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Find the solution : $ x^2y'=1-xy$

$(a)\;y=\frac{\log x+c}{x} \\ (b)\;y=x \log x+c \\ (c)\;y=\frac{\log x}{x}+c \\ (d)\;y= \frac{\log x+x^2}{c} $
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$\large\frac{dy}{dx}+\frac{1}{x}$$y=\large\frac{1}{x^2}$
Integrating factor $=e^{\int (1/x)dx}$
$\qquad= e^{\log x}=x$
$y.x=\int \large\frac{1}{x^2}$$.x dx$
$y.x=\log x+c$
$y= \large\frac{\log x}{x}+\frac{c}{x}$
Hence a is the correct answer.
answered Feb 6, 2014 by meena.p
 

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