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)

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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)
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)

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 =1
C2+1/2 =1
得 C1=C2=1/2
所以f(x) =(cosx +sinx +e^x)/2

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