Question

fx在0到正无穷内有定义,f1阶导=1,当想要属于0到正无穷时满足fxy=yfx xfy,证明？

fx在0到正无穷内可导，且fx导=fx/x +1

Answer 1

f'(x)=f(x)/x+1,
设f(x)=xc(x),则f'(x)=c(x)+xc'(x),代入方程得
c'(x)=1/x,
c(x)=ln|x|+c,
所以f(x)=x(ln|x|+c),
yf(x)+xf(y)=y[x(ln|x|+c)]+x[y(ln|y|+c]
=xyln|xy|+c(x+y),
f(xy)=xy(ln|xy|+c),
命题不成立。