九州大学 システム情報科学府 共通 2015年8月実施 微分方程

标签：

• Kyushu-University
• First-Order-Ordinary-Differential-Equation

Author

思齐塾, 祭音Myyura

Description

$x>0$ で定義された関数 $y(x)$ に関する以下の微分方程式において,変数変換 $x = e^t$ を用いて一般解を求めよ.なお, $y'$ は関数 $y(x)$ の $x$ に関する1階導関数を表している.

(1) $x^2y'' - xy' - x - 3y = 0$ (2) $x^3y''' + 6x^2y'' + 4xy' - 4y = 0$

题目描述

设 $y(x)$ 定义在 $x>0$ 上。使用变量代换

$$x=e^t$$

求下列微分方程的通解；其中 $y',y'',y'''$ 分别表示 $y(x)$ 关于 $x$ 的一、二、三阶导数：

$$x^2y''-xy'-x-3y=0.$$

$$x^3y'''+6x^2y''+4xy'-4y=0.$$

Kai

Solution to (1): $x^2y'' - xy' - x - 3y = 0$

Let $x = e^t$ , then $t = \ln x$ . We have:

$\frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx} = \frac{dy}{dt} \cdot \frac{1}{x} = e^{-t} \frac{dy}{dt}$

$y' = e^{-t} \frac{dy}{dt}$

$\frac{d^2y}{dx^2} = \frac{d}{dx} (e^{-t} \frac{dy}{dt}) = \frac{d}{dt} (e^{-t} \frac{dy}{dt}) \cdot \frac{dt}{dx} = (-e^{-t} \frac{dy}{dt} + e^{-t} \frac{d^2y}{dt^2}) \cdot e^{-t} = e^{-2t} (\frac{d^2y}{dt^2} - \frac{dy}{dt})$

$y'' = e^{-2t} (\frac{d^2y}{dt^2} - \frac{dy}{dt})$

Substitute into the original equation:

$e^{2t} \cdot e^{-2t} (\frac{d^2y}{dt^2} - \frac{dy}{dt}) - e^t \cdot e^{-t} \frac{dy}{dt} - e^t - 3y = 0$

$\frac{d^2y}{dt^2} - \frac{dy}{dt} - \frac{dy}{dt} - e^t - 3y = 0$

$\frac{d^2y}{dt^2} - 2\frac{dy}{dt} - 3y = e^t$

Let $D = \frac{d}{dt}$ . The characteristic equation is $D^2 - 2D - 3 = 0$ , so $(D-3)(D+1) = 0$ . Thus $D = 3$ or $D = -1$ .

The homogeneous solution is $y_h = c_1e^{3t} + c_2e^{-t}$ .

For the particular solution, let $y_p = Ae^t$ . Then $y_p' = Ae^t$ and $y_p'' = Ae^t$ .

$Ae^t - 2Ae^t - 3Ae^t = e^t$

$-4Ae^t = e^t$ , so $A = -\frac{1}{4}$ . Thus $y_p = -\frac{1}{4}e^t$ .

The general solution is $y = c_1e^{3t} + c_2e^{-t} - \frac{1}{4}e^t = c_1x^3 + \frac{c_2}{x} - \frac{1}{4}x$ .

Solution to (2): $x^3y''' + 6x^2y'' + 4xy' - 4y = 0$

Let $x = e^t$ , then $t = \ln x$ . We have:

$y' = e^{-t} \frac{dy}{dt}$

$y'' = e^{-2t} (\frac{d^2y}{dt^2} - \frac{dy}{dt})$

$y''' = e^{-3t} (\frac{d^3y}{dt^3} - 3\frac{d^2y}{dt^2} + 2\frac{dy}{dt})$

Substitute into the original equation:

$e^{3t} e^{-3t} (\frac{d^3y}{dt^3} - 3\frac{d^2y}{dt^2} + 2\frac{dy}{dt}) + 6e^{2t} e^{-2t} (\frac{d^2y}{dt^2} - \frac{dy}{dt}) + 4e^t e^{-t} \frac{dy}{dt} - 4y = 0$

$\frac{d^3y}{dt^3} - 3\frac{d^2y}{dt^2} + 2\frac{dy}{dt} + 6\frac{d^2y}{dt^2} - 6\frac{dy}{dt} + 4\frac{dy}{dt} - 4y = 0$

$\frac{d^3y}{dt^3} + 3\frac{d^2y}{dt^2} + 0\frac{dy}{dt} - 4y = 0$

$\frac{d^3y}{dt^3} + 3\frac{d^2y}{dt^2} - 4y = 0$

Let $D = \frac{d}{dt}$ . The characteristic equation is $D^3 + 3D^2 - 4 = 0$ , so $(D-1)(D+2)^2 = 0$ . Thus $D = 1$ or $D = -2$ .

The general solution is $y = c_1e^{t} + c_2e^{-2t} + c_3te^{-2t} = c_1x + \frac{c_2}{x^2} + \frac{c_3\ln x}{x^2}$ .