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Solve: $(3xy+y^2)dx+(x^2+xy)dy=0 \qquad y(1)=1$

$(a)\;x^2(2x+y)=3 \\ (b)\;y^2(2x+y)=3 \\ (c)\;x^2y(2x+y)=3 \\ (d)\;x^2y(2x+3)=9 $
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$\large\frac{dy}{dx}=-\large\frac{3xy+y^2}{x^2+xy}$
$y=vx$
$v+xv'= -\large\frac{3v+v^2}{1+v}$
$xv'= -[-v+\large\frac{3v+v^2}{1+v}]$
$\qquad =\large\frac{-2v (2+v)}{1+v}$
Separating variables ,
$\large\frac{(1+v)dv}{v(2+v)}=\large\frac{-2}{x} $$dx$
$\large\frac{1}{2} \bigg[\large\frac{1}{v}+\frac{1}{2+v}\bigg]dv=-\large\frac{-2}{x}$$dx$
Integrate $\log (v+2+v))+4 \log x =2 \log c$
$v (2+v)x^4=c^2$
$v=\large\frac{y}{x}$
$x^2y (2x+y)=c^2=k$
=> $y(1)=1$
$x^2y (2x+y)=3$
Hence c is the correct answer.
answered Feb 7, 2014 by meena.p
 
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