A) \[\sin y=\frac{1}{x}{{e}^{x}}+c\]
B) \[\sin y+{{e}^{x}}\log x+c=0\]
C) \[\sin y={{e}^{x}}\log x+c\]
D) None of these
Correct Answer: C
Solution :
\[x\cos ydy=(x{{e}^{x}}\log x+{{e}^{x}})dx\] Þ \[\cos ydy=\left( {{e}^{x}}\log x+\frac{{{e}^{x}}}{x} \right)\text{ }dx\] On integrating, \[\sin y={{e}^{x}}\log x+c\].
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