求解微分方程:(1) 2yy‘‘=(y‘)^2+y^2 (2) yy‘‘+(y‘)^2+2x=0

来源:学生作业帮助网 编辑:作业帮 时间:2024/05/03 22:15:09
求解微分方程:(1) 2yy‘‘=(y‘)^2+y^2 (2) yy‘‘+(y‘)^2+2x=0

求解微分方程:(1) 2yy‘‘=(y‘)^2+y^2 (2) yy‘‘+(y‘)^2+2x=0
求解微分方程:(1) 2yy‘‘=(y‘)^2+y^2 (2) yy‘‘+(y‘)^2+2x=0

求解微分方程:(1) 2yy‘‘=(y‘)^2+y^2 (2) yy‘‘+(y‘)^2+2x=0
这两题都可以化成全微分求解 .


点击放大:


1)令p=y'
则 y"=pdp/dy
代入得:2ypdp/dy=p^2+y^2
2dp/dy=p/y+y/p
设p/y=u,p=yu. p'=u+yu'代入:
2u+2yu'=u+1/u,2yu'=-u+1/u
-2udu/(1-u^2)+dy/y=0
得:y(1-u^2)=C1 1-C1/y=p^2/y^2
p=±√(y^2...

全部展开

1)令p=y'
则 y"=pdp/dy
代入得:2ypdp/dy=p^2+y^2
2dp/dy=p/y+y/p
设p/y=u,p=yu. p'=u+yu'代入:
2u+2yu'=u+1/u,2yu'=-u+1/u
-2udu/(1-u^2)+dy/y=0
得:y(1-u^2)=C1 1-C1/y=p^2/y^2
p=±√(y^2-C1y)
dy/√(y^2-C1y)=±dx,
dy/√[(y-C1/2)^2-(C1^2/4)]=±dx
积分:arcsin[(y-C1/2)/(C1/2)]=±x+C2

2)
(yy')'=-2x
积分:yy'=-x^2+C1
2yy'=-2x^2+2C
(y^2)'=-2x^2+C1
积分:
y^2=-2x^3/3+C1x+C2

收起

1.
y'=p y''=pdp/dy
2ypdp/dy=p^2+y^2
2dp/dy=p/y+y/p
设p/y=u,p=yu. p'=u+yu'代入:
2u+2yu'=u+1/u,2yu'=-u+1/u
-2udu/(1-u^2)+dy/y=0
解得:y(1-u^2)=C1 1-C1/y=p^2/y^2
p=±√(y^2-C1...

全部展开

1.
y'=p y''=pdp/dy
2ypdp/dy=p^2+y^2
2dp/dy=p/y+y/p
设p/y=u,p=yu. p'=u+yu'代入:
2u+2yu'=u+1/u,2yu'=-u+1/u
-2udu/(1-u^2)+dy/y=0
解得:y(1-u^2)=C1 1-C1/y=p^2/y^2
p=±√(y^2-C1y)
dy/√(y^2-C1y)=±dx,
dy/√[(y-C1/2)^2-(C1^2/4)]=±dx
积分得通arcsin[(y-C1/2)/(C1/2)]=±x+C2
2:
(yy')'+2x=0
积分得:yy'+x^2=C1
(y^2)'/2+x^2=C1
积分得通
y^2/2+x^3/3=C1x+C2

收起