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Exact Differential Equation – Grouping Differentials | JEE

JEE Maths question with a full step-by-step solution.

Question
The solution of {y(1+x−1)+sin⁡y} dx+(x+ln⁡x+xcos⁡y) dy=0\big\{y(1 + x^{-1}) + \sin y\big\}\,dx + (x + \ln x + x\cos y)\,dy = 0 is:
Ay+xln⁡x+xsin⁡y=Cy + x\ln x + x\sin y = C
Bxy+yln⁡x+xsin⁡y=Cxy + y\ln x + x\sin y = Ccorrect
Cxy+xln⁡x+ysin⁡y=Cxy + x\ln x + y\sin y = C
Dx+yln⁡x+ysin⁡x=Cx + y\ln x + y\sin x = C
Solution
Expand the equation:
y dx+yx dx+sin⁡y dx+x dy+ln⁡x dy+xcos⁡y dy=0.y\,dx + \frac{y}{x}\,dx + \sin y\,dx + x\,dy + \ln x\,dy + x\cos y\,dy = 0.
 Group into exact differentials:
(y dx+x dy)⏟d(xy)+(yx dx+ln⁡x dy)⏟d(yln⁡x)+(sin⁡y dx+xcos⁡y dy)⏟d(xsin⁡y)=0.\underbrace{(y\,dx + x\,dy)}_{d(xy)} + \underbrace{\left(\frac{y}{x}\,dx + \ln x\,dy\right)}_{d(y\ln x)} + \underbrace{(\sin y\,dx + x\cos y\,dy)}_{d(x\sin y)} = 0.
 Integrating,
xy+yln⁡x+xsin⁡y=C.xy + y\ln x + x\sin y = C.
 Correct answer: (2)
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