2. Let $\mathbb{R}_{+}$ denote the set of all positive numbers. Find all functions $f: \mathbb{R}_{+} \rightarrow \mathbb{R}$ such that $x f(y)-y f(x)=$ $f(y / x)$ for all $x, y \in \mathbb{R}_{+}$.
2. Solution. Substitute $x=y=1: f(1)=0$. Now substitute $y=1: f(x)=-f\left(\frac{1}{x}\right)$. Now consider the pairs $(x, y)$ and $\left(\frac{1}{y}, \frac{1}{x}\right)$. We get $x f(y)-y f(x)=\frac{1}{y} f\left(\frac{1}{x}\right)-\frac{1}{x} f\left(\frac{1}{y}\right)=\frac{1}{x} f(y)-\frac{1}{y} f(x)$. From this, $f(x)=\frac{f(y)\left(x-\frac{1}{x}\right)}{y-\frac{1}{y}}$. Therefore, the function has the form $f(x)=c\left(x-\frac{1}{x}\right)$, where $c-$ is some number.