求三阶微分方程 $y^{\prime \prime \prime}$ $+$ $y^{\prime \prime}$ $-$ $y^{\prime}$ $-$ $y$ $=$ $0$ 满足指定初值的特解 $y^{*}$

一、题目

解题思路简图

 graph TB
A(判断方程类型) --> B(三阶常系数齐次) --> C(求出特征值) --> D(根据公式确定通解的形式) --> E(代入条件求出待定常数) --> F(写出特解)


二、解析

$$y^{\prime \prime \prime}(0) + y^{\prime \prime} – y^{\prime} – y = 0 \Rightarrow$$

$$y^{\prime \prime \prime}(0) + 0 – 4 – 4 = 0 \Rightarrow$$

$$y^{\prime \prime \prime}(0) = 8$$

Next

$$\lambda^{3} + \lambda^{2} – \lambda – 1 = 0 \Rightarrow$$

$$\left\{\begin{matrix} \lambda_{1} = 1; \\ \lambda_{2} = \lambda_{3} = -1. \end{matrix}\right.$$

Next

$$Y = C_{1} e^{\lambda_{1} x} + (C_{2} + C_{3} x)e^{\lambda_{2} x} \Rightarrow$$

$$Y = C_{1} e^{\lambda_{1} x} + C_{2} e^{\lambda_{2} x} + C_{3} x e^{\lambda_{2} x} \Rightarrow$$

$$Y = C_{1} e^{x} + C_{2} e^{- x} + C_{3} x e^{- x} \Rightarrow$$

Next

$$Y^{\prime} = C_{1} e^{x} – C_{2}e^{-x} + C_{3} e^{-x} – C_{3} x e^{-x}$$

$$Y^{\prime \prime} = C_{1} e^{x} + C_{2} e^{-x} – C_{3} e^{-x} – C_{3} e^{-x} + C_{3} x e^{-x}$$

$$Y^{\prime \prime \prime} = C_{1} e^{-x} – C_{2} e^{-x} + C_{3} e^{-x} + C_{3} e^{-x} + C_{3} e^{-x} – C_{3}x e^{-x}$$

Next

$$\left\{\begin{matrix} Y = C_{1} e^{x} + C_{2} e^{- x} + C_{3} x e^{- x}; \\ Y^{\prime} = C_{1} e^{x} – C_{2}e^{-x} + C_{3} e^{-x} – C_{3} x e^{-x}; \\ Y^{\prime \prime} = C_{1} e^{x} + C_{2} e^{-x} – 2 C_{3} e^{-x} + C_{3} x e^{-x} \\ Y^{\prime \prime \prime} = C_{1} e^{-x} – C_{2} e^{-x} + 3 C_{3} e^{-x} – C_{3}x e^{-x} \end{matrix}\right.$$

$$\left\{\begin{matrix} C_{1} + C_{2} = 4; \\ C_{1} – C_{2} + C_{3} = 4; \\ C_{1} + C_{2} – 2 C_{3} = 0; \\ C_{1} – C_{2} + 3 C_{3} = 8 \end{matrix}\right. \Rightarrow$$

$$\left\{\begin{matrix} C_{1} = 4 – C_{2}; \\ 2 C_{1} – C_{3} = 4 \\ C_{1} – C_{2} + 3 C_{3} = 8 \end{matrix}\right. \Rightarrow$$

$$\left\{\begin{matrix} 2 (4 – C_{2}) – C_{3} = 4 \\ 4 – 2C_{2} + 3 C_{3} = 8 \end{matrix}\right. \Rightarrow$$

$$\left\{\begin{matrix} 8 – 2C_{2} – C_{3} = 4 \\ 4 – 2C_{2} + 3 C_{3} = 8 \end{matrix}\right. \Rightarrow$$

$$4 – 4 C_{3} = -4 \Rightarrow C_{3} = 2 \Rightarrow$$

$$\left\{\begin{matrix} C_{1} = 3\\ C_{2} = 1\\ C_{3} = 2 \end{matrix}\right.$$

Next

$$Y = C_{1} e^{x} + C_{2} e^{- x} + C_{3} x e^{- x} \Rightarrow$$

$$y^{*} = 3 e^{x} + e^{- x} + 2 x e^{- x}.$$