ODE Guidance Needed This isn't from any homework assignment. I'm just trying to find the solution (either explicit or implicit) for this ODE.
$$(x+2y)\,dx + (y)\,dy = 0$$
The first obvious step to take is to check if this is exact, and if not, find an integrating factor $\mu(x,y)$ such that it will become one when multiplied by the function $\mu$. It isn't exact (can be quickly checked), so here's what I did.
Multiply by $\mu$ and assume that the equation is then exact:
$$ \frac{\partial}{\partial y}(\mu(x,y)(x+2y))= \frac{\partial}{\partial x}(\mu(x,y)(y)) $$
$$ \mu_y(x,y)(x+2y) + 2\mu(x,y) = \mu_x(x,y)y $$
Now we assume that $\mu(x,y)$ is either an equation of only $x$ or only $y$. If we assume the former, we will inevitably end up with $\mu'$ multiplied by $y$, and so the solution ($\mu$(x)) would have to have a $y$ in it, which would contradict our original assumption. If we go with the latter ($\mu = \mu(y)$), then it's clear that we would end up with the same problem/contradiction. Thus, our integrating function/factor $\mu = \mu(x,y)$, but now that would be out of the scope of ordinary differential equations.
In addition to that, whenever I try to make some nifty substitution to simplify this and end up with an explicit solution, it never works.
Could you give me some guidance please?
P.S: Don't just give me the solution (either implicit or explicit). I can easily get that on Wolframalpha
 A: in this type of equation it may be useful to try the substitution $y=vx$ so that 
$$
dy = vdx + x dv
$$
this may help you "separate the variables" and obtain an integrable form
A: Rewrite as $2y+yy'+x=0$, then define $y:=x \lambda(x)$, separate $\lambda, \lambda'$ on the LHS and $x$s on the RHS. It's easy from there.
Was this in the exact equations chapter of your book? I tend to prefer other methods since they're faster, so I see no reason to transform this into an exact equation.
A: As already said in answers, define $y=x z$ and replace. You end with $$x z \frac{dz}{dx}+(z+1)^2=0$$ where you can separate the variables $$-\frac{dx}{x}=\frac{z}{(1+z)^2}dz=\frac{1+z-1}{(1+z)^2}dz=\frac{dz}{1+z}-\frac{dz}{(1+z)^2}$$ Integrating both sides leads to $$-\log(x)+c=\log(1+z)+\frac{1}{1+z}$$ For more simplicity, define now $t=\frac{1}{1+z}$ and $a=-\log(x)+c$ and the equation write $$a=-\log(t)+t$$ for which the solution is $$t=-W\left(-e^{-a}\right)$$ where $W$ is Lambert function and then, after a series of replacements and simplifications $$y=e^{W\left(-e^{-c} x\right)+c}-x$$
