Forced Linear Differential Equations | 强迫线性微分方程

📚 Forced Linear Differential Equations | 强迫线性微分方程

In IB Mathematics Analysis and Approaches HL, a forced linear differential equation is a differential equation in which the dependent variable and its derivatives appear linearly, while the right-hand side contains a non-zero forcing function. These non-homogeneous equations model systems subjected to external influences, such as a driven mass-spring system, an RLC circuit with an applied voltage, or population growth with harvesting. Understanding how to construct the general solution by combining a complementary function (homogeneous solution) with a particular integral is a core skill in the syllabus.

在IB数学分析与方法(HL)课程中，强迫线性微分方程是指因变量及其导数线性出现，而等号右侧含有非零强迫函数的微分方程。这类非齐次方程描述的是受到外部影响的系统，例如受驱动质量-弹簧系统、施加电压的RLC电路或具有收获项的种群增长。掌握如何通过将余函数(齐次解)与特解积分相结合来构建通解，是考纲中的一项核心技能。

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1. What Is a Forced Linear DE? | 什么是强迫线性微分方程

A linear differential equation is called forced (or non-homogeneous) when it can be written as a_n(x) y⁽ⁿ⁾ + ... + a₁(x) y' ally zero. The function f(x) is the forcing term, representing an external input or disturbance. In IB, attention is focused on first-order and second-order cases with constant coefficients, where the forcing function may be a polynomial, exponential, sine, cosine, or a combination of these.

当一个线性微分方程可以写成 a_n(x) y⁽ⁿ⁾ + ... + a₁(x) y' + a₀(x) y = f(x) 且 f(x) 不恒为零时，我们称其是强迫的(或非齐次的)。函数 f(x) 被称为强迫项，代表外部输入或扰动。在IB课程中，重点集中在常系数的一阶和二阶情形，其中强迫函数可以是多项式、指数函数、正弦、余弦或它们的组合。

The general solution to a forced linear DE is y = y_h + y_p, where y_h is the general solution of the associated homogeneous equation (right-hand side zero) and y_p is any particular solution of the full non-homogeneous equation. This superposition principle is fundamental and always tested.

强迫线性微分方程的通解为 y = y_h + y_p，其中 y_h 是对应齐次方程(右侧为零)的通解，而 y_p 是整个非齐次方程的任一特解。这一叠加原理是基本且必考的内容。

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2. First-Order Linear Equations with Forcing | 一阶强迫线性微分方程

The standard form of a first-order forced linear equation is y' + P(x) y = Q(x), where Q(x) is the forcing function. If Q(x) ≡ on. The method of integrating factor avoids guessing and works for any continuous P and Q.

一阶强迫线性方程的标准形式为 y' + P(x) y = Q(x)，其中 Q(x) 是强迫函数。若 Q(x) ≡ 0，则方程为齐次且可分离；否则我们必须求出一个特解。积分因子法无需猜测，适用于任意连续的 P 和 Q。

For example, consider y' + 2x y = x. This models a first-order system with a forcing term x. The homogeneous part y' + 2x y = 0 gives y_h = C e^−x². To find the particular integral we multiply through by the integrating factor μ(x) = e^{∫ 2x dx} = e^x².

考虑例子 y' + 2x y = x。这模拟了一个带有强迫项 x 的一阶系统。齐次部分 y' + 2x y = 0 给出 y_h = C e^−x²。为求特解积分，我们乘以积分因子 μ(x) = e^{∫ 2x dx} = e^x²。

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3. Integrating Factor Method | 积分因子法

The integrating factor for y' + P(x)y = Q(x) is defined as μ(x) = e^{∫ P(x) dx} ng both sides with respect to x then yields the general solution y = (1/μ) ∫ μ Q dx + C/μ.

对于 y' + P(x)y = Q(x)，积分因子定义为 μ(x) = e^{∫ P(x) dx}。将整个方程乘以 μ(x) 就把左边变成了 μ(x)y 的精确导数：d(μ y)/dx = μ Q。然后对 x 积分两边即得通解 y = (1/μ) ∫ μ Q dx + C/μ。

In the example above, after multiplication we have d/dx (e^x² y) = x e^x². Integrating the right-hand side requires substituting u = x², giving ∫ x e^x² dx = ½ e^x² + constant. Thus the particular solution is y_p = ½, and the general solution becomes y = C e^−x² + ½. Notice that y_p is a constant, even though the forcing was linear in x.

