2022考研数二第18题解析：一阶线性微分方程、曲线的弧长

一、题目

设函数 $y \left( x \right)$ 是微分方程 $2 x y^{\prime} – 4 y = 2 \ln x – 1$, 满足条件 $y\left( 1 \right) = \frac{1}{4}$ 的解，求曲线 $y = y\left(x\right) \left( 1 \leqslant x \leqslant e \right)$ 的弧长.

二、解析

要计算曲线 $y = y\left(x\right) \left( 1 \leqslant x \leqslant e \right)$ 的弧长，我们首先需要求解出曲线 $y = y\left(x\right) \left( 1 \leqslant x \leqslant e \right)$ 的表达式.

又因为：

$$
\begin{aligned}
& \ 2 x y^{\prime} – 4 y = 2 \ln x – 1 \\ \\
\textcolor{lightgreen}{ \leadsto } & \ y^{\prime} – \frac{2}{x} \cdot y = \frac{2 \ln x – 1}{2x}
\end{aligned}
$$

观察可知，微分方程 $y^{\prime} – \frac{2}{x} \cdot y = \frac{2 \ln x – 1}{2x}$ 是一个一阶线性非齐次微分方程，由一阶线性微分方程的求解公式，得：

$$
\begin{aligned}
y & = \left[\int \frac{2 \ln x – 1}{2 x} \mathrm{e}^{-\int \frac{2}{x} \mathrm{~d}x} \mathrm{~d}x + C\right] \mathrm{e}^{\int \frac{2}{x} \mathrm{~d}x} \\ \\
& = \left[\int \frac{2 \ln x – 1}{2 x} \mathrm{e}^{-2\int \frac{1}{x} \mathrm{~d}x} \mathrm{~d}x + C\right] \mathrm{e}^{2\int \frac{1}{x} \mathrm{~d}x} \\ \\
& = \left[\int \frac{2 \ln x – 1}{2 x} \mathrm{e}^{-2 \ln x} \mathrm{~d}x + C\right] \mathrm{e}^{2 \ln x} \\ \\
& = \left[\int \frac{2 \ln x – 1}{2 x} \mathrm{e}^{\ln x^{-2}} \mathrm{~d}x + C\right] \mathrm{e}^{\ln x^{2}} \\ \\
& = \left[\int \frac{2 \ln x – 1}{2 x} x^{-2} \mathrm{~d}x + C\right] x^{2} \\ \\
& = \left[\int \frac{2 \ln x – 1}{2 x^{3}} \mathrm{~d}x + C\right] x^{2} \\ \\
& \textcolor{lightgreen}{ \leadsto } \textcolor{gray}{\left( \frac{- \ln x}{2x^{2}} \right)^{\prime} = \frac{-1}{2} \left( \frac{\ln x}{x^{2}} \right)^{\prime} = \frac{-1}{2} \left( \frac{\frac{1}{x} \cdot x^{2} – 2x \ln x}{x^{4}} \right) = \frac{-1}{2} \left( \frac{1}{x^{3}} – \frac{2 \ln x}{x^{3}} \right) = \frac{2 \ln x – 1}{2 x^{3}} } \\ \\
& = \left[ \frac{- \ln x}{2 x^{2}} + C \right] x^{2} \\ \\
& = \textcolor{lightgreen}{ \frac{-1}{2} \ln x + C x^{2} }
\end{aligned}
$$

将 $x = 1$, $y \left( 1 \right) = \frac{1}{4}$ 代入 $y = \frac{-1}{2} \ln x + C x^{2}$, 得：

$$
\textcolor{lightgreen}{
C = \frac{1}{4}
}
$$

因此：

$$
\textcolor{lightgreen}{
y = \frac{-1}{2} \ln x + \frac{1}{4} x^{2}
}
$$

进而：

$$
\textcolor{lightgreen}{
y ^{\prime} = \frac{-1}{2 x} + \frac{1}{2} x
}
$$

于是，根据曲线弧长的计算公式，得：

$$
\begin{aligned}
s & = \int_{1}^{e} \sqrt{1 + \left(-\frac{1}{2x} + \frac{x}{2}\right)^{2}} \mathrm{~d}x \\ \\
& = \int_{1}^{e} \sqrt{1 + \frac{1}{4}\left(x – \frac{1}{x}\right)^{2}} \mathrm{~d}x \\ \\
& = \int_{1}^{e} \sqrt{1 + \frac{1}{4}\left(x^{2} – 2 + \frac{1}{x^{2}}\right)} \mathrm{~d}x \\ \\
& = \int_{1}^{e} \sqrt{\textcolor{orangered}{ \frac{1}{4} \cdot 4 } + \frac{1}{4}\left(x^{2} \textcolor{orangered}{ – 2 } + \frac{1}{x^{2}}\right)} \mathrm{~d}x \\ \\
& = \int_{1}^{e} \sqrt{\frac{1}{4}\left(x^{2} \textcolor{orangered}{ + 2 } + \frac{1}{x^{2}}\right)} \mathrm{~d}x \\ \\
& = \int_{1}^{e} \sqrt{\frac{1}{4}\left(x + \frac{1}{x}\right)^{2}} \mathrm{~d}x \\ \\
& \textcolor{lightgreen}{ \leadsto } \textcolor{gray}{x \in \left[ 1, e \right] \leadsto x + \frac{1}{x} > 0 } \\ \\
& = \int_{1}^{e} \frac{1}{2}\left|x + \frac{1}{x}\right| \mathrm{~d}x \\ \\
& = \int_{1}^{e} \left(\frac{x}{2} + \frac{1}{2x}\right) \mathrm{~d}x \\ \\
& = \left[\frac{x^{2}}{4} + \frac{1}{2}\ln x\right]_{1}^{e} \\ \\
& = \left(\frac{e^{2}}{4} + \frac{1}{2}\ln e\right)
– \left(\frac{1}{4} + \frac{1}{2}\ln 1\right) \\ \\
& = \frac{e^{2}}{4} + \frac{1}{2} – \frac{1}{4} \\ \\
& = \textcolor{lightgreen}{ \frac{1}{4}e^{2} + \frac{1}{4} }
\end{aligned}
$$

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[01]. 2023年考研数二第12题解析：曲线弧长计算、凑微分、挖掘隐含条件
[02]. 2021年考研数二第19题解析：曲线的弧长、旋转曲面的面积
[03]. 2019年考研数二第12题解析：曲线弧长的计算
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[05]. 计算平面曲线的弧长需要知道积分上下限，但如果这个积分上线限题目中没有给出该怎么办？

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