C.9 · ปัจจัยอินทิเกรตและสมการดิฟเฟอเรนเชียลไม่ Homogeneous
GRE Subject Test · GRE Mathematics หัวข้อ 25 2:54 การบรรยายภาษาอังกฤษ · คำบรรยายภาษาอังกฤษ + 中文 ลอยตัวบนภาพ
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This unit solves first-order linear equations with an integrating factor, builds the homogeneous and particular parts of a constant-coefficient solution, and handles resonance, verifying the result in the original equation.
本单元用积分因子解一阶线性方程，构造常系数方程解的齐次部分和特解部分，处理共振情形，并在原方程中验证结果。
Set mu as e to the integral of p.
令 μ = e^{∫p}。
Multiplying makes the left side the derivative of mu y.
乘上它后，左边成为 (μy)′。
For y prime plus 2 x y equals x with y of 0 equals 1, mu is e to the x squared, and integrating gives y equals 1 half plus C e to the minus x squared, with C equal to 1 half.
对 y′ + 2xy = x，y(0) = 1，μ = e^{x²}，积分得 y = 1/2 + Ce^{−x²}，C = 1/2。
Solve the characteristic equation for the homogeneous part: distinct roots, a repeated root, or complex roots.
先解特征方程得到齐次部分：不同实根、重根或复根。
The complete solution is y h plus y p.
完整的解是 y_h + y_p。
A polynomial, exponential or sine forcing suggests the same family, with enough coefficients for differentiation.
多项式、指数或正弦型的外力项提示用同类函数作试探解，并留足系数以应对求导。
If a trial duplicates a homogeneous solution, it cannot produce the forcing.
若试探解与齐次解重复，就不能产生外力项。
For y double prime minus 3 y prime plus 2y equals e to the x, the naive K e to the x is annihilated.
对 y″ − 3y′ + 2y = e^x，朴素的 Ke^x 被消成零。
Trying K x e to the x gives minus K e to the x, so y p is minus x e to the x.
试 Kxe^x 得到 −Ke^x，所以 y_p = −xe^x。
The characteristic roots are 1 and 2.
特征根是 1 和 2。
A constant particular solution 2 gives 2 times 2 equals 4, so y equals C1 e to the x plus C2 e to the 2x plus 2.
常数特解 2 给出 2 × 2 = 4，所以 y = C₁e^x + C₂e^{2x} + 2。
The first-order equation, solved separately, gives 1 plus e to the minus x squared, over 2.
一阶方程单独求解，得 (1 + e^{−x²})/2。
Keep the particular solution.
保留特解。
A resonant trial needs an x factor.
共振情形的试探解需要乘以 x。
And the integrating-factor derivation works on a valid coefficient interval without dividing by y.
积分因子的推导在系数有效的区间上进行，不需要除以 y。
The integrating factor is e to the 2x, giving the derivative of e to the 2x times y equal to e to the 3x.
积分因子是 e^{2x}，得到 (e^{2x}y)′ = e^{3x}。
Integrating gives e to the x over 3 plus C e to the minus 2x, with C equal to 2 thirds.
积分得 y = e^x/3 + Ce^{−2x}，C = 2/3。
Differentiating and adding 2y returns e to the x.
求导再加 2y，回到 e^x。
The homogeneous root 1 is double, so both e to the x and x e to the x are annihilated.
齐次方程的根 1 是二重根，所以 e^x 和 xe^x 都被消成零。
Substitute y equals e to the x times v: the equation becomes v double prime equals 1, so v is x squared over 2 plus constants, and the initial conditions make both constants zero.
代入 y = e^x v：方程变为 v″ = 1，所以 v = x²/2 加常数，初始条件使两个常数都为零。
Check yourself.
自我检测。
The integrating factor is e to the x squared.
积分因子是 e^{x²}。
A constant forcing 4 gives particular solution 2.
常数外力 4 给出特解 2。
Resonance with e to the x gives minus x e to the x.
与 e^x 共振时特解是 −xe^x。
A double root needs x squared.
二重根需要乘以 x²。
And initial values alone never verify the equation.
只检查初值永远不能验证方程。