\[({{x}^{2}}+{{y}^{2}})dy=xydx\]. If \[y({{x}_{0}})=e\], \[y(1)=1\], then value of \[{{x}_{0}}=\] [IIT Screening 2005]

\[\sqrt{{{e}^{2}}-\frac{1}{2}}\]

\[\sqrt{\frac{{{e}^{2}}-1}{2}}\]

\[\sqrt{\frac{{{e}^{2}}+1}{2}}\]

\[({{x}^{2}}+{{y}^{2}})dy=xydx\]. If \[y({{x}_{0}})=e\], \[y(1)=1\], t

1.	\[({{x}^{2}}+{{y}^{2}})dy=xydx\]. If \[y({{x}_{0}})=e\], \[y(1)=1\], then value of \[{{x}_{0}}=\] [IIT Screening 2005]
A.	\[\sqrt{3}e\]
B.	\[\sqrt{{{e}^{2}}-\frac{1}{2}}\]
C.	\[\sqrt{\frac{{{e}^{2}}-1}{2}}\]
D.	\[\sqrt{\frac{{{e}^{2}}+1}{2}}\]
Answer» B. \[\sqrt{{{e}^{2}}-\frac{1}{2}}\]

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\[({{x}^{2}}+{{y}^{2}})dy=xydx\]. If \[y({{x}_{0}})=e\], \[y(1)=1\], then value of \[{{x}_{0}}=\] [IIT Screening 2005]

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