设f(x)连续,且满足f(x)=e^x+∫x上0下（t-x）f（t）dt 求f（x）
人气：423 ℃ 时间：2019-10-23 08:35:13
解答
∵f(x)=e^x+∫(t-x)f(t)dt
∴f'(x)=e^x-∫f(t)dt
f''(x)=e^x-f(x)
f(0)=f'(0)=1
故 解此微分方程得 f(x)=C1e^x+C2e^(-x)+(x/2)e^x (C1,C2是积分常数).
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