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

难度评级：

二、解析

由题知：

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

且可知：

• $y$ 是 $t$ 的函数；
• $t$ 是 $x$ 的函数。

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

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

$$
\frac{d}{\mathrm{~d} x} \cdot \frac{\mathrm{~d} y}{\mathrm{~d} t} \cdot \frac{1}{x}+\frac{\mathrm{~d} y}{\mathrm{~d} t} \cdot \frac{d}{\mathrm{~d} x}\left(\frac{1}{x}\right)=
$$

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

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

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

于是：

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

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

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

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

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