【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
$$\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{\mathrm{d} y}{\mathrm{~d} t} / \frac{\mathrm{d} x}{\mathrm{~d} t}=\frac{\mathrm{d} y}{\mathrm{~d} t} \cdot \frac{1}{\cos t},$$

\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}

\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}