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Q.The solution of dydx+ycot⁡x=4xcsc⁡x(x≠0)\dfrac{dy}{dx} + y \cot x = 4x \csc x (x \ne 0) is

a
ysin⁡x=2x2+Cy \sin x = 2x^2 + C
b
ysin⁡x=x2+Cy \sin x = x^2 + C
c
ycos⁡x=2x2+Cy \cos x = 2x^2 + C
d
ysin⁡x=log⁡∣x∣+Cy \sin x = \log|x| + C

Correct Answer: Option A

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

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