Question

已知定义在(-1,+∞)上的连续函数f满足f(x)(f(t)dt+1)=xe^x/2(1+x)^2,求f(x)表达式

Answer 1

简单计首肢算一下即可，答者首世芹历案如图所示

Answer 2

整个记迟宏差为码皮y，分离变量求出y后再开方求导可得。要绝轮注意计算

Answer 3

f(x)[∫<0,x>f(t)dt+1]=xe^x/[2(x+1)^2],①腔野扰伍旦
f(0)=0,
对①求导得f'(x)[∫<0,x>f(t)dt+1]+[f(x)]^2=(1+x)e^x/[2(x+1)^2]-xe^x/(x+1)^3
=(x^2+1)e^x/[2(x+1)^3],②
f'(0)=1/2.
把①代入②脊早*f(x),得f'(x)xe^x/[2(x+1)^2]+[f(x)]^3=f(x)(x^2+1)e^x/[2(x+1)^3],
设y=f(x),去分母得x(x+1)e^x*y'+2(x+1)^3*y^3-(x^2+1)e^x*y=0,待续.

Answer 4

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追答