g(x)=e^x+∫ug(u)du-x∫ug(u)du,其中g(x)是连续函数,求g(x).
问题描述：
g(x)=e^x+∫ug(u)du-x∫ug(u)du,其中g(x)是连续函数,求g(x).
答
g(x)=f(x)/x
f(x)/x=e^x+(1-x)∫[0,x]f(u)du
f(x)/(x-x^2)-e^x/(1-x)=∫[0,x]f(u)du
f'(x)/(x-x^2)+f(x)(2x-1)/(x-x^2)^2-xe^x/(1-x)^2=f(x)
f(x)=Ce^x
f'(x)=C'e^x+Ce^x
(C'+C)/(x-x^2)+C(2x-1)/(x-x^2)^2-x/(1-x)^2=C
(C'+C)+C(2x-1)/(x-x^2)-x^2/(1-x)=C(x-x^2)
C'=C[(x-x^2)-(2x-1)/(x-x^2)-1]-x^2/(1-x) 难解