The solution of the differential equation $(3xy+y^2)dx+(x^2+xy)dy=0$ is

The correct option is A $x^2(2xy+y^2)=c^2$
Homogeneous equation can be written in the form of
$\frac{dy}{dx}=-\frac{3xy+y^2}{x^2+xy}$
Put $y=vx$ and $\frac{dy}{dx}=v+x\frac{dv}{dx}$, we get
$v+x\frac{dv}{dx}=-\frac{3x^{2}v+x^2{v}^2}{x^2+x^{2}v}$
$\Rightarrow x\frac{dv}{dx}=\frac{-2v(v+2)}{v+1}$
$\Rightarrow \frac{1}{x}dx=-\frac{(v+1)}{2v(v+2)}dv$
$\Rightarrow -\frac{2}{x}=-[\frac{1}{2(v+2)}+\frac{1}{2v}]dv$
On integrating, we get
$-2{\log }_{e}x=\frac{1}{2}\log (v+2)+\frac{1}{2}\log v-\log c$
$\Rightarrow v(v+2)x^4=c^2$
$\Rightarrow \frac{y}{x}(\frac{y}{x}+2)x^4=c^2(\because v=\frac{y}{x})$
Hence, required solution is $(y^2+2xy)x^2=c^2$.