# 二阶非齐次微分方程满足一定条件特解的计算步骤你还记得吗？

## 二、解析

### 1. 求解“齐通”

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

$$\lambda=\frac{-1 \pm \sqrt{1+8}}{2} \Rightarrow \lambda_{1}=1, \ \lambda_{2}=-2$$

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

### 2. 求解“非齐特”

$$y^{*}=x^{\textcolor{springgreen}{k}}(A x+B) e^{\textcolor{orangered}{1} \cdot x} \Rightarrow$$

Tips:

$e^{\textcolor{orangered}{1} \cdot x}$ 来自原式的右端项中的 $e^{x}$, 由于 $\textcolor{orangered}{1} = \lambda_{1} \neq \lambda_{2}$, 因此 $\textcolor{springgreen}{k} = 1$

$$y^{*}=x(A x+B) e^{x} \Rightarrow$$

$$y^{*}=\left(A x^{2}+B x\right) e^{x} \Rightarrow$$

$$y^{* \prime}=(2 A x+B) e^{x}+\left(A x^{2}+B x\right) e^{x} \Rightarrow$$

$$y^{* \prime}=\left(A x^{2}+B x+2 A x+B\right) e^{x} \Rightarrow$$

$$y ^{* \prime \prime}=(2 A x+B+2 A) e^{x}+\left(A x^{2}+B x+2 A x+B\right) e^{x}$$

$$y^{* \prime}+y^{* \prime}-2 y^{*}=(6 x+2) e^{x}$$

$$2 A x+B+2 A+A x^{2}+B x+2 A x+B+$$

$$A x^{2}+B x+2 A x+B-2 A x^{2}-2 B x=6 x+2 \Rightarrow$$

$$\left\{\begin{array}{l}2 A+B+2 A+B+2 A-2 B=6 \\ B+2 A+B+B=2\end{array} \Rightarrow\right.$$

$$\left\{\begin{array}{l}6 A=6 \\ 2 A+3 B=2\end{array} \Rightarrow\left\{\begin{array}{l}A=1 \\ B=0\end{array} \Rightarrow\right.\right.$$

$$y^{*}=x^{2} e^{x}$$

### 3. 求解“齐通+非齐特”

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

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

$$\left\{\begin{array}{l}x=0 \Rightarrow y=3 \\ x=0 \Rightarrow y^{\prime}=0\end{array} \Rightarrow\left\{\begin{array}{l}C_{1}+C_{2}=3 \\ C_{1}-2 C_{2}=0\end{array} \Rightarrow\right.\right.$$

$$\left\{\begin{array}{l}C_{1}=2 \\ C_{2}=1\end{array} \Rightarrow\right.$$

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

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