计算微分方程 $y^{\prime \prime}$ $+$ $2 m y^{\prime}$ $+$ $n^{2} y$ $=$ $0$ 满足一定条件特解的无穷限反常积分

一、题目题目 - 荒原之梦

设 $y$ $=$ $y(x)$ 是二阶常系数线性微分方程 $\textcolor{orange}{y^{\prime \prime}}$ $\textcolor{orange}{+}$ $\textcolor{orange}{2 m y^{\prime}}$ $\textcolor{orange}{+}$ $\textcolor{orange}{n^{2} y}$ $\textcolor{orange}{=}$ $\textcolor{orange}{0}$ 满足 $\textcolor{orange}{y(0)}$ $\textcolor{orange}{=}$ $\textcolor{orange}{a}$ 与 $\textcolor{orange}{y^{\prime}(0)}$ $\textcolor{orange}{=}$ $\textcolor{orange}{b}$ 的特解,其中 $m$ 和 $n$ 为常数,且 $\textcolor{orange}{m}$ $\textcolor{orange}{>}$ $\textcolor{orange}{n}$ $\textcolor{orange}{>}$ $\textcolor{orange}{0}$, 则 $\textcolor{orange}{\int_{0}^{+ \infty}}$ $\textcolor{orange}{y(x)}$ $\textcolor{orange}{\mathrm{d} x}$ $\textcolor{orange}{=}$ $\textcolor{orange}{?}$

二、解析 解析 - 荒原之梦

观察可知,题目所给的微分方程是 线 ,要求解的是该微分方程满足给定条件的特解的

Next - 荒原之梦 Next Next - 荒原之梦

首先,由微分方程 $\textcolor{cyan}{y^{\prime \prime}}$ $\textcolor{cyan}{+}$ $\textcolor{cyan}{2 m y^{\prime}}$ $\textcolor{cyan}{+}$ $\textcolor{cyan}{n^{2} y}$ $\textcolor{cyan}{=}$ $\textcolor{cyan}{0}$ 可知,其 为:

$$
\textcolor{orange}{
\lambda^{2} + 2m \lambda + n^{2} = 0
} \quad ①
$$

$$
\lambda = \frac{-2m \pm \sqrt{4m^{2} – 4n^{2}}}{2} \Rightarrow
$$

$$
\lambda = \frac{-2m \pm 2 \sqrt{m^{2} – n^{2}}}{2} \Rightarrow
$$

$$
\textcolor{orange}{\lambda_{1}} = \textcolor{tan}{- m + \sqrt{m^{2} – n^{2}}} \quad ②
$$

$$
\textcolor{orange}{\lambda_{2}} = \textcolor{tan}{- m – \sqrt{m^{2} – n^{2}}} \quad ③
$$

Next - 荒原之梦 Next Next - 荒原之梦

由于 $\textcolor{cyan}{\lambda_{1}}$ $\textcolor{red}{\neq}$ $\textcolor{cyan}{\lambda_{2}}$, 于是可知,该齐次微分方程 可设为:

$$
\textcolor{orange}{
y = C_{1} e^{\lambda_{1} x} + C_{2} e^{\lambda_{2} x}} \quad ④
$$

Next - 荒原之梦 Next Next - 荒原之梦

由于在题目中并 要求解的目标单元 $\textcolor{cyan}{\int_{0}^{+ \infty}}$ $\textcolor{cyan}{y(x)}$ $\textcolor{cyan}{\mathrm{d} x}$, 因此,我们需要通过接下来的步骤,将 $\textcolor{cyan}{\int_{0}^{+ \infty}}$ $\textcolor{cyan}{y(x)}$ $\textcolor{cyan}{\mathrm{d} x}$ 出来。

由微分方程 $y^{\prime \prime}$ $+$ $2 m y^{\prime}$ $+$ $n^{2} y$ $=$ $0$

$$
\textcolor{orange}{
y^{\prime \prime} + 2 m y^{\prime} + n^{2} y = 0} \Rightarrow
$$

为所有式子 无穷限反常积分运算 $\textcolor{cyan}{\int_{0}^{+ \infty}}$:

$$
\textcolor{cyan}{\int_{0}^{+ \infty}} \textcolor{orange}{y^{\prime \prime}} \textcolor{cyan}{\mathrm{d} x} + \textcolor{cyan}{\int_{0}^{+ \infty}} \textcolor{orange}{2 m y^{\prime}} \textcolor{cyan}{\mathrm{d} x} + \textcolor{cyan}{\int_{0}^{+ \infty}} \textcolor{orange}{n^{2} y} \textcolor{cyan}{\mathrm{d} x} = \textcolor{cyan}{\int_{0}^{+ \infty}} \textcolor{orange}{0} \textcolor{cyan}{\mathrm{d} x} \Rightarrow
$$

提取到积分运算符号前:

$$
\int_{0}^{+ \infty} y^{\prime \prime} \mathrm{d} x + 2 m \int_{0}^{+ \infty} y^{\prime} \mathrm{d} x + n^{2} \int_{0}^{+ \infty} y \mathrm{d} x = \int_{0}^{+ \infty} 0 \mathrm{d} x \Rightarrow
$$

