一、题目
已知 $y = \frac{x}{\ln x}$ 是微分方程 $y ^{\prime} = \frac{y}{x} + \phi \left( \frac{x}{y} \right)$ 的唯一解，则函数 $\phi \left(\frac{x}{y}\right)$ 的显式表达式为 $\underline{\quad \quad \quad}$.
难度评级：
二、解析 
解题思路图：
由题可知：
$$
\begin{aligned}
& y ^{\prime} = \frac{y}{x} + \phi \left( \frac{x}{y} \right) \\ \\
\textcolor{lightgreen}{ \leadsto } \ & \phi \left( \frac{x}{y} \right) = y ^{\prime} – \frac{y}{x} \\ \\
\textcolor{lightgreen}{ \leadsto } \ & \textcolor{gray}{y = \frac{x}{\ln x}} \\ \\
\textcolor{lightgreen}{ \leadsto } \ & \phi \left( \frac{x}{y} \right) = \left( \frac{x}{\ln x} \right) ^{\prime} – \frac{1}{x} \cdot \frac{x}{\ln x} \\ \\
\textcolor{lightgreen}{ \leadsto } \ & \phi \left( \frac{x}{y} \right) = \frac{\ln x – 1}{\ln^{2} x} – \frac{1}{\ln x} \\ \\
\textcolor{lightgreen}{ \leadsto } \ & \phi \left( \frac{x}{y} \right) = \frac{\ln x – 1}{\ln^{2} x} – \frac{\ln x}{\ln^{2} x} \\ \\
\textcolor{lightgreen}{ \leadsto } \ & \textcolor{orange}{ \phi \left( \frac{x}{y} \right) = \frac{- 1}{\ln^{2} x} } \\ \\
\textcolor{lightgreen}{ \leadsto } \ & \textcolor{gray}{y = \frac{x}{\ln x}} \\ \\
\textcolor{lightgreen}{ \leadsto } \ & \phi \left( \frac{x}{y} \right) = – y^{2} \cdot \frac{1}{x^{2}} \\ \\
\textcolor{lightgreen}{ \leadsto } \ & \textcolor{lightgreen}{ \phi \left( \frac{x}{y} \right) = – \left( \frac{y}{x} \right)^{2} }
\end{aligned}
$$
Next 
由 $\phi \left( \frac{x}{y} \right)$ 可知，其自变量为 $\frac{x}{y}$, 因此，将 $\phi \left( \frac{x}{y} \right)$ 的表达式写作 $\textcolor{orange}{ \phi \left( \frac{x}{y} \right) = \frac{- 1}{\ln^{2} x} }$ 是不对的，因为这个表达式中没有显含自变量 $\frac{x}{y}$, 必须写成下面这样的形式：
$$
\textcolor{lightgreen}{ \phi \left( \frac{x}{y} \right) = – \left( \frac{y}{x} \right)^{2} }
$$
或者至少是下面的形式：
$$
\textcolor{yellow}{ \phi \left( \frac{x}{y} \right) = \frac{- y^{2}}{x^{2}} }
$$
