设f(x)连续,且f(x)一2∫0xf(x—t)dt=ex,则f(x)=_______.

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问题 设f(x)连续,且f(x)一2∫0xf(x—t)dt=ex,则f(x)=_______.

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答案f(x)=2e2x-ex

解析 由∫0xf(x—t)dt0xf(u)(一du)=∫0xf(u)du得f(x)一2∫0xf(u)du=ex
求导得f’(x)一2f(x)=ex,解得
f(x)=[∫ex.e∫-2dxdx+C]e-∫-2dx=(一e-x+C)e2x=Ce2x-ex
由f(0)=1得C=2,故f(x)=2e2x一ex
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