A) \[{{\log }_{e}}({{x}^{2}}+{{y}^{2}})+2{{\tan }^{-1}}\frac{y}{x}+c=0\]
B) \[\frac{{{y}^{2}}}{2}+xy=xy-\frac{{{x}^{2}}}{2}+c\]
C) \[\left( 1+\frac{x}{y} \right)\text{ }y=\left( 1-\frac{x}{y} \right)\text{ }x+c\]
D) \[y=x-2{{\log }_{e}}y+c\]
Correct Answer: A
Solution :
Put \[y=vx\]Þ\[\frac{dy}{dx}=v+x\frac{dv}{dx}\] \[\therefore v+x\frac{dv}{dx}=\frac{v-1}{v+1}\] Þ \[x\frac{dv}{dx}=\frac{v-1}{v+1}-v\] Þ \[x\frac{dv}{dx}=-\frac{{{v}^{2}}+1}{v+1}\] Þ \[\int_{{}}^{{}}{\frac{dx}{x}}=-\int_{{}}^{{}}{\frac{v+1}{{{v}^{2}}+1}}\text{ }dv\] Þ \[-{{\log }_{e}}x=\frac{1}{2}\int_{{}}^{{}}{\frac{2v}{{{v}^{2}}+1}}dv+\int_{{}}^{{}}{\frac{1}{{{v}^{2}}+1}}dv\] Þ \[-{{\log }_{e}}x=\frac{1}{2}\log ({{v}^{2}}+1)+{{\tan }^{-1}}v+c\] Þ \[-2{{\log }_{e}}x=\log \left( \frac{{{x}^{2}}+{{y}^{2}}}{{{x}^{2}}} \right)+2{{\tan }^{-1}}\left( \frac{y}{x} \right)+c\] Þ \[{{\log }_{e}}({{x}^{2}}+{{y}^{2}})+2{{\tan }^{-1}}\frac{y}{x}+c=0\].
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