微分方程(siny+y^2sinx)dx+(xcosy-2ycosx)dy=0. 求详解。谢谢

2025-03-07 06:36:13
推荐回答(2个)
回答1:

(sinydx+xcosydy)+(y^2sinxdx-2ycosx)dy=0
[sinydx+xd(siny)]+[y^2d(-cosx)-cosx(dy^2)]=0
d(xsiny)+d(-y^2cosx)=0
d(xsiny-y^2cosx)=0
xsiny-y^2cosx=C,C为任意常数.

回答2:

丫,