f(x+y)=f(x)+f(y)+xy(x+y),且f'(0)=1,求f(x)的解析式.
问题描述：
f(x+y)=f(x)+f(y)+xy(x+y),且f'(0)=1,求f(x)的解析式.
答
∵f(x+y)=f(x)+f(y)+xy(x+y)∴f(x+y)-f(x)=f(y)+xy(x+y),两边同时除以y[f(x+y)-f(x)]/y=f(y)/y+x(x+y),取极限y→0lim(y→0)[f(x+y)-f(x)]/y=lim(y→0)f(y)/y+x(x+y)即f'(x)=f'(0)+x²=x²+1∴f(x)=x³/3...