∵xy'-y-√(y²-x²)=0 ==>y'-y/x-√(y²/x²-1)=0
∴设y=xt,则y'=xt'+t
代入方程得xt'-√(t²-1)=0 ==>dt/√(t²-1)=dx/x
==>ln(t+√(t²-1))=ln│x│+ln│C│ (C是积分常数)
==>t+√(t²-1)=Cx
==>y/x+√(y²/x²-1)=Cx
==>y+√(y²-x²)=Cx²
故原方程的通解是y+√(y²-x²)=Cx² (C是积分常数).