A) \[{{y}^{2}}=1+\cos x\]
B) \[y={{\sin }^{2}}x\]
C) \[{{y}^{2}}\sin x=1+\cos x\]
D) \[{{y}^{2}}=\sin x\]
Correct Answer: D
Solution :
\[2y\frac{dy}{dx}+{{y}^{2}}.\cot x=2\cos x\] put \[{{y}^{2}}=t\Rightarrow \frac{dt}{dx}+(\cot x)t=2\cos x\] \[I.F.={{e}^{\ln (\sin x)}}=\sin x\] \[\therefore \] we get\[{{y}^{2}}=\sin x\]
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