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Question

Solve the following differential equation :
[yxcosyx]dy+[ycosyx2xsinyx]dx=0.

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Solution

concept [Extract differential equation]
M dx+N dy=0 is given difference equation
If My=Nx Then given equation is exact And solution is given by
M dx+(terms of N notcontaining x) dy=C
Given question
[yxcosyx]=Ndy+[ycosyx2xsinyx]=Mdx=0
So, cheching whether give difference equation is exact or not.
My=y[sin(yx)]×1x+cos(yx)2xcos(yx)×1x
=yxsin(yx)+cos(yx)2cos(yx)
=yxsin(yx)cos(yx)....(1)
Nx=0[cos(yx)+x(sin(yx))×yx2]
=[cos(yx)+yxsin(yx)]
=yxsin(yx)cos(yx)....(2)
(1) & (2), My=Nx\to$ So, given equation is exact
solution is given by y constantM dx+(term of N dy=C notcontaining)
y constant[ycosyx2xsinyx]dx+y dy=C
y constant[ycosyx2xsin(yx)dx+y22=C]

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