f(x)的导数为f'(x).且x^2f'(x)-(2x-1)f(x)=1.求f(x)
问题描述：
f(x)的导数为f'(x).且x^2f'(x)-(2x-1)f(x)=1.求f(x)
答
解:整理原式y'=[(2x-1)/(x^2)]y+1/(x^2)即dy/dx=[(2x-1)/(x^2)]y∴y=c(x^2)e^(1/x),常数变易c=c(x)代入,得c'=e^(-1/x)/x^4c=(1/x^2+2/x+2)e^(-1/x)+C∴y=1+2x+2*x^2+C(x^2)e^(1/x) 显然f(0)=0∴y=1+2x+2*x^2...