∵yy'=-x ==>ydy=-xdx
==>y^2=C-x^2 (C是常数)
∴原方程的通解是y^2=C-x^2
∵y(0)=1,则代入通解,得C=1
∴原方程满足所给初始条件的特解是y^2=1-x^2.