一、题目
求微分方程 $x^{2} y^{\prime \prime}$ $-$ $2xy^{\prime}$ $-$ $(y^{\prime})^{2}$ $=$ $0$ $(x > 2)$ 满足条件 $y|_{x=3} = \frac{1}{2}$, $y^{\prime}|_{x=3} = -9$ 的解.
二、解析
解法 1
观察可知，该方程属于二阶可降阶微分方程中的显含 $x$, 但不显含 $y$ 的类型，只显含 $y$ 的一阶导函数和二阶导函数.
但由于我们要求解出（零阶导函数）$y$ 的表达式，所以，就需要利用 $p = y ^{\prime}$ 这样的代换，将一阶导函数 $y ^{\prime}$ 降阶成某种零阶导函数 $p$, 即：
$$
\begin{aligned}
p & = y^{\prime} \\ \\
y^{\prime \prime} & = \frac{\mathrm{d} p}{\mathrm{d} x} = p^{\prime}
\end{aligned}
$$
将上面的式子，代入原微分方程 $x^{2} y^{\prime \prime}$ $-$ $2xy^{\prime}$ $-$ $(y^{\prime})^{2}$ $=$ $0$, 可得：
$$
\begin{align}
& \ x^{2} p^{\prime} – 2xp – p^{2} = 0 \notag \\ \notag \\
\textcolor{lightgreen}{ \leadsto } & \ p^{\prime} – \frac{2}{x} p = x^{-2} p^{2} \notag \\ \notag \\
\textcolor{lightgreen}{ \leadsto } & \ \frac{p ^{\prime}}{p^{2}} – \frac{2}{x} \cdot \frac{1}{p} = x^{-2} \notag \\ \notag \\
\textcolor{lightgreen}{ \leadsto } & \ – \frac{p ^{\prime}}{p^{2}} + \frac{2}{x} \cdot \frac{1}{p} = – x^{-2} \tag{1}
\end{align}
$$
接着，令 $z = \frac{1}{p}$, 则：
$$
z^{\prime} = -p^{-2} \cdot \frac{\mathrm{d} p}{\mathrm{d} x} = – \frac{p^{\prime}}{p ^{2}}
$$
将上面有关 $z$ 和 $z ^{\prime}$ 的式子代入 $(1)$ 式，得：
$$
z^{\prime} + \frac{2}{x} z = -x^{-2}
$$
上面的式子就是一个一阶线性微分方程，根据一阶线性微分方程的求解公式，可得：
$$
\begin{aligned}
z & = \left( \int (-x^{-2}) \cdot \mathrm{e}^{\int \frac{2}{x} \mathrm{~d}x} \mathrm{~d}x + C_{1} \right) \cdot \mathrm{e}^{-\int \frac{2}{x} \mathrm{~d}x} \\ \\
& = \left( \int (-x^{-2}) \cdot \mathrm{e}^{2 \ln x} \mathrm{~d}x + C_{1} \right) \cdot \mathrm{e}^{-2 \ln x} \\ \\
& = \left( \int (-x^{-2}) \cdot \mathrm{e}^{\ln x^{2}} \mathrm{~d}x + C_{1} \right) \cdot \mathrm{e}^{\ln x^{-2}} \\ \\
& = \left( \int (-x^{-2}) \cdot x^{2} \mathrm{~d}x + C_{1} \right) \cdot x^{-2} \\ \\
& = \left( \int -1 \mathrm{~d}x + C_{1} \right) \cdot x^{-2} \\ \\
& = \left( -x + C_{1} \right) \cdot x^{-2} \\ \\
& = -\frac{1}{x} + \frac{C_{1}}{x^{2}}
\end{aligned}
$$
又由题可知：
$$
z|_{x=3} = \frac{1}{y^{\prime}(3)} = -\frac{1}{9}
$$
将 $z|_{x=3}$ 的结果代入 $z=-\frac{1}{x} + \frac{C_{1}}{x^{2}}$ 可得：
$$
C_{1} = 2
$$
于是：
$$
\begin{aligned}
& \ z = \frac{1}{y^{\prime}} = -\frac{1}{x} + \frac{2}{x^{2}} \\ \\
\textcolor{lightgreen}{ \leadsto } & \ y^{\prime} = \frac{x^{2}}{2 – x}
