# 用两种不同的思路解决一道隐函数变量替换的题目

## 一、题目

$$\left(1-x^2\right) \frac{\mathrm{d}^2 y}{\mathrm{d} x^2}-x \frac{\mathrm{d} y}{\mathrm{d} x}+a^2 y=0$$

### 解题思路简图

graph TB
A1(隐函数)
A2(二阶导数)
A3(变量替换)
A1 --> B1(用新的变量替换掉隐函数所满足的方程中原有的变量)
A2 --> B1
A3 --> B1
B1 --> C1(复合函数求导)
C1 --> D1(方法一)
C1 --> D2(方法二)
D1 --> E1(用 dy/dt 表示隐函数方程中的 dy/dx)
D2 --> E2(用 dy/dx 表示 dy/dt)


## 二、解析

### 思路一：用 $\frac{\mathrm{d} y}{\mathrm{d} t}$ 表示 $\frac{\mathrm{d} y}{\mathrm{d} x}$

Next

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

$$t(x) = \arcsin x$$

Next

$$\frac{\mathrm{d} y}{\mathrm{d} x} = \frac{\mathrm{d} y}{\mathrm{d} t} \cdot \frac{1}{\sqrt{1 – x^{2}}}$$

$$\frac{\mathrm{d}^{2} y}{\mathrm{d} x^{2}} = \Big( \frac{\mathrm{d} y}{\mathrm{d} x} \Big) \frac{\mathrm{d}}{\mathrm{d} x} = \Big( \frac{\mathrm{d} y}{\mathrm{d} t} \cdot \frac{1}{\sqrt{1 – x^{2}}} \Big) \frac{\mathrm{d}}{\mathrm{d} x} \Rightarrow$$

$$\frac{\mathrm{d}^{2} y}{\mathrm{d} x^{2}} = \Big( \textcolor{orange}{\frac{\mathrm{d} y}{\mathrm{d} t} } \times \textcolor{red}{ \frac{1}{\sqrt{1 – x^{2}}} } \Big) \textcolor{white}{ \frac{\mathrm{d}}{\mathrm{d} t} \frac{\mathrm{d} t}{\mathrm{d} x} } \Rightarrow$$

$$\frac{\mathrm{d}^{2} y}{\mathrm{d} x^{2}} = \textcolor{orange}{ \Big( \frac{\mathrm{d} y}{\mathrm{d} t} \Big) } \textcolor{white}{ \frac{\mathrm{d}}{\mathrm{d} t} \frac{\mathrm{d} t}{\mathrm{d} x} } \times \textcolor{red}{ \frac{1}{\sqrt{1 – x^{2}}} } + \textcolor{orange}{ \frac{\mathrm{d} y}{\mathrm{d} t} } \times \textcolor{red}{ \Big( \frac{1}{\sqrt{1 – x^{2}}} \Big) } \textcolor{white}{ \frac{\mathrm{d}}{\mathrm{d} t} \frac{\mathrm{d} t}{\mathrm{d} x} } \Rightarrow$$

Next

$$\frac{\mathrm{d}^{2} y}{\mathrm{d} x^{2}} = \frac{\mathrm{d} ^{2} y}{\mathrm{d} t^{2}} \cdot \frac{\mathrm{d} t}{\mathrm{d} x} \times \frac{1}{\sqrt{1 – x^{2}}} + \frac{\mathrm{d} y}{\mathrm{d} t} \big[ ( 1 – x^{2} )^{-\frac{1}{2}} \big]^{\prime}_{x} \Rightarrow$$

$$\frac{\mathrm{d}^{2} y}{\mathrm{d} x^{2}} = \frac{\mathrm{d} ^{2} y}{\mathrm{d} t^{2}} \cdot \frac{1}{1 – x^{2}} + \frac{\mathrm{d} y}{\mathrm{d} t} \times \frac{-1}{2} ( 1 – x^{2} )^{-\frac{3}{2} } \cdot (-2x) \Rightarrow$$

$$\frac{\mathrm{d}^{2} y}{\mathrm{d} x^{2}} = \frac{\mathrm{d} ^{2} y}{\mathrm{d} t^{2}} \cdot \frac{1}{1 – x^{2}} + \frac{\mathrm{d} y}{\mathrm{d} t} \times \frac{x}{( 1 – x^{2} )^{\frac{3}{2}}}.$$

