A) \[y(\log ex+cx)=1\]
B) \[y(\log ex+cx)\]
C) \[y(\log ex-(x))=-1\]
D) None of the above
Correct Answer: A
Solution :
The given differential equation can be rewritten as, \[x\frac{dy}{dx}+y={{y}^{2}}\log x\] ?(i) \[\Rightarrow \] \[\frac{1}{{{y}^{2}}}.\frac{dy}{dx}+\frac{1}{xy}=\frac{1}{x}\log x\](divide by\[x{{y}^{2}}\]) Let \[\frac{1}{y}=v\Rightarrow \frac{-1}{{{y}^{2}}}.\frac{dy}{dx}=\frac{dv}{dx}\] So that, \[\frac{dv}{dx}-\frac{1}{x}v=\frac{-1}{x}\log x\] ?(ii) \[\therefore \] \[\text{IF}={{e}^{\int_{{}}^{{}}{-\frac{1}{x}dx}}}=\frac{1}{x}\] \[\therefore \] Solution is \[v.\frac{1}{x}=\int_{{}}^{{}}{\frac{1}{x}}\left( \frac{-1}{x}\log x \right)dx=-\int_{{}}^{{}}{\frac{\log }{{{x}^{2}}}}dx\] \[\Rightarrow \] \[\frac{1}{y}=\log ex+cx\] \[\Rightarrow \] \[y(\log ex+cx)=1\]
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