设可微函数f(x)满足f(x)=x+定积分x~0f(u)du。求f(x)
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f(x)= x+∫(0->x) f(u) dx
f(0) =0
f'(x) = 1+ f(x)
dy/dx = 1+ y
∫dy/(1+y) = ∫dx
ln(1+y) =x + C
y(0) =0
=>C=0
ln(1+y) =x + C
1+y = e^x
y = -1+ e^x
ie
f(x) =-1+e^x