Next

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

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

$$\frac{\mathrm{d} y}{\mathrm{d} t} \times \frac{x}{( 1 – x^{2} )^{\frac{3}{2}}} \Big)-x \Big( \frac{\mathrm{d} y}{\mathrm{d} t} \cdot \frac{1}{\sqrt{1 – x^{2}}} \Big) + a^2 y = 0 \Rightarrow$$

$$\frac{\mathrm{d} ^{2} y}{\mathrm{d} t^{2}} + \frac{\mathrm{d} y}{\mathrm{d} t} \cdot \frac{x}{\sqrt{1 – x^{2}}} – \frac{\mathrm{d} y}{\mathrm{d} t} \cdot \frac{x}{\sqrt{1 – x^{2}}} + a^2 y = 0 \Rightarrow$$

$$\frac{\mathrm{d} ^{2} y}{\mathrm{d} t^{2}} + a^2 y = 0$$

Next

$$\frac{\mathrm{d} ^{2} y}{\mathrm{d} t^{2}} + a^2 y = 0.$$

### 思路二：用 $\frac{\mathrm{d} y}{\mathrm{d} x}$ 表示 $\frac{\mathrm{d} y}{\mathrm{d} t}$

Next

$$\frac{\mathrm{d} y}{\mathrm{d} t} = \frac{\mathrm{d} y}{\mathrm{d} x} \frac{\mathrm{d} x}{\mathrm{d} t} \Rightarrow$$

$$\frac{\mathrm{d} y}{\mathrm{d} t} = \frac{\mathrm{d} y}{\mathrm{d} x} \cos t.$$

Next

$$\frac{\mathrm{d}^{2} y}{\mathrm{d} t^{2}} = \Big( \frac{\mathrm{d} y}{\mathrm{d} x} \times \cos t \Big) \frac{\mathrm{d}}{\mathrm{d} t} \Rightarrow$$

$$\frac{\mathrm{d}^{2} y}{\mathrm{d} t^{2}} = \Big( \frac{\mathrm{d} y}{\mathrm{d} x} \Big) \frac{\mathrm{d}}{\mathrm{d} t} \times \cos t + \frac{\mathrm{d} y}{\mathrm{d} x} \times \cos t \frac{\mathrm{d}}{\mathrm{d} t} \Rightarrow$$

$$\frac{\mathrm{d}^{2} y}{\mathrm{d} t^{2}} = \Big( \frac{\mathrm{d} y}{\mathrm{d} x} \Big) \frac{\mathrm{d}}{\mathrm{d} x} \frac{\mathrm{d} x}{\mathrm{d} t} \times \cos t + \frac{\mathrm{d} y}{\mathrm{d} x} \times (- \sin t) \Rightarrow$$

$$\frac{\mathrm{d}^{2} y}{\mathrm{d} t^{2}} = \frac{\mathrm{d} ^{2} y}{\mathrm{d} x ^{2}} \cos ^{2} t – \frac{\mathrm{d} y}{\mathrm{d} x} \sin t \Rightarrow$$

$$\frac{\mathrm{d}^{2} y}{\mathrm{d} t^{2}} = \frac{\mathrm{d} ^{2} y}{\mathrm{d} x ^{2}} (1 – \sin ^{2} t) – \frac{\mathrm{d} y}{\mathrm{d} x} \sin t \Rightarrow$$

Next

$$\frac{\mathrm{d}^{2} y}{\mathrm{d} t^{2}} = \frac{\mathrm{d} ^{2} y}{\mathrm{d} x ^{2}} (1 – x ^{2}) – \frac{\mathrm{d} y}{\mathrm{d} x} x \Rightarrow$$

$$\textcolor{red}{ \frac{\mathrm{d}^{2} y}{\mathrm{d} t^{2}} } = \textcolor{orange}{ (1 – x ^{2}) \frac{\mathrm{d} ^{2} y}{\mathrm{d} x ^{2}} – x \frac{\mathrm{d} y}{\mathrm{d} x} } \Rightarrow$$

Next

$$\textcolor{orange}{ \left(1-x^2\right) \frac{\mathrm{d}^2 y}{\mathrm{d} x^2} – x \frac{\mathrm{d} y}{\mathrm{d} x} } + a^2 y = 0 \Rightarrow$$

$$\textcolor{red}{ \frac{\mathrm{d}^{2} y}{\mathrm{d} t^{2}} } + a^2 y = 0.$$

Next

$$\frac{\mathrm{d} ^{2} y}{\mathrm{d} t^{2}} + a^2 y = 0.$$