2. 考研
  3. 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)= _________．

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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考研数学三

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