九州大学 システム情報科学府 情報理工学専攻・電気電子工学専攻 2017年度 微分方程式

Author

Zero

Description

次の微分方程式の一般解を求めよ．

(1) \(\frac{dy}{dx} + \frac{y}{x} = \frac{1}{1 + x^2}\)

(2) \((\sqrt{xy} - x) \frac{dy}{dx} = -y\)

Kai

(1)

\[ \begin{aligned} \frac{dy}{dx} + \frac{y}{x} &= \frac{1}{1 + x^2} \\ xy' + y &= \frac{x}{1 + x^2} \\ \frac{d}{dx}(xy) &= \frac{x}{1 + x^2} \end{aligned} \]

両辺を積分

\[ \begin{aligned} xy &= \int\frac{x}{1 + x^2}dx \\ xy &= \frac{1}{2}\log(1 + x^2) + C \\ \therefore y &= \frac{1}{2x}\log(1 + x^2) + Cx^{-1} \end{aligned} \]

(2)

\[ \begin{align} (\sqrt{xy} - x)\frac{dy}{dx} &= -y \notag \\ (\sqrt{\frac{y}{x}} - 1)\frac{dy}{dx} &= -\frac{y}{x} \tag{*} \\ \end{align} \]

\(u = \frac{y}{x} \Leftrightarrow y = ux\) とおく

\(\frac{dy}{dx} = u + x\frac{du}{dx}\)

\((*)\) に代入

\[ \begin{aligned} (\sqrt{u} - 1)(u + x\frac{du}{dx}) &= -u \\ u + x\frac{du}{dx} &= -\frac{u}{\sqrt{u} - 1} \\ x\frac{du}{dx} &= -u - \frac{u}{\sqrt{u} - 1} \\ x\frac{du}{dx} &= - \frac{u \cdot \sqrt{u}}{\sqrt{u} - 1} \\ -\frac{\sqrt{u} - 1}{u\sqrt{u}}du &= \frac{1}{x}dx \end{aligned} \]

\[ \begin{aligned} \int-\frac{u^{\frac{1}{2}} - 1}{u^{\frac{3}{2}}}du &= \int\frac{1}{x}dx \\ \int-\frac{1}{u}du + \int u^{-\frac{3}{2}}du &= \log x + C \\ -\log u - 2u^{-\frac{1}{2}} &= \log x + C \\ -2u^{-\frac{1}{2}} &= \log ux + C \\ -2u^{-\frac{1}{2}} &= \log y + C \end{aligned} \]

\(u = \frac{y}{x}\) より、

\[ \begin{aligned} -2\sqrt{\frac{x}{y}} &= \log y + C \\ \log y + 2\sqrt{\frac{x}{y}} &= -C \\ \log y + \log e^{2\sqrt{\frac{x}{y}}} &= -C \\ \log (ye^{2\sqrt{\frac{x}{y}}}) &= -C \\ \end{aligned} \]