# 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

## 二、解析

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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$$0 < \sqrt{m^{2} – n^{2}} < m.$$

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

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

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

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

Next

$$y(+ \infty) =$$

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

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

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

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

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

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

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

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

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

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

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

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