2025年考研数二第15题解析：可分离变量的微分方程

一、题目

难度评级：

二、解析

解法 1

由题可知：

$$
\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{2y-3x}{5y-2x}=\frac{2\frac{y}{x}-3}{5\frac{y}{x}-2} \tag{1}
$$

接着，若设 $u=\frac{y}{x}$, 则：

$$
\begin{aligned}
& \ y = ux \\ \\
\textcolor{lightgreen}{ \leadsto } & \ y^{\prime} = u+xu^{\prime} \\ \\
\textcolor{lightgreen}{ \leadsto } & \ \frac{\mathrm{d} y}{\mathrm{d} x} = u + x \frac{\mathrm{d} u}{\mathrm{d} x}
\end{aligned}
$$

于是，结合上面的 $(1)$ 式，可知：

$$
\begin{aligned}
& \ u + x \frac{\mathrm{d} u}{\mathrm{d} x} = \frac{2u-3}{5u-2} \\ \\
\textcolor{lightgreen}{ \leadsto } & \ \frac{5u-2}{5u^{2}-4u+3}\mathrm{~d}u=-\frac{1}{x}\mathrm{~d}x \\ \\
\textcolor{lightgreen}{ \leadsto } & \ \int \frac{5u-2}{5u^{2}-4u+3}\mathrm{~d}u=-\int \frac{1}{x}\mathrm{~d}x \\ \\
\textcolor{lightgreen}{ \leadsto } & \ \textcolor{gray}{\text{为了去掉接下来的 } x \text{ 的绝对值，等号两边同时乘以 }2} \\ \\
\textcolor{lightgreen}{ \leadsto } & \ 2 \int \frac{5u-2}{5u^{2}-4u+3}\mathrm{~d}u = -2\int \frac{1}{x}\mathrm{~d}x \\ \\
\textcolor{lightgreen}{ \leadsto } & \ \int \frac{10u-4}{5u^{2}-4u+3}\mathrm{~d}u=-2\ln \left| x \right|+\ln \left| C \right| \\ \\
\textcolor{lightgreen}{ \leadsto } & \ \textcolor{gray}{\ln} \frac{\left| C \right|}{x^{2}} = \textcolor{gray}{ \int \frac{1}{5u^{2}-4u+3}\mathrm{~d}\left( 5u^{2}-4u+3 \right) } = \textcolor{gray}{ \ln } \left| 5u^{2}-4u+3 \right| \\ \\
\textcolor{lightgreen}{ \leadsto } & \ \left| 5u^{2}-4u+3 \right| = \frac{\left| C \right|}{x^{2}} \\ \\
\textcolor{lightgreen}{ \leadsto } & \ 5u^{2}-4u+3 = \frac{C}{x^{2}} \\ \\
\textcolor{lightgreen}{ \leadsto } & \ \textcolor{gray}{u=\frac{y}{x}} \\ \\
\textcolor{lightgreen}{ \leadsto } & \ 5y^{2}-4xy+3x^{2}=C \\ \\
\textcolor{lightgreen}{ \leadsto } & \ \textcolor{yellow}{y\left( 1 \right)=1 \leadsto C = 4} \\ \\
\textcolor{lightgreen}{ \leadsto } & \ \textcolor{lightgreen}{5y^{2}-4xy+3x^{2}=4}
\end{aligned}
$$

综上可知，微分方程 $\left( 2y-3x \right)\mathrm{d} x + \left( 2x-5y \right) \mathrm{d} y = 0$ 满足条件 $y \left( 1 \right) = 1$ 的解为：

$$
\textcolor{lightgreen}{5y^{2}-4xy+3x^{2}=4}
$$

解法 2

由题可知：

$$
\begin{aligned}
& \ \left( 2y – 3x \right)\mathrm{~d}x + \left( 2x – 5y \right)\mathrm{~d}y = 0 \\ \\
\textcolor{lightgreen}{ \leadsto } & \ 2y\mathrm{~d}x + 2x\mathrm{~d}y – 3x\mathrm{~d}x – 5y\mathrm{~d}y = 0 \\ \\
\textcolor{lightgreen}{ \leadsto } & \ \mathrm{d}\left( 2xy \right) – \mathrm{d}\left( \dfrac{3}{2} x^{2} \right) – \mathrm{d}\left( \dfrac{5}{2} y^{2} \right) = 0 \\ \\
\textcolor{lightgreen}{ \leadsto } & \ \mathrm{d}\left( 2xy – \dfrac{3}{2} x^{2} – \dfrac{5}{2} y^{2} \right) = 0 \\ \\
\textcolor{lightgreen}{ \leadsto } & \ 2xy – \dfrac{3}{2} x^{2} – \dfrac{5}{2} y^{2} = C \\ \\
\textcolor{lightgreen}{ \leadsto } & \ \textcolor{yellow}{y\left( 1 \right) = 1} \\ \\
\textcolor{lightgreen}{ \leadsto } & \ \textcolor{yellow}{2 – \dfrac{3}{2} – \dfrac{5}{2} = C \leadsto C = -2} \\ \\
\textcolor{lightgreen}{ \leadsto } & \ 2xy – \dfrac{3}{2} x^{2} – \dfrac{5}{2} y^{2} = -2 \\ \\
\textcolor{lightgreen}{ \leadsto } & \ \textcolor{lightgreen}{3x^{2} – 4xy + 5y^{2} = 4} \\ \\
\textcolor{lightgreen}{ \leadsto } & \ \textcolor{lightgreen}{y = \dfrac{2x + \sqrt{20 – 11x^{2}}}{5}}
\end{aligned}
$$

综上可知，微分方程 $\left( 2y-3x \right)\mathrm{d} x + \left( 2x-5y \right) \mathrm{d} y = 0$ 满足条件 $y \left( 1 \right) = 1$ 的解为：

$$
\textcolor{lightgreen}{
y = \dfrac{2x + \sqrt{20 – 11x^{2}}}{5}
}
$$

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