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

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

Grade:12th pass

4 Answers

Aditya Gupta
2081 Points
4 years 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 :)

Vikas TU
14149 Points
4 years 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.
Mnadeem
13 Points
2 years ago
Select the no. of arbitrary constants in the particular solution of differential equation of second order is .
K K Harjeeth
26 Points
one year ago
xy²sin(xy)dx+ycos(xy)+x²ysin(xy)dy-xcos(xy)dy=0
xy²sin(xy)dx + x²ysin(xy)dy=[cos(xy)][xdy-ydx]
[xysin(xy)][ydx+xdy]=[cos(xy)][xdy-ydx]
Divide Both sides by 'xy'
[sin(xy)][ydx+xdy]  = [cos(xy)][d(log(y/x))]
[tan(xy)]d(xy) = d(log(y/x))
Integrate both sides
log(sec(xy)) = log(y/x) + logc
log(sec(xy))= log(cy/x)
xsec(xy)=cy

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