在上例中，相乘后我们得到 d/dx (e^x² y) = x e^x²。对右边积分需令 u = x²，得到 ∫ x e^x² dx = ½ e^x² + 常数。于是特解为 y_p = ½，通解变为 y = C e^−x² + ½。注意尽管强迫项是 x 的线性函数，特解却是一个常数。

C·1 + ½, so C = ½, and the particular solution satisfying the initial condition is y = ½ e^−x² + ½.

IB试题常常要求从初始条件如 y(0) = 1 求出常数 C。代入得 1 = C·1 + ½，故 C = ½，满足初始条件的特解为 y = ½ e^−x² + ½。

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4. Second-Order Constant Coefficient Equations | 二阶常系数微分方程

The second-order forced linear DE with constant coefficients has the form a y'' + b y' and f(t) is the forcing function. In physical systems t usually represents time. The complementary function y_h solves a y'' + b y' + c y = 0 and is found from the roots of the characteristic equation a r² + b r + c = 0.

常系数二阶强迫线性微分方程的形式为 a y'' + b y' + c y = f(t)，其中 a, b, c 为常数，f(t) 是强迫函数。在物理系统中 t 通常表示时间。余函数 y_h 满足 a y'' + b y' + c y = 0，可通过特征方程 a r² + b r + c = 0 的根求得。

The standard cases are: two distinct real roots r₁, r₂ giving y_h = A e^{r₁ t} + B e^{r₂ t}; a repeated root r giving y_h = (A + B t) e^{r t}; and complex conjugate roots α ± iβ giving y_h = e^{α t} (A cos βt + B sin βt). All three appear in IB HL papers.

标准情形为：两个不同实根 r₁, r₂ 给出 y_h = A e^{r₁ t} + B e^{r₂ t}；一个重根 r 给出 y_h = (A + B t) e^{r t}；共轭复根 α ± iβ 给出 y_h = e^{α t} (A cos βt + B sin βt)。这三种情况在IB HL试题中都会出现。

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5. Homogeneous Solution and Characteristic Equation | 齐次解与特征方程

To illustrate, solve y'' + 3y' + 3r + 2 = 0, factorising to (r+1)(r+2)=0, so r = −1, −2. The complementary solution is y_h = C₁ e^−t + C₂ e^−2t. This represents the transient behaviour of the system if it is stable.

举例说明，求解 y'' + 3y' + 2y = 0。特征方程为 r² + 3r + 2 = 0，因式分解得 (r+1)(r+2)=0，故 r = −1, −2。余解为 y_h = C₁ e^−t + C₂ e^−2t。这在系统稳定时代表瞬态行为。

When a forcing term is added, say y'' + 3y' + 2y = 10 sin t, the complete response consists of the transient y_h plus a steady-state particular solution y_p that reflects the forcing. The coefficients C₁ and C₂ are determined from the initial conditions after adding y_p.

当加入强迫项，例如 y'' + 3y' + 2y = 10 sin t，完整响应由瞬态 y_h 加上一个反映强迫作用的稳态特解 y_p 构成。系数 C₁ 和 C₂ 在加入 y_p 后根据初始条件确定。

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6. Method of Undetermined Coefficients | 待定系数法

For constant-coefficient equations where the forcing function is a polynomial, exponential, sine, cosine, or a sum/product of these, the method of undetermined coefficients provides an efficient way to find y_p. You make an educated guess for the form of y_p with unknown coefficients, substitute into the DE, and equate coefficients.

对于强迫函数为多项式、指数、正弦、余弦或它们的和与积的常系数方程，待定系数法提供了一种求 y_p 的高效途径。你需要猜测含有待定系数的 y_p 形式，代入微分方程并比较系数。

The table below lists common forcing functions and the corresponding trial y_p. If any term in the trial solution already appears in the complementary function, multiply by t (or t² if repeated) to achieve independence.

下表列出了常见强迫函数及相应的试探 y_p。若试探解中的任何项已在余函数中出现，需乘以 t(若重复则乘 t²)以保证线性独立。

Forcing f(t) | 强迫项 f(t)	Trial yp | 试探特解 yp
Polynomial a₀ + a₁t + a₂t²	A₀ + A₁t + A₂t²
ekt	A ekt
sin ωt or cos ωt	A sin ωt + B cos ωt
ekt sin ωt or ekt cos ωt	ekt (A sin ωt + B cos ωt)
Polynomial × exponential	(A₀ + A₁t + ...) ekt

Consider y'' + 3y' + 2y = 10 sin t. Since sin t is not a solution of the homogeneous equation, we try y_p B +3A) cos t = 10 sin t. Hence A −3B = 10 and B +3A = 0, giving A = 1, B = −3. Therefore y_p = sin t − 3 cos t.

