∵(x-ydx/dy)²+(y-xdy/dx)²=1
==>(x-y/y')²+(y-xy')²=1
==>(y-xy')²/y'²+(y-xy')²=1
==>(y-xy')²(1/y'²+1)=1
==>(y-xy')²=y'²/(1+y'²)
∴y=xy'±y'/√(1+y'²).(1)
∵方程(1)是克来洛方程
∴由克来洛定理得,方程(1)的通解是y=Cx±C/√(1+C²) (C是积分常数)
故原方程的通解是y=Cx±C/√(1+C²) (C是积分常数).