- 设f(x)在x>0时连续,f(1)=3,且∫xy1f(t)dt=
- 设f(x)在x>0时连续,f(1)=3,且∫xy 1 f(t)dt=x∫y 1 f(t)dt +y∫x 1 f(t)dt (x〉0,y〉0),则f(x)=
详细解答,谢谢
- ∫(1,xy) f(t)dt=x∫(1,y) f(t)dt +y∫(1,x) f(t)dt .
固定y,两边对x求导,
f(xy)*y=∫(1,y) f(t)dt+yf(x).
x=1--->yf(y)= ∫(1,y) f(t)dt+3y
f(y)连续,所以∫(1,y) f(t)dt可导--->f(y)可导,两边对y求导
f(y)+yf'(y)=f(y)+3
f'(y)=3/y,
f(y)=3ln|y|+C
f(1)=3--->c=3
所以f(x)=3lnx+3 (x>0)