函数f(x)满足f'(x)是一次函数,设f '(x)= 2ax+b,则f(x) = ax^2 +bx +c,a、b、c为常数
由x^2 * f '(x)-(2x-1)f(x)=1得,
x^2 * (2ax+b) -(2x-1)( ax^2 +bx +c)
= 2ax^3+bx^2 - 2ax^3- 2bx^2-2cx + ax^2+bx+c
=(a-b)^x^2+(b-2c)x+c =1
则c=1,b-2c=0,a-b=0,则a=b=2c=2
即f(x)的解析式为f(x) = 2x^2 + 2x+1