Definition:
We say $f(x,y)$ is homogenous of degree $n$ if for each $\lambda$, $f(\lambda x, \lambda y)=\lambda^n f(x,y)$
We call a differential equation of form $P(x,y)dx+Q(x,y)dy=0$ homogenous, If $P$ and $Q$ are two homogenous functions of degree $n$.
I've read in a book that all of the homogenous ODE's can be solved by these substitutions:
$y=vx$ , $dy=vdx+xdv$
Mysteriously, That seems to be working!
I have two questions regarding this way of solving homogenous ODE's.
$1$) How did they come up with this solution? They could do many other substitutions. Why this one?
$2$) Why can we assume that there exists something like $v$ that $y=vx$? Maybe there is not a linear relation between $x$ and $y$! How are we sure?
Note: I want to understand this method. That's the point.
Note that $\nu$ is not a constant, i.e. $y=\nu x$ is not a linear relation, but a change of variables $(x,y)\mapsto(x,\nu)$ $$ \begin{cases} x=x,\\ \nu=\frac{y}{x} \end{cases} $$ Since $P$ and $Q$ are homogeneous one can factor out e.g. $x$ to get $$ x^nP\left(1,\frac{y}{x}\right)dx+x^nQ\left(1,\frac{y}{x}\right)dy=0. $$ Then one can cancel $x^n$ and it is obviously worth to try $\frac{y}{x}$ as a new variable $\nu$. The diferential of the product is $dy=d(\nu x)=\nu dx+x d\nu$ and we get $$ [P(1,\nu)+\nu Q(1,\nu)]dx=-xQ(1,\nu)d\nu\quad\Leftrightarrow\quad \frac{dx}{x}=-\frac{Q(1,\nu)}{P(1,\nu)+\nu Q(1,\nu)}d\nu. $$ Hence, we can separate the variables, which makes the equation easier to integrate, in general.