方企勤 第五章 多元函数微分学 第5题
📝 题目
例 5 取 $y$ 为因变量,解方程
$$ {\left( \frac{\partial z}{\partial y}\right) }^{2}\frac{{\partial }^{2}z}{\partial {x}^{2}} - 2\frac{\partial z}{\partial x}\frac{\partial z}{\partial y}\frac{{\partial }^{2}z}{\partial x\partial y} + {\left( \frac{\partial z}{\partial x}\right) }^{2}\frac{{\partial }^{2}z}{\partial {y}^{2}} = 0. $$
💡 答案解析
解 由上题启发, $z = z\left( {x,y}\right)$ 中把 $x,z$ 看成自变量,对 $x$ 求偏导数, 得
$$ 0 = \frac{\partial z}{\partial x} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial x},\;\text{ 解出 }\;\frac{\partial y}{\partial x} = - \frac{\partial z}{\partial x}/\frac{\partial z}{\partial y}. $$
再对 $x$ 求偏导,得
$$ 0 = \frac{{\partial }^{2}z}{\partial {x}^{2}} + 2\frac{{\partial }^{2}z}{\partial x\partial y}\frac{\partial y}{\partial x} + \frac{{\partial }^{2}z}{\partial {y}^{2}}{\left( \frac{\partial y}{\partial x}\right) }^{2} + \frac{\partial z}{\partial y}\frac{{\partial }^{2}y}{\partial {x}^{2}}. $$
将 $\frac{\partial y}{\partial x}$ 代入上式,有
$$ 0 = \frac{{\left( \frac{\partial z}{\partial y}\right) }^{2}\frac{{\partial }^{2}z}{\partial {x}^{2}} - 2\frac{\partial z}{\partial x}\frac{\partial z}{\partial y}\frac{{\partial }^{2}z}{\partial x\partial y} + {\left( \frac{\partial z}{\partial x}\right) }^{2}\frac{{\partial }^{2}z}{\partial {y}^{2}}}{{\left( \frac{\partial z}{\partial y}\right) }^{2}} + \frac{\partial z}{\partial y}\frac{{\partial }^{2}y}{\partial {x}^{2}}. $$
利用条件得出 $\frac{\partial z}{\partial y}\frac{{\partial }^{2}y}{\partial {x}^{2}} = 0,y$ 可取为因变量隐含条件 $\frac{\partial z}{\partial y} \neq 0$ ,所以 $\frac{{\partial }^{2}y}{\partial {x}^{2}}$ $= 0$ ,由此解出 $y = {x\varphi }\left( z\right) + \psi \left( z\right)$ .