A) \[\frac{x}{y}+\frac{1}{xy}=k\]
B) \[\log \left( \frac{x}{y} \right)=\frac{1}{xy}+k\]
C) \[\frac{x}{y}+\frac{1}{xy}=k\]
D) \[\log \left( \frac{x}{y} \right)=xy+k\]
Correct Answer: B
Solution :
\[ydx+xdy+x{{y}^{2}}dx-{{x}^{2}}ydy=0\] \[\frac{ydx+xdy}{{{x}^{2}}{{y}^{2}}}+\frac{dx}{x}-\frac{dy}{y}=0\]. On integrating, we get \[-\frac{1}{xy}+\log x-\log y=k\] Þ \[\log \frac{x}{y}=\frac{1}{xy}+k\].
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