f(x)=x^2+∫[0~x]e^(x-t)f '(t)dt 怎么变到 f '(x)=2x+f '(x)+∫[0~x]e^(x-t)f '(t)dt

问题描述:

f(x)=x^2+∫[0~x]e^(x-t)f '(t)dt 怎么变到 f '(x)=2x+f '(x)+∫[0~x]e^(x-t)f '(t)dt

f(x)=x²+∫[0→x] e^(x-t) f '(t) dt=x²+e^x∫[0→x] e^(-t) f '(t) dt两边同时对x求导:f '(x)=2x+e^x∫[0→x] e^(-t) f '(t) dt+e^x*e^(-x) f '(x)=2x+e^x∫[0→x] e^(-t) f '(t) dt+f '(x)=2x+∫[0→x]...