A) \[y={{\tan }^{-1}}\left( \log \frac{c}{x} \right)\]
B) \[y=x{{\tan }^{-1}}\left( \log \frac{x}{c} \right)\]
C) \[y=x{{\tan }^{-1}}\left( \log \frac{c}{x} \right)\]
D) None of these
Correct Answer: C
Solution :
We have,\[\frac{dy}{dx}=\frac{y}{x}-{{\cos }^{2}}\left( \frac{y}{x} \right)\] Putting\[y=vx\], so that \[\frac{dy}{dx}=v+x\frac{dv}{dx}\], \[\therefore \] \[v+x\frac{dv}{dx}=v-{{\cos }^{2}}v\] \[\Rightarrow \] \[\frac{dv}{{{\cos }^{2}}v}=-\frac{dx}{x}\] \[\Rightarrow \] \[{{\sec }^{2}}v\,\,dv=-\frac{1}{x}dx\] on integration, we get \[\tan v=-\log x+\log c\] \[\Rightarrow \] \[\tan \left( \frac{y}{x} \right)=-\log x+\log c\] this passes through\[(1,\,\,\pi /4)\], therefore, \[1=\log c\] So; \[\tan \left( \frac{y}{x} \right)=-\log x+1\] \[\Rightarrow \] \[\tan \left( \frac{y}{x} \right)=-\log x+\log c\] \[\Rightarrow \] \[y=x{{\tan }^{-1}}\left( \log \frac{c}{x} \right)\].
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