# 你知道怎么确定已知解的哪部分是非齐次微分方程的特解吗？

## 二、解析

### 1. 挖掘条件

$$y_{1}-y_{2}=x e^{x}+e^{2 x}-x e^{x}-e^{-x}=e^{2 x}-e^{-x}$$

$$y_{1}-y_{3}=x e^{x}+e^{2 x}-x e^{x}-e^{2 x}+e^{-x}=e^{-x}$$

$$\left(y_{1}-y_{2}\right)+\left(y_{1}-y_{3}\right)=e^{2 x}$$

$$y_{0}^{*}=C_{1} e^{2 x}+C_{2} e^{-x}$$

$$\lambda_{1}=-1, \ \lambda_{2}=2 \Rightarrow \lambda_{1} \neq \lambda_{2}$$

$$\lambda_{1}=\lambda_{2}$$

$$y^{*}=\left(C_{1}+C_{2} x\right) e^{\lambda_{1} x} = C_{1} e^{\lambda_{1} x} + C_{2} \cdot \textcolor{orangered}{x e^{\lambda_{1} x}}$$

$$Y^{*} = \textcolor{orangered}{ x e^{x} }$$

### 2. 解法一

$$\lambda_{1}=-1, \ \lambda_{2}=+2 \Rightarrow$$

$$(\lambda + 1)(\lambda – 2)=0 \Rightarrow$$

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

$$y^{\prime \prime}-y^{\prime}-2 y=f(x)$$

$$Y^{*}=x e^{x} \Rightarrow$$

$$Y^{* \prime}=e^{x}+x e^{x}$$

$$Y^{* \prime \prime}=e^{x}+e^{x}+x e^{x} \Rightarrow$$

$$2 e^{x}+x e^{x}-e^{x}-x e^{x}-2 x e^{x}=$$

$$e^{x}-2 x e^{x}=(1-2 x) e^{x} \Rightarrow$$

$$y^{\prime \prime}-y^{\prime}-2 y=(1-2 x) e^{x}$$

### 3. 解法二

$$Y=y^{*}+Y^{*} \Rightarrow$$

$$Y=C_{1} e^{2 x}+C_{2} e^{-x}+x e^{x} \Rightarrow$$

$$Y^{\prime}=2 C_{1} e^{2 x}-C_{2} e^{-x}+e^{x}+x e^{x}$$

$$Y^{\prime \prime}=4 C_{1} e^{2 x}+C_{2} e^{-x}+e^{x}+e^{x}+x e^{x}$$

$$y^{\prime \prime}-y^{\prime}-2 y=f(x)$$

$$y^{\prime \prime}-y^{\prime} \Rightarrow$$

$$4 C_{1} e^{2 x}+C_{2} e^{-x}+2 e^{x}+x e^{x}-2 C_{1} e^{2 x}+$$

$$C_{2} e^{-x}-e^{x}-x e^{x} \Rightarrow$$

$$2 C_{1} e^{2 x}+2 C_{2} e^{-x}+e^{x}=$$

$$2\left(C_{1} e^{2 x}+C_{2} e^{-x}+x e^{x}\right)-2 x e^{x}+e^{x}=$$

$$2 y+(1-2 x) e^{x} \Rightarrow$$

$$2 y+(1-2 x) e^{x}=2 y+f(x) \Rightarrow$$

$$f(x)=(1-2 x) e^{x} \Rightarrow$$

$$y^{\prime \prime}-y^{\prime}-2 y=(1-2 x) e^{x}$$