A) \[-\frac{1}{xy}=c\]
B) \[-\frac{1}{xy}+\log y=c\]
C) \[\frac{1}{xy}+\log y=c\]
D) \[\log y=cx\]
Correct Answer: B
Solution :
\[y\,dx+(x+{{x}^{2}}y)dy=0\] \[\Rightarrow \] \[y\,dx+x\,dy=-{{x}^{2}}y\,dy\] \[\Rightarrow \] \[\frac{y\,dx+x\,dy}{{{x}^{2}}{{y}^{2}}}=-\frac{1}{y}dy\] \[\Rightarrow \] \[d\left( -\frac{1}{xy} \right)=-\frac{1}{y}dy\] On integrating, we get \[-\frac{1}{xy}=-\log y+c\] \[\Rightarrow \] \[-\frac{1}{xy}+\log y=c\]
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