A) \[{{\tan }^{-1}}\left( \frac{y}{x} \right)=\log x+c\]
B) \[{{\tan }^{-1}}\left( \frac{y}{x} \right)=-\log x+c\]
C) \[{{\sin }^{-1}}\left( \frac{y}{x} \right)=\log x+c\]
D) \[{{\tan }^{-1}}\left( \frac{x}{y} \right)=\log x+c\]
Correct Answer: A
Solution :
It is homogeneous equation which can be written in the form \[\frac{dy}{dx}=\frac{{{x}^{2}}+xy+{{y}^{2}}}{{{x}^{2}}}\] Now put \[y=vx\]and \[\frac{dy}{dx}=v+x\frac{dv}{dx}\] Therefore, \[v+x\frac{dv}{dx}=\frac{{{x}^{2}}+v{{x}^{2}}+{{v}^{2}}{{x}^{2}}}{{{x}^{2}}}=1+v+{{v}^{2}}\] Þ \[x\frac{dv}{dx}=1+{{v}^{2}}\] Þ \[\frac{dv}{1+{{v}^{2}}}=\frac{dx}{x}\] Now integrating both sides, we get \[{{\tan }^{-1}}v=\log x+c\] Þ \[{{\tan }^{-1}}\left( \frac{y}{x} \right)=\log x+c\] \[\{\because \,\,y=vx\Rightarrow v=y/x\}\]
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