Question

已知函数f(x)可导,且f(1)＝1 若f(x)满足方程f(x)＋xf'(x)＝0,求f(2)

Answer 1

f(x)＋xf'(x)＝0

df(x)/f(x)=-1/x

两边积分,得

ln|f(x)|=-ln|x|+ln|c|

f(x)=c/x

f(1)=1

所以

1=c/1

c=1

所以

f(x)=1/x

f(2)=1/2