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

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答案[*]

解析 f(x)=∫₀^xf(x—t)dt+∫₀¹f²(t)dt(第—个积分令x—t=u)
=一∫_x⁰f(u)du+∫₀¹f²(t)dt=∫₀^xf(u)du+∫₀¹f²(t)dt．
令∫₀¹f²(t)dt=a，于是
f(x)=∫₀^xf(u)du+a，f’(x)=f(x)，f(0)=a，
解得f(x)=Ce^x．由f(0)=a，得f(x)=ae^x，代入∫₀¹f²(t)dt=a中，得
a=∫₀¹f²(t)dt=a²∫₀¹e^2tdt=