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

求三阶微分方程 $y^{\prime \prime \prime}$ $+$ $y^{\prime \prime}$ $-$ $y^{\prime}$ $-$ $y$ $=$ $0$ 满足指定初值 $y(0)$ $=$ $4$, $y^{\prime}(0)$ $=$ $4$, $y^{\prime \prime}(0)$ $=$ $0$ 的特解 $y^{*}$.

难度评级：

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

首先，将 $y(0)$ $=$ $4$, $y^{\prime}(0)$ $=$ $4$, $y^{\prime \prime}(0)$ $=$ $0$ 代入原式可得：

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

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

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

Next

又有特征方程：

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

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

Next

即，该常系数齐次微分方程有一个一重特征根和一个二重特征根，于是，可设特解为：

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

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

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

Next

进而有：

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

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

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

Next

整理可得：

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

将 $y(0)$ $=$ $4$, $y^{\prime}(0)$ $=$ $4$, $y^{\prime \prime}(0)$ $=$ $0$, $y^{\prime \prime \prime}(0)$ $=$ $8$ 代入上面这组式子，可得：

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

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

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

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

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

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

Next

综上可知，满足该微分方程指定初值的特解为：

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

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

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