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Q.The solution of dydx=1−y2y\dfrac{dy}{dx} = \dfrac{\sqrt{1 - y^2}}{y} is

a
sin⁡−1y=log⁡∣x∣+C\sin^{-1} y = \log |x| + C
b
sin⁡−1y=log⁡∣x+1∣+C\sin^{-1} y = \log |x + 1| + C
c
cos⁡−1y=log⁡∣x∣+C\cos^{-1} y = \log |x| + C
d
tan⁡−1y=log⁡∣x∣+C\tan^{-1} y = \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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The solution of \dfrac{dy}{dx} = \dfrac{\sqrt{1 - y^2}}{y}... | ParikshaNiti