A) \[y\sin x+\cos 2x=2c\]
B) \[2y\sin x+\cos x=c\]
C) \[y\sin x+\cos x=c\]
D) \[2y\sin x+\cos 2x=c\]
Correct Answer: D
Solution :
\[\frac{dy}{dx}+y\cot x=2\cos x\] It is linear equation of the form \[\frac{dy}{dx}+Py=Q\] So, I.F. \[={{e}^{\int_{{}}^{{}}{Pdx}}}={{e}^{\int_{{}}^{{}}{\cot xdx}}}={{e}^{\log \sin x}}=\sin x\] Hence the solution is \[y\sin x=\int_{{}}^{{}}{2\sin x\cos xdx+c}\] Þ \[y\sin x=-\frac{1}{2}\cos 2x+c\]Þ\[2y\sin x+\cos 2x=c\].
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