IB Mathematics: Special Cases of Particular Solutions in Differential Equations | IB数学：微分方程特解的特殊情形

📚 IB Mathematics: Special Cases of Particular Solutions in Differential Equations | IB数学：微分方程特解的特殊情形

ts and f(x) is a polynomial, an exponential, a sine/cosine function, or a product of these. The key skill in IB Mathematics Analysis & Approaches HL is to choose the correct form for a particular solution, especially when the obvious guess collides with the complementary function.

本文研究常系数线性二阶微分方程：a y″ + b y′ + c y = f(x)，其中 a、b、c 为实数常数，f(x) 为多项式、指数函数、正弦/余弦函数，或它们的乘积。IB 数学分析与方法（HL）中的关键技能是选出正确的特解形式，尤其是当“看起来显然”的猜测恰好与补函数相撞时。

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1. General Solution and Particular Integral | 通解与特解

mogeneous equation a y″ + b y′ + c y = f(x), the general solution is the sum of two parts:

对于任意线性非齐次方程 a y″ + b y′ + c y = f(x)，通解由两部分相加得到：

y = y_c + y_p

The complementary function y_c + b y′ + c y = 0. It contains two arbitrary constants C₁ and C₂. The particular integral y_p is any single solution of the full equation and contains no arbitrary constants.

补函数 y_c 是对应齐次方程 a y″ + b y′ + c y = 0 的通解，它含有两个任意常数 C₁ 和 C₂。特解 y_p 是完整方程的任意一个解，不含任意常数。

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

The method of undetermined coefficients uses the form of f(x) to suggest a trial expression for y_p. The table below shows the standard guesses before any resonance adjustment.

待定系数法根据 f(x) 的形式来猜测 y_p 的表达式。下表给出在考虑“共振”调整之前的标准试探形式。

f(x)	Trial yp
Polynomial of degree n	aₙxⁿ + aₙ₋₁xⁿ⁻¹ + … + a₀
ekx	A ekx
sin kx or cos kx	A sin kx + B cos kx
ekxPₙ(x)	ekxQₙ(x), where Qₙ is a polynomial of degree n
eαx sin βx or eαx cos βx	eαx(A sin βx + B cos βx)

These guesses are valid only if no term in the trial y_p already appears in the complementary function. In that situation the special cases below apply.

这些猜测只有在试探函数中的任何一项都不出现在补函数中时才有效。如果出现重叠，就需要使用下面的特殊情形。

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3. Why the Obvious Trial Solution Fails | 为什么“显然”的试解会失效

Consider y″ − y = e^x. The auxiliary equation is λ² − 1 = 0, so λ = 1 and λ = −1. Therefore y_c = C₁e^x + C₂e^−x.

来看 y″ − y = e^x。特征方程为 λ² − 1 = 0，于是 λ = 1 与 λ = −1，所以 y_c = C₁e^x + C₂e^−x。

If we try y_p = A e^x, substitution gives y_p″ − y_p = Ae^x − Ae^x = 0. The left-hand side becomes zero because e^x is already a solution of the homogeneous equation. No value of A can make it equal e^x.

如果我们尝试 y_p = A e^x，代入后得到 y_p″ − y_p = Ae^x − Ae^x = 0。因为 e^x 已经是齐次方程的解，所以左端变为零，无论 A 取什么值都不能使它等于 e^x。

y_p = A x e^x gives y_p″ − y_p = 2Ae^x, so A = ½.

The cure is to multiply the original guess by x. This removes the overlap with the complementary function.

解决办法是把原来的猜测乘以 x，这样就可以消除与补函数的重叠。

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4. Special Case 1: Repeated Exponential Roots | 特殊情形一：指数型重根

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