Question

这道题f（x）怎么求

Answer 1

简单分析一下，答案如图所示

解法一

解法二

Answer 2

f(x) - [1/(x+1)]∫(0->x) tf(t) dt = 1

∫(0->x) tf(t) dt = (x+1).[f(x) -1]

xf(x) = (x+1).f'(x)  +[f(x) -1]

(x+1)f'(x) +(1-x)f(x) = 1

f'(x) +[(1-x)/(1+x)] f(x) = 1/(x+1)

• p(x) =(1-x)/(1+x) = -1 +1/(1+x)

• e^[∫ p(x) dx] =  (1+x)e^(-x)

(1+x)e^(-x) .[ f'(x) +[(1-x)/(1+x)] f(x) ]=(1+x)e^(-x)  [1/(x+1)]

d/dx { (1+x)e^(-x) .f(x) } = e^(-x)

(1+x)e^(-x) .f(x) = -e^(-x) + C

f(0)=2

2 = -1 + C

C=3

(1+x)e^(-x) .f(x) = -e^(-x) + 3

f(x) = -1/(1+x) + 3e^x/(1+x)