设f(x)一ex一∫0x(x一t)f(t)dt,其中f(x)连续,求f(x).

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问题 设f(x)一ex一∫0x(x一t)f(t)dt,其中f(x)连续,求f(x).

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答案由f(x)=ex一∫0x(x一t)f(t)dt,得f(x)=ex一x∫0xf(t)dt+∫0xtf(t)dt, 两边对x求导,得f’(x)=ex一f(t)dt,两边再对x求导得f"(x)+f(x)=ex,其通解为f(x)=C1cosx+C2sinx+[*]ex.在f(x)=ex0x(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=[*],C2=[*],故f(x)=[*](cos+sinx)+[*]ex

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