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

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=0

log(secxy) +logx-logy=logc

xsecxy=cy

or x= cycos(xy)

kindly approve :)

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.