The solution of the differential equation \[({{x}^{2}}-y{{x}^{2}})\frac{dy}{dx}+{{y}^{2}}+x{{y}^{2}}=0\] is [Pb. CET 2003]

\[\log \left( \frac{y}{x} \right)=\frac{1}{x}+\frac{1}{y}+c\]

\[\log \left( \frac{x}{y} \right)=\frac{1}{x}+\frac{1}{y}+c\]

\[\log \left( \frac{y}{x} \right)=\frac{1}{x}+\frac{1}{y}+c\]

\[\log \left( xy \right)=\frac{1}{x}+\frac{1}{y}+c\]

\[\log \left( xy \right)+\frac{1}{x}+\frac{1}{y}=c\]

The solution of the differential equation \[({{x}^{2}}-y{{x}^{2}})\fra

1.	The solution of the differential equation \[({{x}^{2}}-y{{x}^{2}})\frac{dy}{dx}+{{y}^{2}}+x{{y}^{2}}=0\] is [Pb. CET 2003]
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.	\[\log \left( xy \right)=\frac{1}{x}+\frac{1}{y}+c\]
D.	\[\log \left( xy \right)+\frac{1}{x}+\frac{1}{y}=c\]
Answer» B. \[\log \left( \frac{y}{x} \right)=\frac{1}{x}+\frac{1}{y}+c\]

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