# 不要被这道题题目中所用的变量名迷惑了哦

## 一、题目

(A) $y^{*}=x \mathrm{e}^{x}(A \cos 2 x+B \sin 2 x)$
(B) $y^{*}=\mathrm{e}^{x}(A \cos 2 x+B \sin 2 x)$
(C) $y^{*}=A x \mathrm{e}^{x} \cos 2 x$
(D) $y^{*}=A x \mathrm{e}^{x} \sin 2 x$

## 二、解析

$$y^{\prime \prime}-2 y^{\prime}+5 y=e^{x} \cos 2 x \Rightarrow$$

$$\lambda^{2}-2 \lambda+5=0 \Rightarrow$$

$$\lambda = \frac{2 \pm \sqrt{4-20}}{2} \Rightarrow \lambda=1 \pm 2 i$$

$$Y^{*}=x^{k} e^{2 x}\left[W_{n}(x) \cos \beta x+Q_{n}(x) \sin \beta x\right] \Rightarrow$$

$$\alpha \pm i \beta=\lambda=1 \pm 2 i \Rightarrow k=1$$

$$Y^{*} = x e^{x} [\textcolor{springgreen}{a} \cos 2x + \textcolor{springgreen}{b} \sin 2x]$$

$$y_{1}=e^{x} \cos 2 x, \quad y_{2}=e^{x} \sin 2 x$$

$$Y^{*}=x\left(a y_{1}+b y_{2}\right)$$

$$Y^{* \prime}=\left(a y_{1}+b y_{2}\right)+x\left(a y_{1}^{\prime}+b y_{2}^{\prime}\right)$$

$$Y^{* \prime \prime}=2\left(a y_{1}^{\prime}+b y_{2}^{\prime}\right)+x\left(a y_{1}^{\prime \prime}+b y_{2}^{\prime \prime}\right)$$

$$Y^{* \prime \prime}-2 Y^{* \prime}+5 Y^{*}=$$

$$2\left(a y_{1}^{\prime}+b y_{2}^{\prime}\right)+x\left(a y_{1}^{\prime \prime}+b y_{2}^{\prime \prime}\right)-$$

$$2\left(a y_{1}+b y_{2}\right)-2 x\left(a y_{1}^{\prime}+b y_{2}^{\prime}\right)+$$

$$5 x\left(a y_{1}+b y_{2}\right)=$$

$$x\left[a\left(y_{1}^{\prime \prime}-2 y_{1}^{\prime}+5 y_{1}\right)+b\left(y_{2}^{\prime \prime}-2 y_{2}^{\prime}+5 y_{2}\right)\right] +$$

$$2 a\left(y_{1}^{\prime}-y_{1}\right)+2 b\left(y_{2}^{\prime}-y_{2}\right) \Rightarrow$$

$$y_{1}^{\prime \prime}-2 y_{1}^{\prime}+5 y_{1}=0 \quad y_{2}^{\prime \prime}-2 y_{2}+5 y_{2}=0$$

$$\text { 原式 }=2 a\left(y_{1}^{\prime}-y_{1}\right)+2 b\left(y_{2}^{\prime}-y_{2}\right)=$$

$$2 a\left(e^{x} \cos 2 x-2 e^{x} \sin 2 x-e^{x} \cos 2 x\right)+$$

$$2 b\left(e^{x} \sin 2 x+2 e^{x} \cos 2 x-e^{x} \sin 2 x\right)=$$

$$-4 a e^{x} \sin 2 x+4 b e^{x} \cos 2 x = e^{x} \cos 2 x \Rightarrow$$

$$a=0, b=\frac{1}{4} \Rightarrow$$

$$Y^{*}=\frac{1}{4} x e^{x} \sin 2 x$$