令p=y'
则y"=pdp/dy
代入方程得: pdp/dy=1/√y
pdp=dy/√y
积分：p^2/2=2√y +C1
即p^2=4√y+C
p=dy/dx=±√(4√y+C)
dy/√(4√y+c)=±dx
令4√y+c=u
则y=(u-c)^2/16
dy=(u-c)du/8
代入得：
(u-c)du/(8u)=±dx
(1-c/u)du=±8dx
积分：u-cln|u|=±8x+c2
故有4√y+c-cln|4√y+c|=±8x+c2
则y"=pdp/dy
代入方程得: pdp/dy=1/√y
pdp=dy/√y
积分：p^2/2=2√y +C1
即p^2=4√y+C
p=dy/dx=±√(4√y+C)
dy/√(4√y+c)=±dx
令4√y+c=u
则y=(u-c)^2/16
dy=(u-c)du/8
代入得：
(u-c)du/(8u)=±dx
(1-c/u)du=±8dx
积分：u-cln|u|=±8x+c2
故有4√y+c-cln|4√y+c|=±8x+c2
温馨提示：答案为网友推荐，仅供参考
相似回答