A) \[\sqrt{3}e\]
B) \[\sqrt{{{e}^{2}}-\frac{1}{2}}\]
C) \[\sqrt{\frac{{{e}^{2}}-1}{2}}\]
D) \[\sqrt{\frac{{{e}^{2}}+1}{2}}\]
Correct Answer: A
Solution :
\[{{x}^{2}}dy+{{y}^{2}}dy=xydx\] Þ \[x(xdy-ydx)=-{{y}^{2}}dy\] Þ \[x\frac{(ydx-xdy)}{{{y}^{2}}}=dy\] Þ \[\frac{x}{y}d\left( \frac{x}{y} \right)=\frac{dy}{y}\] Integrating, \[\frac{{{x}^{2}}}{2{{y}^{2}}}={{\log }_{e}}y+c\] Given \[y(1)=1\] Þ \[c=\frac{1}{2}\] Þ \[\frac{{{x}^{2}}}{2{{y}^{2}}}={{\log }_{e}}y+\frac{1}{2}\] Now \[y({{x}_{0}})=e\] Þ \[\frac{x_{0}^{2}}{2{{e}^{2}}}-{{\log }_{e}}e-\frac{1}{2}=0\] Þ \[x_{0}^{2}=3{{e}^{2}}\] Þ \[{{x}_{0}}=\pm \sqrt{3}e\]
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