\end{aligned}
$$
又因为：
$$
\begin{aligned}
\frac{x^{2}}{2 – x} & = \frac{x^{2} – 4 + 4}{2 – x} = \frac{x^{2}-4}{2 – x} + \frac{4}{2 – x} \\ \\
& = \frac{(x+2) (x-2)}{2 – x} + \frac{4}{2 – x} \\ \\
& = – (x+2) + \frac{4}{2 – x} \\ \\
& = -x – 2 – \frac{4}{x – 2}
\end{aligned}
$$
于是：
$$
\begin{aligned}
y & = \int y ^{\prime} \mathrm{~d} x = \int \frac{x^{2}}{2 – x} \mathrm{~d} x \\ \\
& = \int \left( -x – 2 – \frac{4}{x – 2} \right) \mathrm{~d} x \\ \\
& = -\frac{1}{2} x^{2} – 2x – 4 \ln(x – 2) + C_{2}
\end{aligned}
$$
又由题可知：
$$
y|_{x=3} = \frac{1}{2}
$$
于是，将 $y|_{x=3}$ 的结果代入 $y = -\frac{1}{2} x^{2} – 2x – 4 \ln(x – 2) + C_{2}$ 可得：
$$
C_{2} = 11
$$
综上可知，微分方程满足条件 $y|_{x=3} = \frac{1}{2}$, $y^{\prime}|_{x=3} = -9$ 的解为：
$$
y = -\frac{1}{2} x^{2} – 2x – 4 \ln(x – 2) + 11
$$
解法 2
观察可知，该方程属于二阶可降阶微分方程中的显含 $x$, 但不显含 $y$ 的类型，只显含 $y$ 的一阶导函数和二阶导函数.
但由于我们要求解出（零阶导函数）$y$ 的表达式，所以，就需要利用 $p = y ^{\prime}$ 这样的代换，将一阶导函数 $y ^{\prime}$ 降阶成某种零阶导函数 $p$, 即：
$$
\begin{aligned}
p & = y^{\prime} \\ \\
y^{\prime \prime} & = \frac{\mathrm{d} p}{\mathrm{d} x} = p^{\prime}
\end{aligned}
$$
将上面的式子，代入原微分方程 $x^{2} y^{\prime \prime}$ $-$ $2xy^{\prime}$ $-$ $(y^{\prime})^{2}$ $=$ $0$, 可得：
$$
\begin{align}
& \ x^{2} p^{\prime} – 2xp – p^{2} = 0 \notag \\ \notag \\
\textcolor{lightgreen}{ \leadsto } & \ p ^{\prime} – 2\frac{p}{x} – \frac{p^{2}}{x^{2}} = 0 \tag{2}
\end{align}
$$
上面的 $(2)$ 式是一个一阶齐次微分方程，若令 $u=\frac{p}{x}$, 则：
$$
\begin{aligned}
& p = ux \textcolor{lightgreen}{ \leadsto } p ^{\prime} = \frac{\mathrm{d} u}{\mathrm{d} x} \cdot x + u \\ \\
& u ^{2} = \frac{p ^{2}}{u ^{2}}
\end{aligned}
$$
于是，$(2)$ 式可改写为：
$$
\begin{aligned}
& \ p ^{\prime} -2u – u^{2} = 0 \\ \\
\textcolor{lightgreen}{ \leadsto } & \ \frac{\mathrm{d} u}{\mathrm{d} x} x = u + u^{2} \\ \\
\textcolor{lightgreen}{ \leadsto } & \ \frac{\mathrm{d} u}{\mathrm{d} x} = \frac{u + u^{2}}{x} \\ \\
\textcolor{lightgreen}{ \leadsto } & \ \frac{\mathrm{d} u}{\mathrm{d} x} = \frac{u(u+1)}{x} \\ \\
\textcolor{lightgreen}{ \leadsto } & \ \frac{\mathrm{d} u}{u (u+1)} = \frac{\mathrm{d} x}{x} \\ \\
\textcolor{lightgreen}{ \leadsto } & \ \int \frac{\mathrm{d} u}{u (u+1)} = \int \frac{\mathrm{d} x}{x} \\ \\
