令dy/dx=p(x),则有d^2y/dx^2=dp/dx,题目可化为,（1+x^2)dp/dx-2px=0,即dp/p=2x/(1+p)dx,
两边积分：in|p|=in（1+x^2)+in|C1|
p=C1(1+x^2)或者dy/dx=C1(1+x^2)
再次积分可得.y=C1（x+x^3/3)+C2