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Find the solution : $-x dy +(y+\sqrt {xy})dx=0$

$(a)\;y= 2x (\log x+c) \\ (b)\;y= \frac{x}{4} (\log x+c) \\ (c)\;y= \frac{x}{4} (\log x+c)^2 \\ (d)\;y= x(\log x+c)^2 $
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$(y+ \sqrt {xy}) dx= x dy$
$\large\frac{dx}{dy}=\frac{x}{y+ \sqrt {xy}}$
$x= vy$
$\large\frac{dx}{dy}=y \large\frac{dv}{dy}$$+v$
$y \large\frac{dv}{dy}$$+v=\large\frac{v}{1+\sqrt v}$
$y \large\frac{dv}{dy}=\large\frac{-v \sqrt v}{1+\sqrt v}$
$\int \large\frac{1+\sqrt v}{v \sqrt v}$$dv=\int -\large\frac{1}{y} $$dy$
$-2 v^{-1/2}=- \log |v|- \log y+c_1$
$2 \bigg( \large\frac{x}{y} \bigg)^{-1/2}$$=\log x +c$
$2 \bigg( \large\frac{y}{x} \bigg)^{1/2}$$=\log x +c$
$4y=x(\log x+c)^2$
Hence c is the correct answer.
answered Feb 6, 2014 by meena.p
 

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