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Q.The solution of d2ydx2+4y=sin⁡2x\dfrac{d^2 y}{dx^2} + 4y = \sin 2x is

a
y=Acos⁡2x+Bsin⁡2x−xcos⁡2x4y = A \cos 2x + B \sin 2x - \dfrac{x \cos 2x}{4}
b
y=Acos⁡2x+Bsin⁡2x+xsin⁡2x4y = A \cos 2x + B \sin 2x + \dfrac{x \sin 2x}{4}
c
y=Acos⁡2x+Bsin⁡2x−xsin⁡2x4y = A \cos 2x + B \sin 2x - \dfrac{x \sin 2x}{4}
d
y=Acos⁡2x+Bsin⁡2x+xcos⁡2x4y = A \cos 2x + B \sin 2x + \dfrac{x \cos 2x}{4}

Correct Answer: Option A

The correct solution involves applying the fundamental concept to derive the final value step by step...

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