已知等式 $\left(1-x^2\right) \frac{d^2 y}{d x^2}-x \frac{d y}{d x}+a^2 y=0$ ,对其作变量代 换 $x=\sin t$ ,计算所得 $y$ 关于 $t$ 的导数的等式.
【答案】 【参考解答】: 由 $\frac{\mathrm{d} y}{\mathrm{~d} t}=\frac{\mathrm{d} y}{\mathrm{~d} x} \cdot \frac{\mathrm{d} x}{\mathrm{~d} t}$ 得
$$
\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{\mathrm{d} y}{\mathrm{~d} t} / \frac{\mathrm{d} x}{\mathrm{~d} t}=\frac{\mathrm{d} y}{\mathrm{~d} t} \cdot \frac{1}{\cos t},
$$
于是
$$
\begin{aligned}
\frac{\mathrm{d}^2 y}{\mathrm{~d} x^2} &=\frac{\mathrm{d}}{\mathrm{d} x}\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)=\frac{\frac{\mathrm{d}}{\mathrm{d} t}\left(\frac{\mathrm{d} y}{\mathrm{~d} t} \cdot \frac{1}{\cos t}\right)}{\frac{\mathrm{d} x}{\mathrm{~d} t}} \\
&=\frac{\frac{\mathrm{d}^2 y}{\mathrm{~d} t^2} \cdot \frac{1}{\cos t}+\frac{\mathrm{d} y}{\mathrm{~d} t} \cdot \frac{\sin t}{\cos ^2 t}}{\cos t}=\frac{\mathrm{d}^2 y}{\mathrm{~d} t^2} \cdot \frac{1}{\cos ^2 t}+\frac{\mathrm{d} y}{\mathrm{~d} t} \cdot \frac{\sin t}{\cos ^3 t}
\end{aligned}
$$
代入题中等式,得
$$
\begin{aligned}
\left(1-\sin ^2 t\right)\left(\frac{\mathrm{d}^2 y}{\mathrm{~d} t^2}\right.&\left.\cdot \frac{1}{\cos ^2 t}+\frac{\mathrm{d} y}{\mathrm{~d} t} \cdot \frac{\sin t}{\cos ^3 t}\right) \\
&-\sin t \frac{\mathrm{d} y}{\mathrm{~d} t} \cdot \frac{1}{\cos t}+a^2 y=0
\end{aligned}
$$

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


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