\textcolor{lightgreen}{ \leadsto } & \ \ln \frac{u}{u+1} = \ln x + \ln C_{3} \\ \\
\textcolor{lightgreen}{ \leadsto } & \ \textcolor{gray}{u = \frac{p}{x} = \frac{y ^{\prime} }{x}} \\ \\
\textcolor{lightgreen}{ \leadsto } & \ \ln \frac{\frac{y ^{\prime} }{x}}{\frac{y ^{\prime} }{x} + 1} = \ln x + \ln C_{3}
\end{aligned}
$$
接着，由 $y ^{\prime} |_{x=3} = -9$ 得：
$$
\begin{aligned}
& \ \ln \frac{-3}{-3 + 1} = \ln 3 + \ln C_{3} \\ \\
\textcolor{lightgreen}{ \leadsto } & \ \ln \frac{3}{2} = \ln 3 + \ln C_{3} \\ \\
\textcolor{lightgreen}{ \leadsto } & \ \ln 3 – \ln 2 = \ln 3 + \ln C_{3} \\ \\
\textcolor{lightgreen}{ \leadsto } & \ \ln C_{3} = – \ln 2 \\ \\
\textcolor{lightgreen}{ \leadsto } & \ C_{3}=\frac{1}{2}
\end{aligned}
$$
于是可知：
$$
\begin{aligned}
& \ \ln \frac{\frac{y ^{\prime} }{x}}{\frac{y ^{\prime} }{x} + 1} = \ln x + \ln C_{3} \\ \\
\textcolor{lightgreen}{ \leadsto } & \ \ln \frac{\frac{y ^{\prime} }{x}}{\frac{y ^{\prime} }{x} + 1} = \ln x + \ln \frac{1}{2} \\ \\
\textcolor{lightgreen}{ \leadsto } & \ \ln \frac{\frac{y ^{\prime} }{x}}{\frac{y ^{\prime} }{x} + 1} = \ln \frac{x}{2} \\ \\
\textcolor{lightgreen}{ \leadsto } & \ \frac{\frac{y ^{\prime} }{x}}{\frac{y ^{\prime} }{x} + 1} = \frac{x}{2} \\ \\
\textcolor{lightgreen}{ \leadsto } & \ y ^{\prime} \left( 1 – \frac{2}{x} \right) = -x \\ \\
\textcolor{lightgreen}{ \leadsto } & \ y ^{\prime} = \frac{-x}{1 – \frac{2}{x}} \\ \\
\textcolor{lightgreen}{ \leadsto } & \ y ^{\prime} = \frac{x^{2}}{2-x} \\ \\
\textcolor{lightgreen}{ \leadsto } & \ y = \int \frac{x^{2}}{2-x} \mathrm{~d} x \\ \\
\textcolor{lightgreen}{ \leadsto } & \ y = \int \frac{x^{2} – 4 + 4}{2-x} \mathrm{~d} x \\ \\
\textcolor{lightgreen}{ \leadsto } & \ y = \int \left( -x – 2 – \frac{4}{x – 2} \right) \mathrm{~d} x \\ \\
\textcolor{lightgreen}{ \leadsto } & \ y = -\frac{1}{2} x^{2} – 2x – 4 \ln(x – 2) + C_{4}
\end{aligned}
$$
又由题可知：
$$
y|_{x=3} = \frac{1}{2}
$$
于是，将 $y|_{x=3}$ 的结果代入 $y = -\frac{1}{2} x^{2} – 2x – 4 \ln(x – 2) + C_{4}$ 可得：
$$
C_{4} = 11
$$
综上可知，微分方程满足条件 $y|_{x=3} = \frac{1}{2}$, $y^{\prime}|_{x=3} = -9$ 的解为：
$$
y = -\frac{1}{2} x^{2} – 2x – 4 \ln(x – 2) + 11
$$
高等数学
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线性代数
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