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If $\large\frac{dy}{dx}=\frac{x-y}{x+y}$ then

\[\begin{array}{1 1}(1)2xy+y^{2}+x^{2}=c&(2)x^{2}+y^{2}-x+y=c\\(3)x^{2}+y^{2}-2xy=c&(4)x^{2}-y^{2}-2xy=c\end{array}\]

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$\large\frac{dy}{dx}=\frac{x-y}{x+y}$
$(x+y) dy=(x-y)dx$
$x \;dy +y \;dy = x \; dx -y\; dx$
$x \;dy +y \;dx + y \; dy = x\;dx$
$d(xy) +y\;dy =x\;dx$
$xy +\large\frac{y^2}{2}-\frac{x^2}{2}$$=c_1$
$2xy +y^2 -x^2 =2c_1$
$2xy+y^{2}+x^{2}=c$
Hence 1 is the correct answer.
answered May 22, 2014 by meena.p
 
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