A) \[y={{\tan }^{-1}}\log (e/x)\]
B) \[y={{e}^{1+\cot (y/x)}}\]
C) \[y=x{{\tan }^{-1}}\log (e/x)\]
D) \[y={{e}^{1+\tan (y/x)}}\]
Correct Answer: C
Solution :
Given, \[\frac{dy}{dx}=\frac{y}{x}-{{\cos }^{2}}\left( \frac{y}{x} \right)\] Putting, y = vx so that \[\frac{dy}{dx}=v+x\frac{dv}{dx}\] We get, \[v+x\frac{dv}{dx}=v-{{\cos }^{2}}v\] \[\Rightarrow \]\[\frac{dv}{{{\cos }^{2}}v}=-\frac{dx}{x}\Rightarrow {{\sec }^{2}}vdv=-\frac{dx}{x}\] Integrating, we get \[\tan v=-\ln x+\ln c\tan \left( \frac{y}{x} \right)=-\ln x+\ln c.\] This passes through \[\left( 1,\frac{\pi }{4} \right)\Rightarrow \ln c=1\] \[\therefore \]\[y=x{{\tan }^{-1}}\left( \log \frac{e}{x} \right)\]
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