复合函数二阶导的题目：明确谁是谁的函数，谁是真自变量，谁是中间变量

一、题目

令 $x = e^{t}$ , 则，方程 $a x^{2} \frac{d^{2} y}{d x^{2}} + b x \frac{d y}{d x} + c y = 0$ 可以转换为什么？

难度评级：

由题知：

$x = e^{t} \Rightarrow t = \ln x \Rightarrow t (x) = \ln x .$

且可知：

$\frac{d y}{d x} = \frac{d y}{d t} \cdot \frac{d t}{d x} = \frac{d y}{d t} \cdot \frac{1}{x}$

$\frac{d^{2} y}{d x^{2}} = \frac{d}{d x} (\frac{d y}{d t} \cdot \frac{1}{x}) =$

$\frac{d}{d x} \cdot \frac{d y}{d t} \cdot \frac{1}{x} + \frac{d y}{d t} \cdot \frac{d}{d x} (\frac{1}{x}) =$

$\frac{d}{d t} \cdot \frac{d t}{d x} \cdot \frac{d y}{d t} \cdot \frac{1}{x} + \frac{d y}{d t} \cdot (\frac{- 1}{x^{2}}) =$

$\frac{d t}{d x} \cdot (\frac{d \cdot}{d t} \cdot \frac{d y}{d t}) \cdot \frac{1}{x} + \frac{d y}{d t} \cdot (\frac{- 1}{x^{2}}) =$

$\frac{1}{x^{2}} \frac{d^{2} y}{d t^{2}} - \frac{1}{x^{2}} \frac{d y}{d t} .$

于是：

$a x^{2} \frac{d^{2} y}{d x^{2}} + b x \frac{d y}{d x} + c y = 0 \Rightarrow$

$a x^{2} (\frac{1}{x^{2}} \frac{d^{2} y}{d t^{2}} - \frac{1}{x^{2}} \frac{d y}{d t}) + b x (\frac{d y}{d t} \cdot \frac{1}{x}) + c y = 0 \Rightarrow$

$a \frac{d^{2} y}{d t^{2}} - a \frac{d y}{d t} + b \frac{d y}{d t} + c y = 0 \Rightarrow$

$a \frac{d^{2} y}{d t^{2}} + (b - a) \frac{d y}{d t} + c y = 0$