求微分方程的通解:x/ydy-1/ydx=(2+y)/(1-y-y^2)dx
答案:1 悬赏:80
解决时间 2021-02-18 05:10
- 提问者网友:冷天寄予
- 2021-02-17 21:27
求微分方程的通解:x/ydy-1/ydx=(2+y)/(1-y-y^2)dx
最佳答案
- 二级知识专家网友:两不相欠
- 2021-02-17 22:56
解:∵x/ydy-1/ydx=(2+y)/(1-y-y^2)dx
==>(xdy-dx)(1-y-y^2)=y(2+y)dx
==>x(y^2+y-1)dy+(y+1)dx=0
==>(y^2+y-1)dy/(y+1)+dx/x=0
==>(y-1/(y+1))dy+dx/x=0
==>∫(y-1/(y+1))dy+∫dx/x=0
==>y^2/2-ln│y+1│+ln│x│=ln│C│ (C是非零常数)
==>xe^(y^2/2)/(y+1)=C
==>x=C(y+1)e^(-y^2/2)
∴此方程的通解是x=C(y+1)e^(-y^2/2)。
==>(xdy-dx)(1-y-y^2)=y(2+y)dx
==>x(y^2+y-1)dy+(y+1)dx=0
==>(y^2+y-1)dy/(y+1)+dx/x=0
==>(y-1/(y+1))dy+dx/x=0
==>∫(y-1/(y+1))dy+∫dx/x=0
==>y^2/2-ln│y+1│+ln│x│=ln│C│ (C是非零常数)
==>xe^(y^2/2)/(y+1)=C
==>x=C(y+1)e^(-y^2/2)
∴此方程的通解是x=C(y+1)e^(-y^2/2)。
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