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Find the solution : $y (\log x-1) y dx=x dy $

$(a)\;y=\frac{1}{\log ex+cx} \\ (b)\;y=\frac{1}{\log x+x}+c \\ (c)\;y=\frac{1}{\log x-x}+c \\ (d)\;y=\frac{x}{\log x}+c $
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The given equation can be written as
$x \large\frac{dy}{dx}$$+y =y^2 \log x$
=> $ \large\frac{1}{y^2} \frac{dy}{dx}+\frac{1}{xy}=\frac{1}{x}$$ \log x$
Let $\large\frac{1}{y} $$=t$
=> $\large\frac{-1}{y^2} \frac{dy}{dx}=\frac{dt}{dx}$
So, $\large\frac{dt}{dx}-\frac{1}{x}$$ \times t = \large\frac{-1}{x} \log x$----(1)
(1) is solved linear differential equation $I.f= e^{\int -\large\frac{1}{x}dx}=\frac{1}{x}$
$\large\frac{t.1}{x}= \int \large\frac{1}{x} \bigg(\frac{-1}{x}$$\log x\bigg)dx$
$\qquad= -\int \large\frac{\log x}{x^2} $$dx$
$\qquad= \large\frac{\log x}{x} -\int \large\frac{1}{x}.\frac{1}{x} $$dx$
$\qquad= \large\frac{\log x}{x} +\frac{1}{x}+c$
=> $t= 1+\log x +cx =\log ex +cx$
or $y. (\log ex +cx)=1$
Hence a is the correct answer.
answered Feb 5, 2014 by meena.p

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