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Question: Answered & Verified by Expert
The solution of the differential equation sinyexdx-excosydy=sin2ydx is (where, c is an arbitrary constant)
MathematicsDifferential EquationsJEE Main
Options:
  • A exsiny=x+c
  • B ex=x+csiny
  • C exx=siny+c
  • D exsiny=x2+c
Solution:
2802 Upvotes Verified Answer
The correct answer is: ex=x+csiny
Given equation is
siny·exdx-excosydysin2y=dx
dexsiny=dx
On integrating, we get
exsiny=x+c
or ex=x+csiny

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