原式化为2x*f(x)*f'(x)=[f(x)]^2-x^2
x*{[f(x)]^2}'=f(x)]^2-x^2
令 u(x)=[f(x)]^2
则x*u'(x)=u(x)-x^2
x*u'(x)-u(x)=-x^2
(x*u(x))'=-x^2
x*u(x)=-x^3/3+C
u(x)=-x^2/3+C/x
即 f(x)]^2=-x^2/3+C/x
所以 f(x)=根号(-x^2/3+C/x) 或 f(x)=-根号(-x^2/3+C/x)
原式化为2x*f(x)*f'(x)=[f(x)]^2-x^2
x*{[f(x)]^2}'=f(x)]^2-x^2
令 u(x)=[f(x)]^2
则x*u'(x)=u(x)-x^2
x*u'(x)-u(x)=-x^2
(x*u(x))'=-x^2
x*u(x)=-x^3/3+C
u(x)=-x^2/3+C/x
即 f(x)]^2=-x^2/3+C/x
所以 f(x)=根号(-x^2/3+C/x) 或 f(x)=-根号(-x^2/3+C/x)