通过代换简化微分方程：自变量代换

一、题目

利用代换 $x = \cos t$ $(0 < t < \pi)$ 将原微分方程 $(1-x^{2})y^{\prime \prime} – xy^{\prime} +y = 0$ 化简，并求出该微分方程满足 $y(0) = 1$, $y^{\prime}(0) = 2$ 的特解.

Tip

拓展 1：《通过代换简化微分方程：函数代换》
拓展 2：《一点处函数值的多种表示形式》

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二、解析

由于 $x = \cos t$，所以：

$$
\textcolor{lightgreen}{
\frac{\mathrm{d} x}{\mathrm{d} t} = – \sin t
} \tag{1}
$$

于是，由 $(1)$ 式，可知：

$$
\begin{align}
\textcolor{lightgreen}{ y^{\prime} } & = \frac{\mathrm{d} y}{\mathrm{d} x} = \frac{\mathrm{d} y}{\mathrm{d} t} \cdot \frac{\mathrm{d} t}{\mathrm{d} x} = \frac{\mathrm{d} y}{\mathrm{d} t} \cdot \frac{1}{\frac{\mathrm{d} x}{\mathrm{d} t}} = \textcolor{lightgreen}{ \frac{\mathrm{d} y}{\mathrm{d} t} \cdot \frac{1}{-\sin t} } \tag{2} \\ \notag \\ \notag \\
\textcolor{lightgreen}{ y^{\prime \prime} } & = \frac{\mathrm{d}}{\mathrm{d} x}(y^{\prime} ) = \textcolor{orange}{ \frac{\mathrm{d}}{\mathrm{d} t}\left(-\frac{1}{\sin t} \cdot \frac{\mathrm{d} y}{\mathrm{d} t}\right) } \cdot \frac{\mathrm{d} t}{\mathrm{d} x} \notag \\ \notag \\
& = \textcolor{orange}{ \left(\frac{\cos t}{\sin ^{2} t} \cdot \frac{\mathrm{d} y}{\mathrm{d} t}-\frac{1}{\sin t} \frac{\mathrm{d}^{2} y}{\mathrm{d} t^{2}}\right) } \cdot \left(-\frac{1}{\sin t}\right) \notag \\ \notag \\
& = \textcolor{lightgreen}{ -\frac{\cos t}{\sin ^{3} t} \frac{\mathrm{d} y}{\mathrm{d} t}+\frac{1}{\sin ^{2} t} \cdot \frac{\mathrm{d}^{2} y}{\mathrm{d} t^{2}} } \tag{3}
\end{align}
$$

于是，通过上面的 $(2)$ 式和 $(3)$ 式，可以将上面的微分方程 $(1 – x^{2}) y^{\prime \prime} – xy^{\prime} + y=0$ 转化为：

$$
\begin{align}
& ( \textcolor{orange}{ 1 – \cos ^{2} t } ) \cdot \left(-\frac{\cos t}{\sin ^{3} t} \frac{\mathrm{d} y}{\mathrm{~d} t} + \frac{1}{\sin ^{2} t} \frac{\mathrm{d}^{2} y}{\mathrm{~d} t^{2}} \right) – \cos t \cdot \left(-\frac{1}{\sin t}\right) \frac{\mathrm{d} y}{\mathrm{~d} t} + y \notag \\ \notag \\
\leadsto \ & \textcolor{orange}{ \sin^{2} t } \cdot \left(-\frac{\cos t}{\sin ^{3} t} \frac{\mathrm{d} y}{\mathrm{~d} t} + \frac{1}{\sin ^{2} t} \frac{\mathrm{d}^{2} y}{\mathrm{~d} t^{2}} \right) +\frac{\cos t}{\sin t} \frac{\mathrm{d} y}{\mathrm{~d} t} + y \notag \\ \notag \\
\leadsto \ & \textcolor{yellow}{ – \frac{\cos t}{\sin t} \frac{\mathrm{d} y}{\mathrm{~d} t} } + \frac{\mathrm{d}^{2} y}{\mathrm{~d} t^{2}} + \textcolor{yellow}{ \frac{\cos t}{\sin t} \frac{\mathrm{d} y}{\mathrm{~d} t} } + y \notag \\ \notag \\
\leadsto \ & \frac{\mathrm{d}^{2} y}{\mathrm{~d} t^{2}} + y \notag \\ \notag \\
\leadsto \ & \textcolor{lightgreen}{ y ^{\prime \prime} (t) + y = 0 } \tag{4}
\end{align}
$$

于是可知，$(4)$ 式是一个二阶齐次微分方程，该微分方程对应的特征方程为：

$$
\textcolor{lightgreen}{
\lambda^{2} + 1 = 0
} \tag{5}
$$

解得特征值为：

$$
\textcolor{lightgreen}{
\lambda = \pm i
} \tag{6}
$$

综上，根据《二阶常系数线性齐次微分方程通解的求解公式》可知，微分方程 $y ^{\prime \prime} (t) + y = 0$ 的通解为：

$$
\begin{align}
\textcolor{lightgreen}{ \tilde{y}(t) } & = C_{1} \cos t + C_{2} \sin t \notag \\ \notag \\
& = \textcolor{lightgreen}{ C_{1} \cos t + C_{2} \sqrt{1 – \cos^{2} t} } \tag{7}
\end{align}
$$

因为 $x = \cos t$, 所以：

$$
\textcolor{lightgreen}{
y(x)=C_{1} x+C_{2} \sqrt{1-x^{2}}
} \tag{8}
$$

又因为 $y(0) = 1$，$y^{\prime} (0) = 2$, 代入上面的 $(8)$ 式，可得：

$$
C_{1} = 2, \quad C_{2} = 1
$$

综上可知，满足题意的特解为：

$$
\textcolor{springgreen}{
\boldsymbol{
y = 2 x + \sqrt{1-x^{2}}
}
}
$$

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