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