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Exact Differential Equation with sin 2y | JEE

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

Question
The solution of the differential equation dydx=sin⁡y+xsin⁡2y−xcos⁡y\dfrac{dy}{dx} = \dfrac{\sin y + x}{\sin 2y - x\cos y} is:
Asin⁡2y=xsin⁡y+x22+C\sin^2 y = x\sin y + \dfrac{x^2}{2} + Ccorrect
Bsin⁡2y=xsin⁡y−x22+C\sin^2 y = x\sin y - \dfrac{x^2}{2} + C
Csin⁡2y=x+sin⁡y+x22+C\sin^2 y = x + \sin y + \dfrac{x^2}{2} + C
Dsin⁡2y=x−sin⁡y+x22+C\sin^2 y = x - \sin y + \dfrac{x^2}{2} + C
Solution
Cross-multiply:
(sin⁡2y−xcos⁡y) dy=(sin⁡y+x) dx⇒sin⁡2y dy−xcos⁡y dy−sin⁡y dx=x dx.(\sin 2y - x\cos y)\,dy = (\sin y + x)\,dx \Rightarrow \sin 2y\,dy - x\cos y\,dy - \sin y\,dx = x\,dx.
 Now sin⁡2y dy=2sin⁡ycos⁡y dy=d(sin⁡2y)\sin 2y\,dy = 2\sin y\cos y\,dy = d(\sin^2 y) and sin⁡y dx+xcos⁡y dy=d(xsin⁡y)\sin y\,dx + x\cos y\,dy = d(x\sin y), so
d(sin⁡2y)−d(xsin⁡y)=x dx⇒sin⁡2y−xsin⁡y=x22+C.d(\sin^2 y) - d(x\sin y) = x\,dx \Rightarrow \sin^2 y - x\sin y = \frac{x^2}{2} + C.
∴sin⁡2y=xsin⁡y+x22+C.\therefore \sin^2 y = x\sin y + \frac{x^2}{2} + C.
 Correct answer: (1)
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