repeated real root, and complex conjugate roots.

余函数的形式完全取决于这个二次方程的根。一共有三种情况：两个不同实根、一个重实根和一对共轭复根。

5. Case 1: Distinct Real Roots | 情况一：两个不同实根

If the auxiliary equation has two distinct real roots λ₁ and λ₂, then the two independent solutions are e^λ₁x and e^λ₂x. Therefore the complementary function is:

若辅助方程有两个不同实根 λ₁ 和 λ₂，则两个线性无关解为 e^λ₁x 和 e^λ₂x。因此余函数为：

y_c = C₁e^λ₁x + C₂e^λ₂x

For example, the equation y″ − 3y′ + 2y = 0 has auxiliary equation λ² − 3λ + 2 = 0, with roots λ = 1 and λ = 2. Hence y_c = C₁e^x + C₂e^2x.

例如，方程 y″ − 3y′ + 2y = 0 的辅助方程为 λ² − 3λ + 2 = 0，根为 λ = 1 和 λ = 2。因此 y_c = C₁e^x + C₂e^2x。

6. Case 2: Repeated Root | 情况二：重根

If the auxiliary equation has one repeated real root λ = −a/2, then e^λx is only one solution. A second independent solution is obtained by multiplying by x.

若辅助方程只有一个重实根 λ = −a/2，那么 e^λx 只是一个解。另一个线性无关解可以通过乘以 x 得到。

y_c = (C₁ + C₂x)e^λx

For example, y″ − 4y′ + 4y = 0 has auxiliary equation (λ − 2)² = 0, so λ = 2 is repeated. Therefore y_c = (C₁ + C₂x)e^2x.

例如，y″ − 4y′ + 4y = 0 的辅助方程为 (λ − 2)² = 0，因此 λ = 2 是重根。于是 y_c = (C₁ + C₂x)e^2x。

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

If the auxiliary equation has complex roots λ = α ± βi, where β ≠ 0, then the complementary function can be written in real form using Euler’s formula:

若辅助方程具有复根 λ = α ± βi，其中 β ≠ 0，则可以利用欧拉公式将余函数写成实形式：

y_c = e^αx(C₁cos βx + C₂sin βx)

For example, the equation y″ + 4y = 0 has auxiliary equation λ² + 4 = 0, so λ = ±2i. Here α = 0 and β = 2, giving y_c = C₁cos 2x + C₂sin 2x.

例如，方程 y″ + 4y = 0 的辅助方程为 λ² + 4 = 0，所以 λ = ±2i。此时 α = 0，β = 2，因此 y_c = C₁cos 2x + C₂sin 2x。

8. First-Order Forced Linear Equations | 一阶受迫线性方程的通解

For the first-order forced equation dy/dx + P(x)y = Q(x), we use the integrating factor μ = e^∫P(x)dx. Multiplying both sides by μ gives:

对于一阶受迫方程 dy/dx + P(x)y = Q(x)，我们使用积分因子 μ = e^∫P(x)dx。将方程两边乘以 μ 后得到：

d/dx(μy) = μQ(x)

Therefore the general solution is:

因此通解为：

y = (∫ μQ(x)dx + C) / μ

For example, solve dy/dx + 2y = x. Here P = 2, so μ = e^2x. Then y = e^−2x(∫ x e^2x dx + C) = ½x − ¼ + Ce^−2x.

例如，求解 dy/dx + 2y = x。这里 P = 2，所以 μ = e^2x。于是 y = e^−2x(∫ x e^2x dx + C) = ½x − ¼ + Ce^−2x。

9. Particular Integrals Using Undetermined Coefficients | 待定系数法求特解

To find a particular integral y_p for a forced equation, the IB syllabus focuses on the method of undetermined coefficients. The table below shows the standard trial forms for common forcing terms.

为了求受迫方程的特解 y_p，IB课程重点考查待定系数法。下表给出了常见强迫项对应设出的特解形式。

Forcing term f(x)	Trial particular integral y_p(x)
Polynomial of degree n	Aₙxⁿ + Aₙ₋₁xⁿ⁻¹ + ⋯ + A₁x + A₀
e^kx	Ae^kx
sin kx or cos kx	A sin kx + B cos kx
e^kxsin mx or e^kxcos mx	e^kx(A sin mx + B cos mx)

If the trial form is already a solution of the homogeneous equation, then the tentative y_p must be multiplied

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IB Mathematics: General Solution of Forced Linear Differential Equations | IB数学：受迫线性微分方程的通解