求微分方程 $\frac{\mathrm{d} y}{\mathrm{d} x}$ $=$ $\frac{2xy – y^{2}}{x^{2} – 2xy}$ 满足指定条件的特解

一、题目

求微分方程 $\frac{\mathrm{d} y}{\mathrm{d} x}$ $=$ $\frac{2xy – y^{2}}{x^{2} – 2xy}$ 满足 $y(1)$ $=$ $-2$ 的特解。

难度评级：

解 题 思 路 简 图

graph TB
	A(判断方程类型) --> B(一阶齐次) --> C(构造出 y/x) --> D(代入条件求出待定常数) --> E(写出特解)

二、解析

观察知，这是一个一阶齐次微分方程。

又：

$$
\frac{\mathrm{d} y}{\mathrm{d} x} = \frac{2xy – y^{2}}{x^{2} – 2xy} \Rightarrow
$$

$$
\frac{\mathrm{d} y}{\mathrm{d} x} = \frac{\frac{2y}{x} – (\frac{y}{x})^{2}}{1 – \frac{2y}{x}} \Rightarrow
$$

Next

令 $u(x)$ $=$ $u$ $=$ $\frac{y}{x}$ $\Rightarrow$

$$
\frac{\mathrm{d} y}{\mathrm{d} x} = \frac{2u – u^{2}}{1 – 2u} \Rightarrow
$$

Next

又 $u$ $=$ $\frac{y}{x}$ $\Rightarrow$ $y$ $=$ $ux$ $\Rightarrow$ $y^{\prime}$ $=$ $\frac{\mathrm{d} y}{\mathrm{d} x}$ $=$ $u^{\prime} x$ $+$ $u$ $=$ $x \frac{\mathrm{d} u}{\mathrm{d} x}$ $+$ $u$ $\Rightarrow$

$$
x \frac{\mathrm{d} u}{\mathrm{d} x} + u = \frac{2u – u^{2}}{1 – 2u} \Rightarrow
$$

$$
x \frac{\mathrm{d} u}{\mathrm{d} x} = \frac{2u – u^{2}}{1 – 2u} – u \Rightarrow
$$

$$
x \frac{\mathrm{d} u}{\mathrm{d} x} = \frac{2u – u^{2}}{1 – 2u} – \frac{u(1 – 2u)}{1 – 2u} \Rightarrow
$$

$$
x \frac{\mathrm{d} u}{\mathrm{d} x} = \frac{2u – u^{2}}{1 – 2u} – \frac{u – 2u^{2}}{1 – 2u} \Rightarrow
$$

Next

$$
x \frac{\mathrm{d} u}{\mathrm{d} x} = \frac{2u – u^{2} – u + 2u^{2}}{1 – 2u} \Rightarrow
$$

$$
x \frac{\mathrm{d} u}{\mathrm{d} x} = \frac{u + u^{2}}{1 – 2u} \Rightarrow
$$

$$
\mathrm{d} u = \frac{u + u^{2}}{1 – 2u} \cdot \frac{\mathrm{d} x}{x} \Rightarrow
$$

$$
\frac{1 – 2u}{u + u^{2}} \mathrm{d} u = \frac{\mathrm{d} x}{x} \Rightarrow
$$

$$
\int \frac{1 – 2u}{u + u^{2}} \mathrm{d} u = \int \frac{\mathrm{d} x}{x} \Rightarrow
$$

$$
\int \frac{1 – 2u}{u + u^{2}} \mathrm{d} u = \ln |x| + C \Rightarrow
$$

上面的 $\ln |x|$ $+$ $C$ 也可以写成 $\ln |x|$ $+$ $\ln C_{1}$ $=$ $\ln |C_{1} x|$——只要能表示任意常数即可。

Next

$$
\int \frac{1}{u + u^{2}} \mathrm{d} u – \int \frac{2u}{u + u^{2}} \mathrm{d} u = \ln |x| + C \Rightarrow
$$

$$
\int \frac{1}{u(1 + u)} \mathrm{d} u – 2 \int \frac{1}{1 + u} \mathrm{d} u = \ln |x| + C \Rightarrow
$$

$$
\int \frac{1}{u(1 + u)} \mathrm{d} u – 2 \ln |1 + u| = \ln |x| + C \Rightarrow
$$

$$
\int \big( \frac{1}{u} – \frac{1}{1+u} \big) \mathrm{d} u – 2 \ln |1 + u| = \ln |x| + C \Rightarrow
$$

Next

$$
\ln|u| – \ln |1+u| – 2 \ln |1 + u| = \ln |x| + C \Rightarrow
$$

$$
\ln|u| – 3 \ln |1+u| = \ln |x| + C \Rightarrow
$$

$$
\ln|u| – \ln |(1+u)^{3}| = \ln |x| + C \Rightarrow
$$

$$
\ln \Big|\frac{u}{(1+u)^{3}} \Big| = \ln |x| + C \Rightarrow
$$

Next

$$
\ln \Big|\frac{u}{(1+u)^{3}} \Big| = \ln |x| + \ln |C_{1}| \Rightarrow
$$

$$
\ln \Big|\frac{u}{(1+u)^{3}} \Big| = \ln |C_{1} x| \Rightarrow
$$

$$
\frac{u}{(1+u)^{3}} = C_{1} x \Rightarrow
$$

Next

$$
\frac{\frac{y}{x}}{(1 + \frac{y}{x})^{3}} = C_{1} x \Rightarrow
$$

$$
\frac{\frac{y}{x}}{(\frac{x + y}{x})^{3}} = C_{1} x \Rightarrow
$$

$$
\frac{y}{x} \cdot \frac{x^{3}}{(x + y)^{3}} = C_{1} x \Rightarrow
$$

$$
\frac{y x^{2}}{(x + y)^{3}} = C_{1} x.
$$

Next

又因为，当 $x$ $=$ $1$ 的时候，$y$ $=$ $-2$, 所以：

$$
\frac{y x^{2}}{(x + y)^{3}} = C_{1} x \Rightarrow
$$

$$
\frac{-2}{(-1)^{3}} = C_{1} \Rightarrow
$$

$$
C_{1} = 2.
$$

Next

综上可知，微分方程 $\frac{\mathrm{d} y}{\mathrm{d} x}$ $=$ $\frac{2xy – y^{2}}{x^{2} – 2xy}$ 满足 $y(1)$ $=$ $-2$ 的特解是：

$$
\frac{y x^{2}}{(x + y)^{3}} = 2 x \Rightarrow
$$

$$
\frac{y x}{(x + y)^{3}} = 2 \Rightarrow
$$

$$
xy = 2 (x + y)^{3}.
$$

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