I want to solve
$x\dot y'=y(2x+y)$ given $y=xv(x)$
I have put the given equation in the form $\dot y=2(\frac{y}{x})+(\frac{y}{x})^2$ and then said that $ \dot y=F(\frac{y}{x})=F(v)$. The equation can then be recast as $\frac{1}{F(v)-v}dv=\frac{1}{x}dx$. I am unsure how to begin to integrate this without ending up with some horrible exponentials. Any help would be greatly appreciated!
Considering the DE
$$ x y'=y(2x+y) $$
Making $v = 1/y$ we have
$$ v'+2 v+\frac 1x = 0 $$
which is linear...
NOTE
Being linear this last DE the answer can be written as
$$ v = v_h + v_p\\ v_h'+2v_h = 0\\ v_p'+2v_p + \frac 1x = 0 $$
here $v_h = C_0 e^{-2x}$ and now substituting $y_p = C_0(x)e^{-2x}$ (Lagrange constants variation method) we get
$$ C_0'(x) = -\frac{e^{2x}}{x}\Rightarrow C_0(x) = -\int^x \frac{e^{2t}}{t}dt $$
so
$$ v =e^{-2x} \left(C_0 -\int^x \frac{e^{2t}}{t}dt\right) = \frac 1y $$