homogeneous differential equations $y' = f(y/x)$ There is a weird Theorem that comes about when considering whether a function is homogeneous (in the sense of the title definition).
I was unable to prove it, or to find a proof to it.
Can any one help?
If an equation is of the form $$f(x,y) = \frac{x^{a_1}y^{b_1} + x^{a_2}y^{b_2} \cdots}{x^{c_1}y^{d_1} + x^{c_2}y^{d_2} \cdots}$$  
And we know that:
$$\forall i,j\in \Bbb N \colon \quad a_i+b_i = c_j + d_j $$
Then $$f(x,y) = \frac{(\frac{y}{x})^{b_1} + (\frac{y}{x})^{b_2} + \cdots}{(\frac{y}{x})^{d_1} + (\frac{y}{x})^{d_2} + \cdots}$$
Which is then easier to solve. Now as mentioned this came with no proof, I would like to see one, or at least to get a general idea of why this might be true. 

Solution of the differential equation :  $\quad\frac{dy}{dx}=f(y/x)\quad$  with a given function $f$ :
Let $y(x)=x\:t(x)\quad \frac{dy}{dx}=t+x\frac{dt}{dx}$
$$t+x\frac{dt}{dx}=f(t)$$
$$\frac{dx}{x}=\frac{dt}{f(t)-t}$$
$$\ln(x)=\int \frac{dt}{f(t)-t}+\text{constant}$$
$$x=c\: \exp\left(\int \frac{dt}{f(t)-t}\right)$$
The general solution on parametric form (parameter $t$) is :
$$\begin{cases}
x=c\: \exp\left(\int \frac{dt}{f(t)-t}\right) \\
y=c\: t\:\exp\left(\int \frac{dt}{f(t)-t}\right)\\
\end{cases}$$
