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Solve and get solutions : $ \bigg( x +y \large\frac{dy}{dx}\bigg)=\bigg( y -x \large\frac{dy}{dx}\bigg) \bigg(x^4+\large\frac{y^6}{x^2}$$+3 (x^2y^2+y^4)\bigg)$

$(a)\;\frac{y}{x} -\frac{2}{(x^2+y^2)^2}=c' \\ (b)\;\frac{4y}{x} -\frac{1}{(x^2+y^2)^2}=c' \\ (c)\;2y -\frac{1}{(x^2+y^2)^2}=c' \\ (d)\;2yx -\frac{1}{(x^2+y^2)^2}=c' $

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$\bigg(x+y \large\frac{dy}{dx}\bigg) =\bigg( y- x \large\frac{dy}{dx}\bigg) \frac{(x^2+y^2)^3}{x^2}$
$\large\frac{xdx+ydy}{(x^2+y^2)^3}=\frac{ydx-xdy}{y^2}. \frac{y^2}{x^2}$
=> $\int \large\frac{d(x^2+y^2}{(x^2+y^2)^3}$$=2 \int \large\frac{1}{x^2/y^2} d(\large\frac{x}{y})$
Integrating,
$\large\frac{-1}{2(x^2+y^2)^2}=\large\frac{-2}{x/y}$$+c$
$\large\frac{-1}{2(x^2+y^2)^2}=\large\frac{-4y}{x}$$+c'$
$\large\frac{4y}{x} -\frac{1}{(x^2+y^2)^2}=c' $
Hence b is the correct answer.
answered Feb 5, 2014 by meena.p
 

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