Problem with solving differential equation $(x+2y)dx + ydy = 0$ I have a homework with this equation:
$$(x+2y)dx + ydy = 0$$
However I have no idea how to solve it. 
I tried couple things:


*

*Is it linear equation, or does it have "standard form" : $\frac{dy}{dx}+\frac{x}{y} + 2 = 0$. Well $\frac{1}{y}$ is not a standard form.

*Is it exact? $\frac{∂M}{∂y}=2$ and $\frac{∂N}{∂x}=0$. I tried using $\mu(y)=e^{\int\frac{2}{y}dy}=y^2$ to make it exact, but that didn't work

*Is it homogeneous? In theory yes as $M(tx,ty)=tx+2ty=t(x+2y)=tM()$ and $N(tx,ty)=ty=tN()$ and I reached up to $$x(dx)+2ux(dx)+u^2x(dx)+x^2u(du)=0$$ But I can't  separate it.


Tried searching online and going with subsitution (method #3) was suggested, but they never went beyond that point. Maybe it's just too late and I'm not thinking straight.
If you could point me in the right direction, or list steps instead of giving solution if would be great.
 A: You're almost there with approach $(3)$.  Just combine your $dx$'s to get
$$
x(1+2u+u^2)dx+x^2udu=0.
$$
Now, if you divide through by $x$, you get$$
(1+2u+u^2)dx+xudu=0,
$$
which is separable.  Namely, move the $du$ to the other side to get
$$
\frac{dx}{x}=-\frac{udu}{1+2u+u^2}.
$$
Both sides can be integrated using elementary functions (the denominator on the RHS factors).
A: Given $$(x+2y)dx+y\ dy=0$$
Take $\dfrac yx=u\implies y=ux$
$\dfrac{dy}{dx}=u+x\dfrac{du}{dx}\implies dy=u\ dx+x\ du$
$$(x+2ux)dx+(ux)(udx+xdu)=0$$
$$xdx+2ux\ dx+u^2x\ dx+ux^2\ du=0$$
$$(1+u^2+2u)du+(ux)dx=0$$
$$\dfrac{u\ du}{(1+u^2+2u)}=-\dfrac{dx}{x}$$
$$\dfrac{u\ du}{(1+u)^2}=-\dfrac{dx}{x}$$
Now Integrating on both sides we get,
$$\int\dfrac{u\ du}{(1+u)^2}=-\int\dfrac{dx}{x}$$
$$\ln(1+u)+\dfrac{1}{1+u}=-\ln x+C$$
$$\ln(y+x)-\ln x+\dfrac{x}{x+y}=-\ln x+C$$
$$C=(x+y)e^{\frac{x}{x+y}}$$
A: I can give you a hint:
We can do several things to convert it into a separable equation.
$(x+2y)\mathrm{d}x+y\mathrm{d}y=0 \Rightarrow (x+2y)\mathrm{d}x=-y\mathrm{d}y$
When $y \neq 0$, we have $-\frac{x+2y}{y} = \frac{\mathrm{d}y}{\mathrm{d}x}$; thus, $-2-\frac{1}{\frac{y}{x}} = \frac{\mathrm{d}y}{\mathrm{d}x}$. Let $u = \frac{y}{x}$, then we have $y = ux$ and $\mathrm{d}y = x\mathrm{d}u+u\mathrm{d}x$. Hence, $\frac{\mathrm{d}y}{\mathrm{d}x} = x \frac{\mathrm{d}u}{ \mathrm{d}x}+u$. Substitue $\frac{y}{x} \text{ and } \frac{\mathrm{d}y}{\mathrm{d}x}$, we get $x \frac{\mathrm{d}u}{ \mathrm{d}x}+u = -2-\frac{1}{u}$. 
You can use the method of separation of variables to get $\frac{u\mathrm{d}u}{(1+u)^2} = -x\mathrm{d}x$ when $u\neq -1$ and $\frac{\mathrm{d}u}{\mathrm{d}x}=0$ when $u=-1$. You can solve these equations to get the expression of $u$ and substitute $u$ with $u = \frac{y}{x}$ to get y.
Besides, you need to consider the situation when $y=0$.
