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Solve this non exact equation (Xysinxy+cosxy)ydx+(xysinxy_cosxy)xdy=0

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one year ago

```							Given diff.equ is(xysinxy+cosxy)ydx+(xysinxy-cosxy)xdy=0.Denote M=(xysinxy+cosxy)y and N=(xysinxy-coxy)x.Mx-Ny=2xycosxy.Multiplying given d.e. by 1/(2xycosxy) we get(1/2) (tanxy+1/(xy))ydx+(1/2)(tanxy-1/(xy))xdy=0(xdy+ydx)tanxy+(1/x)dx-(1/y)dy=0.tanxyd(xy)+(1/x)dx-(1/y)dy=0log(secxy) +logx-logy=logcxsecxy=cyor x= cycos(xy)kindly approve :)
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one year ago
```							Dear student Starting with ydx+xdy = d(xy),(xy) sin(xy) d(xy) + cos(xy) (y dx - x dy) = 0,dividing by (xy) cos(xy):sin(xy)/cos(xy) d(xy) + dx/x - dy/y = 0,d(-cos(xy))/cos(xy) + d(log|x| - log|y|) = 0,-log |cos(xy)| + log|x| - log|y| = -log|C|,Cx = y cos(xy).x = 0 is a solution, too.
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one year ago
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