考虑 y'' + 3y' + 2y = 10 sin t。由于 sin t 不是齐次方程的解，我们试探 y_p = A sin t + B cos t。代入并合并 sin t 和 cos t 的系数得到两个方程：(−A −3B +2A) cos t + (−B +3A +2B) sin t = 10 sin t，化简为 (A −3B) sin t + (B +3A) cos t = 10 sin t。于是 A −3B = 10 且 B +3A = 0，解得 A = 1, B = −3。因此 y_p = sin t − 3 cos t。

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7. Complex Exponential Method for Trigonometric Forcing | 三角函数强迫项的复指数法

An elegant alternative, especially for sinusoidal forcing, replaces the forcing e.g. F₀ sin ωt or F₀ cos ωt by the complex exponential F₀ e^iωt. One solves for the complex particular solution Y_p and then takes the real or imaginary part as required. This method avoids simultaneous equations and is favoured when the homogeneous solution already contains trigonometric terms.

一种更优雅的替代方法，尤其适用于正弦强迫，是将强迫项如 F₀ sin ωt 或 F₀ cos ωt 替换为复指数 F₀ e^iωt。我们解出复数特解 Y_p，然后根据需求取实部或虚部。此方法避免了解方程组，当齐次解中已含三角函数项时尤为适用。

For instance, to find a particular solution for y'' + 2y' + 5y = cos 2t, consider the complex equation z'' + 2z' + 5z = e^i2t. Assume z_p = A e^i2t, substitute: (−4A + i4A + 5A) e^i2t = e^i2t gives A(1 + i4) = 1, so A = 1/(1+4i) = (1−4i)/17. Thus z_p = (1/17 − i(4/17)) e^i2t. Taking the real part yields y_p = (1/17) cos 2t + (4/17) sin 2t, after expanding e^i2t = cos 2t + i sin 2t.

例如，为求 y'' + 2y' + 5y = cos 2t 的特解，考虑复数方程 z'' + 2z' + 5z = e^i2t。假设 z_p = A e^i2t，代入得 (−4A + i4A + 5A) e^i2t = e^i2t，即 A(1 + i4) = 1，故 A = 1/(1+4i) = (1−4i)/17。于是 z_p = (1/17 − i(4/17)) e^i2t。取实部，利用 e^i2t = cos 2t + i sin 2t，得 y_p = (1/17) cos 2t + (4/17) sin 2t。

IB exam papers sometimes ask for this technique explicitly, as it links differential equations with complex numbers and Euler's formula, a cross-topic connection highly valued in the analysis and approaches course.

IB试题有时会明确要求使用这一技巧，因为它将微分方程与复数和欧拉公式联系起来，这种跨知识点的联系在分析与方法的课程中极受重视。

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8. Initial Value Problems and Superposition | 初值问题与叠加原理

Once the general solution y = y_h + y_p is found, initial conditions y(t₀) = y₀, y' . In the first-order case one initial condition suffices; for second order two are needed.

一旦求得通解 y = y_h + y_p，就利用初始条件 y(t₀) = y₀, y'(t₀) = v₀ 来确定任意常数。这给出了系统从特定状态出发的唯一响应。一阶情形只需一个初始条件；二阶则需要两个。

Superposition is also useful when the forcing function is a sum: if y_p1 solves L(y) = f₁(t) and y_p2 solves L(y) = f₂(t), then y_p1 + y_p2 is allows piecewise handling of complicated forcing terms made of sums of basic functions.

当强迫函数为和式时叠加原理也很实用：若 y_p1 满足 L(y) = f₁(t)，y_p2 满足 L(y) = f₂(t)，则 y_p1 + y_p2 满足 L(y) = f₁ + f₂。这使得我们可以分段处理由基本函数之和组成的复杂强迫项。

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9. Resonance and Damped Forced Oscillations | 共振与阻尼强迫振荡

In undamped second-order systems a y'' + c y = F₀ cos ωt, if the driving frequency ω equals the natural frequency ω₀ = √c/a, the trial solution must be multiplied by t. The particular solution then contains a term t sin ωt, causing amplitude to grow linearly without bound — this is pure resonance. With damping present (b > 0), the response remains bounded but can exhibit a peak amplitude when ω is close to the damped natural frequency.

