九州大学 システム情報科学府 共通 2014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

次の微分方程式を解け.

(1) $(2xy^3 - e^x \cos y)dx + (3x^2y^2 + e^x \sin y)dy = 0$

(2) $\left(y + \frac{-2y}{x^2 - y^2}\right)dx + \left(x + \frac{2x}{x^2 - y^2}\right)dy = 0$

题目描述

求下列两个一阶微分方程的通解：

$$(2xy^3-e^x\cos y)\,dx +(3x^2y^2+e^x\sin y)\,dy=0.$$

$$\left(y-\frac{2y}{x^2-y^2}\right)\,dx +\left(x+\frac{2x}{x^2-y^2}\right)\,dy=0.$$

Kai

(1) Let $M(x,y) = 2xy^3 - e^x\cos y$ and $N(x,y) = 3x^2y^2 + e^x\sin y$ .

$$\frac{\partial M}{\partial y} = 6xy^2 + e^x \sin y$$

$$\frac{\partial N}{\partial x} = 6xy^2 + e^x \sin y$$

Since $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$ , the equation is exact. Then, we want to find a function $F(x,y)$ such that $\frac{\partial F}{\partial x} = M$ and $\frac{\partial F}{\partial y} = N$ .

$$\frac{\partial F}{\partial x} = 2xy^3 - e^x\cos y$$

$$F(x,y) = \int (2xy^3 - e^x\cos y) dx = x^2y^3 - e^x\cos y + g(y)$$

Now, differentiate $F(x,y)$ with respect to $y$ :

$$\frac{\partial F}{\partial y} = 3x^2y^2 + e^x\sin y + g'(y) = N(x,y) = 3x^2y^2 + e^x\sin y$$

Thus, $g'(y) = 0$ , which means $g(y) = C$ , where $C$ is a constant. Therefore, the solution is $x^2y^3 - e^x\cos y = C$ .

(2) Let $M(x,y) = y - \frac{2y}{x^2 - y^2}$ and $N(x,y) = x + \frac{2x}{x^2 - y^2}$ .

$$\frac{\partial M}{\partial y} = 1 - \frac{2(x^2 - y^2) - 2y(-2y)}{(x^2 - y^2)^2} = 1 - \frac{2x^2 - 2y^2 + 4y^2}{(x^2 - y^2)^2} = 1 - \frac{2x^2 + 2y^2}{(x^2 - y^2)^2}$$

$$\frac{\partial N}{\partial x} = 1 + \frac{2(x^2 - y^2) - 2x(2x)}{(x^2 - y^2)^2} = 1 + \frac{2x^2 - 2y^2 - 4x^2}{(x^2 - y^2)^2} = 1 - \frac{2x^2 + 2y^2}{(x^2 - y^2)^2}$$

Since $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$ , the equation is exact. Then, we want to find a function $F(x,y)$ such that $\frac{\partial F}{\partial x} = M$ and $\frac{\partial F}{\partial y} = N$ .

$$\frac{\partial F}{\partial x} = y - \frac{2y}{x^2 - y^2}$$

$$F(x,y) = \int \left(y - \frac{2y}{x^2 - y^2}\right) dx = xy - y\int \frac{2}{x^2 - y^2} dx = xy - y \int \frac{1}{y}\left(\frac{1}{x-y} - \frac{1}{x+y}\right) dx = xy - \ln|x-y| + \ln|x+y|$$

$$=xy-\ln|x-y|+\ln|x+y|+g(y) =xy+\ln\left|\frac{x+y}{x-y}\right|+g(y).$$

これを $y$ で微分すると

$$\frac{\partial F}{\partial y} =x+\frac{1}{x+y}+\frac{1}{x-y}+g'(y) =x+\frac{2x}{x^2-y^2}+g'(y).$$

$F_y=N$ と比較して $g'(y)=0$ である。したがって一般積分は

$$\boxed{xy+\ln\left|\frac{x+y}{x-y}\right|=C},$$

ただし元の方程式と対数が定義される $x^2\ne y^2$ の各領域で考える。