I would like to understand why the following statement is true:

Let there be a homogeneous differential equation whose unknown function is $y(x)$, with $x\in]0,\infty[$. If we rewrite the equation with the variable change $y(x)=x\cdot v(x)$ then we obtain a separable differential equation for $v(x)$.

This statement seems too general for me. What I have considered so far is successive derivatives of $y(x)$:

$y'(x)=v(x)+x\cdot v'(x)$


$y^{(n)}(x)=n\cdot v^{(n-1)}(x)+x\cdot v^{(n)}(x)$

By looking at these derivatives I am trying to understand whether any combination of these (weighted by coefficients) will be separable.

  • $\begingroup$ The homogeneity is the catch. You won't make any progress without thinking about how that comes into the picture. $\endgroup$ – Ian Jun 6 '16 at 12:54
  • $\begingroup$ @Ian Yes I figured, because you can always bring parts of the equation to the other side. But is there a way to make this absolutely clear? $\endgroup$ – john melon Jun 6 '16 at 12:55
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    $\begingroup$ Can you understand how homogeneity essentially means that the entire equation can be written in terms of $y/x$? Because that's really the point... $\endgroup$ – Ian Jun 6 '16 at 13:00
  • $\begingroup$ Ok.. but I am having trouble since a general homogeneous can have all sorts of non constant coefficients and coupling of terms if I have understood correctly @Ian $\endgroup$ – john melon Jun 6 '16 at 13:08

After having looked at the meaning of homogeneous, I believe that this statement is referring to homogeneous equations of the form:

\begin{equation} \frac{dy}{dx}=F(\frac{y}{x}) \end{equation} Thus by making the variable change $u=\frac{y}{x}$ it can be transformed into the separable equation: \begin{equation} \frac{du}{F(u)-u}=\frac{dx}{x} \end{equation}

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