A) \[y={{x}^{2}}\log (\sin \,x)\]
B) \[y=x\sin (log\,x)\]
C) \[{{y}^{2}}=x{{(x-1)}^{2}}\]
D) \[y=2{{x}^{2}}(x-1)\]
Correct Answer: B
Solution :
[b] The given equation can be rewritten as \[{{x}^{2}}\left[ \frac{xdy-ydx}{{{x}^{2}}} \right]=x\sqrt{1-\frac{{{y}^{2}}}{{{x}^{2}}}}\] \[\Rightarrow {{x}^{2}}\frac{d}{dx}\left( \frac{y}{x} \right)=x\sqrt{1-\frac{{{y}^{2}}}{{{x}^{2}}}}\] Put \[\frac{y}{x}=z,\] we get \[{{\sin }^{-1}}(z)=logx+c\Rightarrow si{{n}^{-1}}\left( \frac{y}{x} \right)=\log x+c\] Apply the boundary value y(1) = 0 \[\Rightarrow {{\sin }^{-1}}\left( \frac{0}{1} \right)=\log 1+c\Rightarrow c=0\] \[\therefore \,\,\,\,{{\sin }^{-1}}\left( \frac{y}{x} \right)=\log x+0\Rightarrow y=x\,\,\sin \,(log\,x)\]
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