在无阻尼二阶系统 a y'' + c y = F₀ cos ωt 中，若驱动频率 ω 等于固有频率 ω₀ = √c/a，试探解必须乘以 t。此时特解包含 t sin ωt 项，导致振幅无界地线性增长——这是纯粹的共振现象。当存在阻尼 (b > 0) 时，响应保持有界，但当 ω 接近有阻尼固有频率时会出现振幅峰值。

re resonance through a sprung mass or an electric circuit. For the differential equation y'' + 4y = cos 2t, the complementary solution is y_h os 2t + C₂ sin 2t, so the forcing cos 2t is a homogeneous solution. The correct trial is y_p = t(A cos 2t + B sin 2t). After substituting, you find A = 0, B = 1/4, giving y_p /4) sin 2t, which describes oscillations whose amplitude increases proportionally to time.

IB问题常通过弹簧质量系统或电路来研究共振。对于微分方程 y'' + 4y = cos 2t，余解为 y_h = C₁ cos 2t + C₂ sin 2t，因此强迫项 cos 2t 是齐次解。正确的试探特解为 y_p = t(A cos 2t + B sin 2t)。代入后可求得 A = 0, B = 1/4，即 y_p = (t/4) sin 2t，它描述的是振幅随时间成比例增加的振荡。

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10. Applications and Modelling | 应用与建模

Forced linear DEs appear in Newton' n economics models where investment is forced by government spending. In mechanics, Newton's second law for a mass-spring-damper system yields m x'' + c x' + k x = F(t), where F(t) is an external force.

强迫线性微分方程出现在带有外部热源的牛顿冷却定律中、恒速静脉输注的药物浓度模型中，以及由政府支出驱动的投资经济学模型中。在力学中，质量-弹簧-阻尼系统的牛顿第二定律导出 m x'' + c x' + k x = F(t)，其中 F(t) 是外力。

In an LCR series circuit, Kirchhoff's voltage law gives L Q'' + R Q' harge and E(t) is the applied electromotive force. Solving this provides the current I = Q' and to find the amplitude of the steady-state current for an AC voltage E(t) = E₀ sin ωt.

在LCR串联电路中，基尔霍夫电压定律给出 L Q'' + R Q' + (1/C) Q = E(t)，其中 Q 为电荷，E(t) 为施加的电动势。求解此方程可得到电流 I = Q'。学生可能会被要求解释瞬态和稳态部分，并求出对 AC 电压 E(t) = E₀ sin ωt 的稳态电流振幅。

IB HL questions often provide a scenario, ask to set up the differential equation, find the general solution using homogeneous and particular solutions, fit initial data, and analyze long-term behaviour. Being fluent with both undetermined coefficients and integrating factor methods is crucial.

IB HL试题常给出一个场景，要求建立微分方程，利用齐次解和特解求通解，拟合初始数据，并分析长期行为。熟练运用待定系数法和积分因子法至关重要。

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11. Summary and Exam Tips | 总结与考试技巧

To succeed in forced linear differential equations on the IB exam, always classify the DE: first-order leads to integrating factor, second-order constant coefficient to characteristic equation plus undetermined coefficients or complex method. Check whether the forcing term overlaps with the complementary function and adjust the trial y_p by multiplying by t if necessary.

要在IB考试中成功解决强迫线性微分方程，务必先分类：一阶引导到积分因子法，二阶常系数则到特征方程加待定系数法或复指数法。检查强迫项是否与余函数重合，必要时将试探 y_p 乘以 t。

Show working clearly: writing the homogeneous DE and its characteristic equation, finding roots, stating y_h, proposing y_p with unknown constants, substituting, equating coefficients, and stating the general solution. Finally, apply initial conditions carefully. Using the complex exponential approach for sine/cosine forcing can save time and reduce algebraic errors.

清晰展示解题过程：写出齐次方程及其特征方程，求出根，给出 y_h，提出带有待定常数的 y_p，代入、比较系数，并给出通解。最后仔细应用初始条件。对正弦/余弦强迫采用复指数方法可以节省时间并减少代数错误。

• Always write the DE in standard form before choosing a method.
• Memorise the trial forms for undetermined coefficients and the modification rule.
• For first-order integrating factors, remember μ(x) = e^∫P(x)dx; do not omit the constant of integration, but it can be absorbed in the final constant.
• Practice interpreting solutions in context: transient vs steady-state, resonance, damping.

这些技巧总结如下：

• 在选择方法之前，务必将微分方程写成标准形式。
• 记住待定系数法的试探形式及其修正规则。
• 对于一阶积分因子，牢记 μ(x) = e^∫P(x)dx；不要遗漏积分常数，但该常数可被最终常数吸收。
• 练习在具体情境中解释解：瞬态与稳态、共振、阻尼。

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