A) \[\log \left( \frac{x}{y} \right)=\frac{1}{x}+\frac{1}{y}+C\]
B) \[\log \left( \frac{y}{x} \right)=\frac{1}{x}+\frac{1}{y}+C\]
C) \[\log \,(xy)=\frac{1}{x}+\frac{1}{y}+C\]
D) \[\log \,(xy)+\frac{1}{x}+\frac{1}{y}=C\]
Correct Answer: A
Solution :
The given differential equation is \[{{x}^{2}}(1-y)\frac{dy}{dx}+{{y}^{2}}(1+x)=0\] \[\Rightarrow \] \[{{x}^{2}}(1-y)dy+{{y}^{2}}(1+x)dx=0\] \[\Rightarrow \] \[\frac{1-y}{{{y}^{2}}}dy+\frac{1+x}{{{x}^{2}}}dx=0\] \[\Rightarrow \] \[\left( \frac{1}{{{y}^{2}}}-\frac{1}{y} \right)dy-\left( \frac{1}{{{x}^{2}}}+\frac{1}{x} \right)dx=0\] On integrating both sides, we get \[-\frac{1}{y}-\log y-\frac{1}{x}+\log x=C\] \[\Rightarrow \] \[\log \left( \frac{x}{y} \right)=\frac{1}{x}+\frac{1}{y}+C\]
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