A) \[({{e}^{y}}+1)\cos x=K\]
B) \[({{e}^{y}}+1)\text{cosec}\,x=K\]
C) \[({{e}^{y}}+1)\sin x=K\]
D) None of these
Correct Answer: C
Solution :
\[\frac{dy}{dx}+\frac{({{e}^{y}}+1)\cot x}{{{e}^{y}}}=0\] Þ \[\int_{{}}^{{}}{\frac{{{e}^{y}}}{{{e}^{y}}+1}}dy+\int_{{}}^{{}}{\cot xdx}=0\] Þ \[\log ({{e}^{y}}+1)+\log \sin x=\log K\]Þ\[({{e}^{y}}+1)\sin x=K\].
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