求微分方程dy/dx=(4x+3Y)/(x+y)的通解
答案:2 悬赏:60
解决时间 2021-02-10 16:43
- 提问者网友:喜遇你
- 2021-02-09 22:00
求微分方程dy/dx=(4x+3Y)/(x+y)的通解
最佳答案
- 二级知识专家网友:嗷呜我不好爱
- 2021-02-09 22:14
dy/dx=(4x+3y)/(x+y)
dy/dx=3+x/(x+y)
y/x=u dy=udx+xdu
u+xdu/dx=3+1/(1+u)
xdu/dx=3-u+1/(1+u)
(1+u)du/(4+2u-u^2)=dx/x
(-1+u)du/(4+2u-u^2)-2du/(4+2u-u^2)=dx/x
(-1/2)dln(4+2u-u^2)-2du/[5-(u-1)^2]=dlnx
du/[√5-(u-1)][√5+(u-1)]=(1/(2√5))[ln(√5+u-1)-ln(√5-u+1)]
(-1/2)ln(4+2u-u^2)-(1/√5)[ln(√5+u-1)-ln(√5-u+1)]=lnx+C0
(-1/2)ln[4+2y/x-(y/x)^2] - (1/√5)[ln(√5+y/x-1) - ln(√5-y/x+1)]=lnx+C0
dy/dx=3+x/(x+y)
y/x=u dy=udx+xdu
u+xdu/dx=3+1/(1+u)
xdu/dx=3-u+1/(1+u)
(1+u)du/(4+2u-u^2)=dx/x
(-1+u)du/(4+2u-u^2)-2du/(4+2u-u^2)=dx/x
(-1/2)dln(4+2u-u^2)-2du/[5-(u-1)^2]=dlnx
du/[√5-(u-1)][√5+(u-1)]=(1/(2√5))[ln(√5+u-1)-ln(√5-u+1)]
(-1/2)ln(4+2u-u^2)-(1/√5)[ln(√5+u-1)-ln(√5-u+1)]=lnx+C0
(-1/2)ln[4+2y/x-(y/x)^2] - (1/√5)[ln(√5+y/x-1) - ln(√5-y/x+1)]=lnx+C0
全部回答
- 1楼网友:恕我颓废
- 2021-02-09 23:23
∵dy/dx=-(4x+3y)/(x+y)=-(4+3y/x)/(1+y/x),
∴可令y/x=u,则:y=ux,∴dy/dx=u+xdu/dx,
∴u+xdu/dx=-(4+3u)/(1+u),
∴xdu/dx=-(5+4u)/(1+u),
∴[(1+u)/(5+4u)]du=-(1/x)dx,
∴∫[(1+u)/(5+4u)]du=-∫(1/x)dx,
∴∫[(5+4u-1)/(5+4u)]du=-4∫d(ln|x|),
∴∫[1-1/(5+4u)]du=-4ln|x|+c,
∴u-(1/4)ln|5+4u|=-4ln|x|+c,
∴y/x-(1/4)ln|5+4y/x|=-4ln|x|+c。
∴原微分方程的通解是:y/x-(1/4)ln|5+4y/x|=-4ln|x|+c。
我要举报
如以上问答内容为低俗、色情、不良、暴力、侵权、涉及违法等信息,可以点下面链接进行举报!
大家都在看
推荐资讯