设y'=p,则y''=pdp/dy
代入原方程得y³pdp/dy-1=0
==>y³pdp/dy=1
==>pdp=dy/y³
==>p²/2=-1/(2y²)+C1/2 (C1是积分常数)
==>p²=C1-1/y²
==>p=±√(C1-1/y²)
==>dy/dx=±√(C1y²-1)/y
==>ydy/√(C1y²-1)=±dx
==>d(C1y²-1)/√(C1y²-1)=±2C1dx
==>2√(C1y²-1)=±2C1x+2C2 (C2是积分常数)
==>√(C1y²-1)=±C1x+C2
==>C1y²-1=(C2±C1x)²
==>C1y²=(C2±C1x)²+1
故原微分方程的通解是 C1y²=(C2±C1x)²+1 (C1,C2是积分常数)