A) \[\frac{{{x}^{2}}+{{y}^{2}}}{2}+{{\tan }^{-1}}\sqrt{\frac{y}{x}=c}\]
B) \[\frac{{{x}^{2}}+{{y}^{2}}}{2}+2{{\tan }^{-1}}\sqrt{\frac{x}{y}=c}\]
C) \[\frac{{{x}^{2}}+{{y}^{2}}}{2}+2{{\cot }^{-1}}\sqrt{\frac{x}{y}=c}\]
D) None of these
Correct Answer: B
Solution :
[b] The given equation is written as \[ydx-xdy+x\sqrt{xy}(x+y)dx+y\sqrt{xy}(x+y)dy=0\]or \[ydx-xdy+(x+y)\sqrt{xy}(xdx+ydy)=0\] Or \[\frac{ydx-xdy}{{{y}^{2}}}+\left( \frac{x}{y}+1 \right)\sqrt{\frac{x}{y}}\left( d\left( \frac{{{x}^{2}}+{{y}^{2}}}{2} \right) \right)=0\]or \[d\left( \frac{{{x}^{2}}+{{y}^{2}}}{2} \right)+\frac{d\left( \frac{x}{y} \right)}{\left( \frac{x}{y}+1 \right)\sqrt{\frac{x}{y}}}=0\] Or \[\frac{{{x}^{2}}+{{y}^{2}}}{2}+2{{\tan }^{-1}}\sqrt{\frac{x}{y}}=c\]
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