How to turn homogenous differential equation $f(x,y)=dy/dx=x+y$ into the form of $F(y/x)$? I have been trying to rewrite this equation $dy/dx=x+y$  according to the property that my book describes:
“... thus homogenous functions are of the form: $$dy/dx=F(x/y)$$”
I am having a lot of difficulty in doing so. The aforementioned definition is used by my book and I want to see if I can determine if it is homogenous this way. 
There are a ton of examples which show you can rewrite a differential equation in that form but there are also a lot of other examples where this feat looks nearly impossible. Now if I use the property that a differential equation is homogenous if $M(x,y)$ and $N(x,y)$ are homogenous functions in the differential equation of form: $$M(x,y)dx+N(x,y)dy = 0$$
then I can determine if it’s a homogenous differential equation. 
 A: A function $f : \mathbb{R}^2 \to \mathbb{R}$ is homogeneous if:
$$f(\alpha x, \alpha y) = \alpha^k f(x, y),$$
where $\alpha >0$ and $k$ is constant called "degree of homogeneity".

A first order ODE is homogeneous if it has the form:
$$f(x,y)dy = g(x,y)dx,$$
provided that both $f$ and $g$ are homogeneous functions with the same degree of homogeneity.

Consider the ODE:
$$\frac{dy}{dx} = x + y.$$
It can be  rewritten as:
$$1 \cdot dy = (x+y) dx.$$
Therefore, let $f(x,y) = 1$ and $g(x,y) = x+ y.$
Notice that $f(x,y) = 1$ is homogeneous. Indeed, we have that $f(\alpha x, \alpha y) = 1 = \alpha^0 f(x, y).$ The degree of homogeneity of $f$ is $k = 0$.
Moreover, notice that $g(x,y) = x+y$ is homogeneous. Indeed, for any $\alpha$, we have that $g(\alpha x, \alpha y) = \alpha(x+y) = \alpha^1 g(x, y).$ The degree of homogeneity of $g$ is $k = 1$.

An homogeneous ODE can be rewritten as:
$$\frac{dy}{dx} = F\left(\frac{y}{x}\right),$$
if the degree of homogeneity of $f$ is equal to the degree of homogeneity of $g$. Thus, your equation doesn't satisfy this requirement, and hence there is no way to write it in the form $\frac{dy}{dx} = F\left(\frac{y}{x}\right)$. 

Bonus - a working example
Consider the following:
$$(2x+y)dx = (4x+5y)dy.$$
It is easy to show that both $f(x,y) = 2x+y$ and $g(x,y) = 4x+5y$ are homogeneous with degree $1$ (try by yourself!).
Notice that the equation can be rewritten as:
$$\frac{dy}{dx} = \frac{f(x,y)}{g(x,y)} = \frac{2x+y}{4x+5y}.$$
Now, observe that:
$$\frac{f(\alpha x, \alpha y)}{g(\alpha x,\alpha y)} = \frac{2\alpha x+\alpha y}{4\alpha x+5 \alpha y)} = \frac{\alpha(2x+y)}{\alpha(4x+5y)} = \frac{2x+y}{4x+5y} = \frac{f(x,y)}{g(x,y)}.$$
Consider $\alpha = \frac{1}{x}$. You get:
$$\frac{f(x,y)}{g(x,y)} = \frac{f(\alpha x, \alpha y)}{g(\alpha x,\alpha y)} = \frac{f\left(1, \frac{y}{x}\right)}{g\left(1, \frac{y}{x}\right)} = F\left(\frac{y}{x}\right).$$
Then:
$$\frac{dy}{dx} = F\left(\frac{y}{x}\right) = \frac{2+\frac{y}{x}}{4+5\frac{y}{x}}.$$
