设f(x)连续,且满足f(x)=e^x+∫x上0下(t-x)f(t)dt 求f(x)

问题描述:

设f(x)连续,且满足f(x)=e^x+∫x上0下(t-x)f(t)dt 求f(x)

∵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).