A) \[x{{\log }_{e}}|y|=2(x-1)\]
B) \[x{{\log }_{e}}|y|=x-1\]
C) \[{{x}^{2}}{{\log }_{e}}|y|=-2(x-1)\]
D) \[x{{\log }_{e}}|y|=-2(x-1)\]
Correct Answer: A
Solution :
given \[\frac{dy}{dx}=\frac{2y}{{{x}^{2}}}\] \[\Rightarrow \]\[\int_{{}}^{{}}{\frac{dy}{2y}=\int_{{}}^{{}}{\frac{dx}{{{x}^{2}}}}}\] \[\Rightarrow \]\[\frac{1}{2}\ell ny=-\frac{1}{x}+c\] passes through centre (1,1) \[\Rightarrow \]\[c=1\] \[\Rightarrow \]\[x\ell ny=2(x-1)\]You need to login to perform this action.
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