由已知方程，对x求导,得:\[f\left ( x \right )=\frac{1}{2}f\left ( x \right )+\frac{1}{2}xf{}'\left ( x \right )\]\[\therefore f\left ( x \right )=xf{}'\left ( x \right )\]\[\therefore f\left ( x \right )-xf{}'\left ( x \right )=0\]\[\frac{f\left ( x \right )-xf{}'\left ( x \right )}{f\left ( x \right )} ={\left ( \frac{x}{f\left ( x \right )} \right )}'=0\]\[\therefore \frac{x}{f\left ( x \right )}=c,c为常数\] \[\therefore 有 f\left ( x \right )=cx\]

真厉害。