展开全部
解：设y'=p，则y''=pdp/dy
代入原方程得pdp/dy=(a²+1/r²)y
==>pdp=(a²+1/r²)ydy
==>p²=(a²+1/r²)y²+C1 (C1是积分常数)
==>p=±√[(a²+1/r²)y²+C1]
==>y'=±√[(a²+1/r²)y²+C1]
==>dy/√[(a²+1/r²)y²+C1]=±dx
==>dy/√[y²+C1/(a²+1/r²)]=±√(a²+1/r²)dx
==>ln│y+√[y²+C1/(a²+1/r²)]│=±x√(a²+1/r²)+ln│C2│ (C2是积分常数)
==>y+√[y²+C1/(a²+1/r²)]=C2e^[±x√(a²+1/r²)]
故原方程的通解是y+√[y²+C1/(a²+1/r²)]=C2e^[±x√(a²+1/r²)] (C1,C2是积分常数)
代入原方程得pdp/dy=(a²+1/r²)y
==>pdp=(a²+1/r²)ydy
==>p²=(a²+1/r²)y²+C1 (C1是积分常数)
==>p=±√[(a²+1/r²)y²+C1]
==>y'=±√[(a²+1/r²)y²+C1]
==>dy/√[(a²+1/r²)y²+C1]=±dx
==>dy/√[y²+C1/(a²+1/r²)]=±√(a²+1/r²)dx
==>ln│y+√[y²+C1/(a²+1/r²)]│=±x√(a²+1/r²)+ln│C2│ (C2是积分常数)
==>y+√[y²+C1/(a²+1/r²)]=C2e^[±x√(a²+1/r²)]
故原方程的通解是y+√[y²+C1/(a²+1/r²)]=C2e^[±x√(a²+1/r²)] (C1,C2是积分常数)
已赞过
已踩过<
评论
收起
你对这个回答的评价是？
推荐律师服务：
若未解决您的问题，请您详细描述您的问题，通过百度律临进行免费专业咨询
|
广告 您可能关注的内容 |
