A) \[y={{\tan }^{-1}}\left( \log \left( \frac{e}{x} \right) \right)\]
B) \[y=x{{\tan }^{-1}}\left( \log \left( \frac{x}{e} \right) \right)\]
C) \[y=x{{\tan }^{-1}}\left( \log \left( \frac{e}{x} \right) \right)\]
D) None of thee
Correct Answer: C
Solution :
[c] we have \[\frac{dy}{dx}=\frac{y}{x}-{{\cos }^{2}}\left( \frac{y}{x} \right)\] Putting \[y=vx\]So that \[\frac{dy}{dx}=v+x\frac{dv}{dx},\]we get \[v+x\frac{dv}{dx}=v-{{\cos }^{2}}v\] Or \[\frac{dv}{{{\cos }^{2}}v}=-\frac{dx}{x}\] Or \[{{\sec }^{2}}u\,du=-\frac{1}{x}dx\] On integration, we get \[\tan u=-\log x+\log C\] Or \[\tan \left( \frac{y}{x} \right)=-\log x+\log C\] This passes through \[(1,\pi /4).\] Therefore, \[1=\log C.\] So, \[\tan \left( \frac{y}{x} \right)=-\log x+1\] \[=-\log x+{{\log }_{e}}\] Or \[y=x{{\tan }^{-1}}\left( \log \left( \frac{e}{x} \right) \right)\]
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