设函数f(x)连续,且满足f(x)=e^x+∫(0.x)uf(u)du-x∫(0.x)f(u)du,求f(x)

f'(x) =e^x + xf(x) -(∫(0.x)f(u)du - xf(x)) = e^x-∫(0.x)f(u)du有 f''(x) = e^x -f(x)有 f''(x)+f(x) =e^x解这个微分方程得通解f(x)=C1cosx + C2 sinx + e^x/2注意到 f(0)=1,f'(0)=1得 C1+1/2 =1C2+1/2 =1得 C1=...