📝 题目
2．求下列齐次方程的通解： （1）$y^{2}+x^{2} \frac{\mathrm{~d} y}{\mathrm{~d} x}=x y \frac{\mathrm{~d} y}{\mathrm{~d} x}$ ； （2）$\frac{\mathrm{d} x}{x^{2}-x y+y^{2}}=\frac{\mathrm{d} y}{2 y^{2}-x y}$ ； （3）$x(\ln x-\ln y) \mathrm{d} y-y \mathrm{~d} x=0$ ； （4）$\left(x+y \cos \frac{y}{x}\right) \mathrm{d} x-x \cos \frac{y}{x} \mathrm{~d} y=0$ ．
💡 答案与解析
[AI解答]
[AI解答]
**（1）** 原方程： $$y^{2}+x^{2} \frac{\mathrm{d} y}{\mathrm{d} x}=x y \frac{\mathrm{d} y}{\mathrm{d} x}$$ 移项得： $$x^{2} \frac{\mathrm{d} y}{\mathrm{d} x} - x y \frac{\mathrm{d} y}{\mathrm{d} x} = -y^{2}$$ $$\frac{\mathrm{d} y}{\mathrm{d} x}(x^{2} - x y) = -y^{2}$$ $$\frac{\mathrm{d} y}{\mathrm{d} x} = \frac{-y^{2}}{x(x-y)}$$ 此为齐次方程，令 $y = ux$，则 $\displaystyle\frac{\mathrm{d} y}{\mathrm{d} x} = u + x\frac{\mathrm{d} u}{\mathrm{d} x}$，代入： $$u + x\frac{\mathrm{d} u}{\mathrm{d} x} = \frac{-u^{2}x^{2}}{x(x-ux)} = \frac{-u^{2}}{1-u}$$ 整理： $$x\frac{\mathrm{d} u}{\mathrm{d} x} = \frac{-u^{2}}{1-u} - u = \frac{-u^{2} - u(1-u)}{1-u} = \frac{-u}{1-u}$$ 分离变量： $$\frac{1-u}{u}\,\mathrm{d}u = -\frac{\mathrm{d}x}{x}$$ $$\left(\frac{1}{u} - 1\right)\mathrm{d}u = -\frac{\mathrm{d}x}{x}$$ 积分： $$\ln|u| - u = -\ln|x| + C$$ $$\ln|u| + \ln|x| - u = C$$ $$\ln|ux| - u = C$$ 回代 $u = \frac{y}{x}$： $$\ln|y| - \frac{y}{x} = C$$ 故通解为： $$\boxed{\ln|y| - \frac{y}{x} = C}$$
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**（2）** 原方程： $$\frac{\mathrm{d} x}{x^{2}-x y+y^{2}} = \frac{\mathrm{d} y}{2y^{2}-x y}$$ 即 $$\frac{\mathrm{d} y}{\mathrm{d} x} = \frac{2y^{2} - xy}{x^{2} - xy + y^{2}}$$ 令 $y = ux$，则 $\displaystyle\frac{\mathrm{d} y}{\mathrm{d} x} = u + x\frac{\mathrm{d} u}{\mathrm{d} x}$，代入： $$u + x\frac{\mathrm{d} u}{\mathrm{d} x} = \frac{2u^{2}x^{2} - u x^{2}}{x^{2} - u x^{2} + u^{2}x^{2}} = \frac{2u^{2} - u}{1 - u + u^{2}}$$ 整理： $$x\frac{\mathrm{d} u}{\mathrm{d} x} = \frac{2u^{2} - u}{1 - u + u^{2}} - u = \frac{2u^{2} - u - u(1 - u + u^{2})}{1 - u + u^{2}}$$ 分子： $$2u^{2} - u - u + u^{2} - u^{3} = 3u^{2} - 2u - u^{3} = -u(u^{2} - 3u + 2) = -u(u-1)(u-2)$$ 所以： $$x\frac{\mathrm{d} u}{\mathrm{d} x} = \frac{-u(u-1)(u-2)}{u^{2} - u + 1}$$ 分离变量： $$\frac{u^{2} - u + 1}{u(u-1)(u-2)}\,\mathrm{d}u = -\frac{\mathrm{d}x}{x}$$ 用部分分式分解：设 $$\frac{u^{2} - u + 1}{u(u-1)(u-2)} = \frac{A}{u} + \frac{B}{u-1} + \frac{C}{u-2}$$ 解得 $A = \frac{1}{2},\; B = -1,\; C = \frac{3}{2}$，于是： $$\left(\frac{1}{2u} - \frac{1}{u-1} + \frac{3}{2(u-2)}\right)\mathrm{d}u = -\frac{\mathrm{d}x}{x}$$ 积分： $$\frac{1}{2}\ln|u| - \ln|u-1| + \frac{3}{2}\ln|u-2| = -\ln|x| + C$$ 乘以2： $$\ln|u| - 2\ln|u-1| + 3\ln|u-2| = -2\ln|x| + C_{1}$$ 即： $$\ln\left|\frac{u(u-2)^{3}}{(u-1)^{2}}\right| = \ln\left|\frac{1}{x^{2}}\right| + C_{1}$$ 去掉对数： $$\frac{u(u-2)^{3}}{(u-1)^{2}} = \frac{C}{x^{2}}$$ 回代 $u = \frac{y}{x}$： $$\frac{\frac{y}{x}\left(\frac{y}{x}-2\right)^{3}}{\left(\frac{y}{x}-1\right)^{2}} = \frac{C}{x^{2}}$$ 化简： $$\frac{y(y-2x)^{3}}{x^{4}} \cdot \frac{x^{2}}{(y-x)^{2}} = \frac{C}{x^{2}}$$ $$\frac{y(y-2x)^{3}}{x^{2}(y-x)^{2}} = \frac{C}{x^{2}}$$ 两边乘 $x^{2}$： $$\frac{y(y-2x)^{3}}{(y-x)^{2}} = C$$ 故通解为： $$\boxed{\frac{y(y-2x)^{3}}{(y-x)^{2}} = C}$$
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**（3）** 原方程： $$x(\ln x - \ln y)\,\mathrm{d}y - y\,\mathrm{d}x = 0$$ 即 $$x\ln\frac{x}{y}\,\mathrm{d}y = y\,\mathrm{d}x$$ $$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{y}{x\ln\frac{x}{y}}$$ 令 $y = ux$，则 $\displaystyle\frac{\mathrm{d}y}{\mathrm{d}x} = u + x\frac{\mathrm{d}u}{\mathrm{d}x}$，代入： $$u + x\frac{\mathrm{d}u}{\mathrm{d}x} = \frac{ux}{x\ln\frac{x}{ux}} = \frac{u}{\ln\frac{1}{u}} = -\frac{u}{\ln u}$$ 整理： $$x\frac{\mathrm{d}u}{\mathrm{d}x} = -\frac{u}{\ln u} - u = -u\left(\frac{1}{\ln u} + 1\right) = -u\frac{1+\ln u}{\ln u}$$ 分离变量： $$\frac{\ln u}{u(1+\ln u)}\,\mathrm{d}u = -\frac{\mathrm{d}x}{x}$$ 令 $t = \ln u$，则 $u = e^{t}$，$\mathrm{d}u = e^{t}\mathrm{d}t$，左边： $$\frac{t}{e^{t}(1+t)} e^{t}\mathrm{d}t = \frac{t}{1+t}\,\mathrm{d}t = \left(1 - \frac{1}{1+t}\right)\mathrm{d}t$$ 积分： $$\int\left(1 - \frac{1}{1+t}\right)\mathrm{d}t = -\int\frac{\mathrm{d}x}{x}$$ $$t - \ln|1+t| = -\ln|x| + C$$ 回代 $t = \ln u = \ln\frac{y}{x}$： $$\ln\frac{y}{x} - \ln\left|1+\ln\frac{y}{x}\right| = -\ln|x| + C$$ $$\ln|y| - \ln|x| - \ln\left|1+\ln\frac{y}{x}\right| = -\ln|x| + C$$ 化简： $$\ln|y| - \ln\left|1+\ln\frac{y}{x}\right| = C$$ 即： $$\boxed{\ln\left|\frac{y}{1+\ln\frac{y}{x}}\right| = C}$$
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**（4）** 原方程： $$\left(x + y\cos\frac{y}{x}\right)\mathrm{d}x - x\cos\frac{y}{x}\,\mathrm{d}y = 0$$ 即 $$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{x + y\cos\frac{y}{x}}{x\cos\frac{y}{x}} = \frac{1}{\cos\frac{y}{x}} + \frac{y}{x}$$ 令 $y = ux$，则 $\displaystyle\frac{\