📝 题目
例 5 设 $y = \frac{x - a}{1 - {ax}}\left( {\left| a\right| < 1}\right)$ . 求证: 当 $\left| x\right| < 1$ 时,有
$$ \frac{\mathrm{d}y}{1 - {y}^{2}} = \frac{\mathrm{d}x}{1 - {x}^{2}}. $$
💡 答案与解析
证法 1 由已知条件得
$$ \mathrm{d}y = \frac{1 - {a}^{2}}{{\left( ax - 1\right) }^{2}}\mathrm{\;d}x, $$
$$ 1 - {y}^{2} = \frac{\left( {1 - {a}^{2}}\right) \left( {1 - {x}^{2}}\right) }{{\left( ax - 1\right) }^{2}}, $$
整理化简得
$$ \frac{\mathrm{d}y}{1 - {y}^{2}} = \frac{\frac{1 - {a}^{2}}{{\left( ax - 1\right) }^{2}}\mathrm{\;d}x}{\frac{\left( {1 - {a}^{2}}\right) \left( {1 - {x}^{2}}\right) }{{\left( ax - 1\right) }^{2}}} = \frac{\mathrm{d}x}{1 - {x}^{2}}. $$
证法 2 先由 $y$ 的表达式,解出 $a = \frac{x - y}{1 - {xy}}$ . 再两边取微分,得
$$ 0 = \frac{\left( {\mathrm{d}x - \mathrm{d}y}\right) \left( {1 - {xy}}\right) + \left( {x - y}\right) \left( {y\mathrm{\;d}x + x\mathrm{\;d}y}\right) }{{\left( 1 - xy\right) }^{2}} $$
$$ \Rightarrow \left( {1 - {y}^{2}}\right) \mathrm{d}x - \left( {1 - {x}^{2}}\right) \mathrm{d}y = 0. $$