搜搜做题作业网求yy''－y'^2－1=0 的通解
来源：学生作业帮 编辑：搜搜做题作业网作业帮 分类：数学作业 时间：2024/06/28 07:54:35
求yy''－y'^2－1=0 的通解
∵令y'=p,则y''=pdp/dy
∴代入原方程,得ypdp/dy-p²-1=0
==>pdp/(p²+1)=dy/y
==>ln(p²+1)=2ln│y│+2ln│C1│ (C1是非零积分常数)
==>p²+1=(C1y)²
==>p=±√[(C1y)²-1]
==>y'=±√[(C1y)²-1]
==>dy/√[(C1y)²-1]=±dx
==>ln{C1y+√[(C1y)²-1]}=±C1x+ln│C2│ (C2是非零积分常数)
==>C1y+√[(C1y)²-1]=C2*e^(±C1x)
故原方程的通解是C1y+√[(C1y)²-1]=C2*e^(±C1x).
∴代入原方程,得ypdp/dy-p²-1=0
==>pdp/(p²+1)=dy/y
==>ln(p²+1)=2ln│y│+2ln│C1│ (C1是非零积分常数)
==>p²+1=(C1y)²
==>p=±√[(C1y)²-1]
==>y'=±√[(C1y)²-1]
==>dy/√[(C1y)²-1]=±dx
==>ln{C1y+√[(C1y)²-1]}=±C1x+ln│C2│ (C2是非零积分常数)
==>C1y+√[(C1y)²-1]=C2*e^(±C1x)
故原方程的通解是C1y+√[(C1y)²-1]=C2*e^(±C1x).