A) \[({{e}^{y}}+1)\cos x=c\]
B) \[({{e}^{y}}-1)\sin x=c\]
C) \[({{e}^{y}}+1)\sin x=c\]
D) None of these
Correct Answer: C
Solution :
\[({{e}^{y}}+1)\cos xdx+{{e}^{y}}\sin xdy=0\] Þ \[\frac{{{e}^{y}}dy}{{{e}^{y}}+1}+\frac{\cos x}{\sin x}dx=0\] On integrating both the functions, we get \[\log ({{e}^{y}}+1)+\log (\sin x)=\log c\] Þ\[({{e}^{y}}+1)\sin x=c\].
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