设f(x)=ex—∫0x(x—t)f(t)dt,其中f(x)连续,求f(x).

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问题 设f(x)=ex—∫0x(x—t)f(t)dt,其中f(x)连续,求f(x).

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答案由f(x)=ex—∫0x(x—t)f(t)dt,得f(x)=ex—x∫0xf(t)dt+∫0xtf(t)dt,两边对x求导,得f’(x)=ex—∫0xf(t)出,两边再对x求导得f"(x)+f(x)=ex,其通解为f(x)=Ccosx+Csinx+[*]e.在f(x)=ex—∫0x(x一t)f(t)dt中,令x=0得f(0)=1,在f’(x)=ex—∫0xf(t)dt中,令x=0得f’(0)=1,于是有C1=[*].

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