设f(x)连续且f(x)≠0,又设f(x)满足f(x)=∫0xf(x—t)dt+∫01f2(t)dt,则f(x)= _________.

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问题 设f(x)连续且f(x)≠0,又设f(x)满足f(x)=∫0xf(x—t)dt+∫01f2(t)dt,则f(x)= _________.

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答案[*]

解析 f(x)=∫0xf(x一t)dt+∫01f2(t)dt
    =一∫x0f(u)du+∫0xf(t)dt=∫0xd(u)du+∫01f(t)dt.
令∫01f2(t)dt=a,于是
    f(x)=∫0xf(u)du+a,f’(x)=f(x),f(0)=a,
解得f(x)=cex.由f(0)=a,得f(x)=aex,代入∫01f2(t)dt=a中,得
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