# 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.

Rewrite $$f(x,y) = \frac{x^{a_1}y^{b_1} + x^{a_2}y^{b_2} \cdots}{x^{c_1}y^{d_1} + x^{c_2}y^{d_2} \cdots}$$ as $$f(x,y) = \frac{x^{a_1} x^{b_1}(y/x)^{b_1} + x^{a_2}x^{b_2}(y/x)^{b_2} \cdots}{x^{c_1}x^{d_1}(y/x)^{d_1} + x^{c_2}x^{d_2}(y/x)^{d_2} \cdots}$$ by multiplying monomials upstairs and downstairs by a suitable power of $x$, and then simplify upstairs and downstairs (as $x^{a_1+b_1}=x^{c_1+d_1}=x^{c_2+d_2}=x^{a_2+b_2}$ etc).
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 : $$\begin{cases} x=c\: \exp\left(\int \frac{dt}{f(t)-t}\right) \\ y=c\: t\:\exp\left(\int \frac{dt}{f(t)-t}\right)\\ \end{cases}$$