令 y'=p,,则y"=p*dp/dy原方程化为：p*dp/dy=1/y^3解得：p^2=C1-1/y^2p=(C1-1/y^2)^(1/2)即：y'=(C1-1/y^2)^(1/2)解得：（x-C2）^-y^2/C1+1/C2^2=0
