∵令y'=p,则y"=pdp/dy
代入原方程,得pdp/dy-2yp^3=0
==>p(dp/dy-2yp^2)=0
∴p=0,或dp/dy-2yp^2=0
∵p=0不满足初始条件,舍去
∴dp/dy-2yp^2=0
==>dp/p^2=2ydy
==>-1/p=y^2-C1 (C1是常数)
==>-1/y'=y^2-C1
==>-dx/dy=y^2-C1
==>dx=-y^2+C1
==>x=C1y-y^3/3+C2 (C2是常数)
∵y(0)=1,y'(0)=-1
∴代入x=C1y-y^3/3+C2,得C1=0,C2=1/3
故原方程满足初始条件的特解是x=(1-y^3)/3.