把当前能够较容易进行积分运算表示的式子进行

$$
\textcolor{orange}{
y^{\prime} \Big|_{0}^{+ \infty} + 2m y \Big|_{0}^{+ \infty} + n^{2} \int_{0}^{+ \infty} y \mathrm{d} x = 0} \Rightarrow
$$

$\textcolor{cyan}{\int_{0}^{+ \infty}}$ $\textcolor{orange}{y^{\prime \prime}}$ $\mathrm{d} x$ $=$ $\textcolor{orange}{y^{\prime}}$ $\textcolor{cyan}{\Big|_{0}^{+ \infty}}$, $2 m$ $\textcolor{cyan}{\int_{0}^{+ \infty}}$ $\textcolor{orange}{y^{\prime}}$ $\mathrm{d} x$ $=$ $2m \textcolor{orange}{y}$ $\textcolor{cyan}{\Big|_{0}^{+ \infty}}$, $\textcolor{cyan}{\int_{0}^{+ \infty}}$ $0$ $\mathrm{d} x$ $=$ $0$.

$$
y^{\prime}(+ \infty) – y^{\prime}(0) + 2m \Big[ y(+ \infty) – y(0) \Big] + n^{2} \int_{0}^{+ \infty} y \mathrm{d} x = 0 \Rightarrow
$$

$$
n^{2} \int_{0}^{+ \infty} y \mathrm{d} x = y^{\prime}(0) – y^{\prime}(+ \infty) + 2m \Big[ y(0) – y(+ \infty) \Big] \Rightarrow
$$

$$
\textcolor{orange}{
\int_{0}^{+ \infty} y \mathrm{d} x = \frac{y^{\prime}(0) – y^{\prime}(+ \infty) + 2m \Big[ y(0) – y(+ \infty) \Big]}{n^{2}}} \quad ⑤
$$

由题目可知,$y$ $=$ $y(x)$, 因此,$\textcolor{cyan}{\int_{0}^{+ \infty}}$ $\textcolor{tan}{y}$ $\textcolor{cyan}{\mathrm{d} x}$ 就是 $\textcolor{cyan}{\int_{0}^{+ \infty}}$ $\textcolor{tan}{y(x)}$ $\textcolor{cyan}{\mathrm{d} x}$.

Next - 荒原之梦 Next Next - 荒原之梦

由前面的运算

$$
\textcolor{cyan}{y} = \textcolor{orange}{C_{1} e^{\lambda_{1} x} + C_{2} e^{\lambda_{2} x}} \quad ⑥
$$

进而

$$
\textcolor{cyan}{y^{\prime}} = \textcolor{orange}{C_{1} \lambda_{1} e^{\lambda_{1} x} + C_{2} \lambda_{2} e^{\lambda_{2} x}} \quad ⑦
$$

Next - 荒原之梦 Next Next - 荒原之梦

又由于,$m$ $>$ $n$ $>$ $0$,

$$
0 < \sqrt{m^{2} – n^{2}} < m.
$$

$$
\lambda_{1} = – m + \sqrt{m^{2} – n^{2}} < 0.
$$

$$
\lambda_{2} = – m – \sqrt{m^{2} – n^{2}} < 0.
$$

Next - 荒原之梦 Next Next - 荒原之梦

进而,当 $x$ $\rightarrow$ $+ \infty$ 时,

$$
e^{\lambda_{1} x} \rightarrow 0
$$

$$
e^{\lambda_{2} x} \rightarrow 0
$$

Next - 荒原之梦 Next Next - 荒原之梦

$$
y(+ \infty) =
$$

$$
\lim_{x \rightarrow + \infty} y(x) =
$$

$$
\lim_{x \rightarrow + \infty} \Big[ C_{1} e^{\lambda_{1} x} + C_{2} e^{\lambda_{2} x} \Big] = 0.
$$

Next - 荒原之梦 Next Next - 荒原之梦

$$
y^{\prime}(+ \infty) =
$$

$$
\lim_{x \rightarrow + \infty} y^{\prime}(x) =
$$

$$
\lim_{x \rightarrow + \infty} \Big[ C_{1} \lambda_{1} e^{\lambda_{1} x} + C_{2} \lambda_{2} e^{\lambda_{2} x} \Big] = 0.
$$

$$
\textcolor{orange}{
y^{\prime}(+ \infty) = 0} \quad ⑧
$$

$$
\textcolor{orange}{
y(+ \infty) = 0} \quad ⑨
$$

Next - 荒原之梦 Next Next - 荒原之梦

又由题目条件

$$
\textcolor{cyan}{
y(0) = a}
$$

$$
\textcolor{cyan}{
y^{\prime} (0) = b}
$$

Next - 荒原之梦 Next Next - 荒原之梦

$$
\textcolor{orange}{
\int_{0}^{+ \infty} y \mathrm{d} x = \frac{y^{\prime}(0) – y^{\prime}(+ \infty) + 2m \Big[ y(0) – y(+ \infty) \Big]}{n^{2}}} \Rightarrow
$$

$$
\textcolor{orange}{
\int_{0}^{+ \infty} y \mathrm{d} x = \frac{b + 2 m a}{n^{2}}
}
$$

Next - 荒原之梦 Next Next - 荒原之梦

$$
\textcolor{red}{
\int_{0}^{+ \infty} y(x) \mathrm{d} x = \frac{2 m a + b}{n^{2}}
}
$$


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