$$y^2dx+(x^2+3xy+4y^2)dy=0$$
Substitution: $$x=vy$$ $$dx=vdy+ydv$$
Simplifying everything:
$$(v+2)^2dy+ydv=0$$
$$\frac{dy}{y}+\frac{dv}{(v+2)^2}=0$$
Integrating both we obtain:
$$\frac{1}{(v+2)}+C=ln|y|$$
$$y=e^\frac{1}{v+2}C$$
That's where I think I have a mistake, since the answer looks pretty messy.
Moreover, I need to find $C$ such that $x=2 , y=1$.
Any help would be appreciated.